discrete_mathematics_with_applications/chapter_4/exercises.md

Page 194

Exercise Set 4.1

In 1-4 justify your answers by using the definitions of even, odd, prime, and composite numbers.

1. Assume that k is a particular integer.

a. Is -17 an odd integer?

 -17 = 2(-9) + 1 

Let k = -9, so our expression becomes by substitution:

 -17 = 2k + 1 

Since -17 can be represented by the form 2k + 1 where k = -9 and k is an integer, by the definition of an odd number, -17 is an odd integer.

b. Is 0 neither even nor odd?

No, 0 can be represented as 0 = 2(0), and let k = 0, so 0 = 2k, where k is an integer, and by definition of an even number, 0 is even.

c. Is 2k - 1 odd?

Yes, 2k - 1 = 2(k - 1) + 1 where k - 1 is an integer by the difference of integers. Let m = k - 1, so our expression becomes 2k - 1 = 2m + 1, and since m is an integer, we can conclude that 2k - 1 is an odd integer by definition of odd integers.

2. Assume that c is a particular integer.

a. Is -6c an even integer?

Yes -6c = 2(-3c), where -3c is an integer by the product of integers, and since -6c can be expressed as 2 \cdot \text{ some integer}, it is even by the definition of even integers.

b. Is 8c + 5 an odd integer?

Yes, 8c + 5 = 8c + 4 + 1 = 2(4c + 2) + 1 where 4c + 2 is an integer by the sum of products of integers. Since 8c + 5 can be expressed as 2(\text{some integer}) + 1, we can conclude that 8c + 5 is an odd integer by the definition of odd integers.

c. Is (c^2 + 1) - (c^2 - 1) - 2 an even integer?

Yes, if evaluate the statement by laws of algebra, we get:

 (c^2 + 1) - (c^2 - 1) - 2 = c^2 + 1 - c^2 + 1 - 2 = 1 + 1 - 2 = 0 

And as established in 1b, 0 is an even integer, so (c^2 + 1) - (c^2 - 1) - 2 can be expressed in the form of 2 \cdot \text{ some integer}, so by the definition of integers, (c^2 + 1) - (c^2 - 1) - 2 is an even integer.

3. Assume that m and n are particular integers?

a. Is 6m + 8n even?

Yes, 6m + 8n = 2(3m + 4n). 3m + 4n is an integer by the sum of products of integers. Since 6m + 8n can be expressed as 2 \cdot \text{ some integer}, by the definition of even integers, 6m + 8n is even.

b. Is 10mn + 7 odd?

10mn + 7 = 10mn + 6 + 1 = 2(5mn + 3) + 1. 5mn + 3 is an integer by the product and sum of integers. Since 10mn + 7 can be expressed as 2(\text{some integer}) + 1, 10mn + 7 is an odd integer.

c. If m > n > 0, is m^2 - n^2 composite?

Not necessarily. Consider m = 3 and n = 2, then m^2 - n^2 = 9 - 4 = 5, which is a prime number.

4. Assume that r and s are particular integers.

a. Is 4rs even?

Yes, 4rs = 2(2rs), where 2rs is an integer by the product of integers. Since 4rs = 2(\text{ some integer}), by the definition of an even integer, 4rs is an even integer.

b. Is 6r + 4s^2 + 3 odd?

6r + 4s^2 + 3 = 6r + 4s^2 + 2 + 1 = 2(3r + 2s^2 + 1) + 1. 3r + 2s^2 + 1 is an integer by product and sum of integers. Let k = 3r + 2s^2 + 1, and so 6r + 4s^2 + 3 = 2k + 1. By definition of an odd integer, 6r + 4s^2 + 3 is an odd integer.

c. If r and s are both positive, is r^2 + 2rs + s^2 composite?

Since r^2 + 2rs + s^2 = (r + s)^2 = (r + s)(r + s) and r + s \geq 2. And since r + s > 1, the product of (r + s)(r + s) is composite.

Prove the statements in 5-11.

5. There are integers m and n such that m > 1 and n > 1 and \dfrac{1}{m} + \dfrac{1}{n} is an integer.

For example, let m = 2 and n = 2, then \dfrac{1}{2} + \dfrac{1}{2} = 1, and 1 is an integer.

6. There are distinct integers m and n such that \dfrac{1}{m} + \dfrac{1}{n} is an integer.

For example, let m = -2, and n = 2, then \dfrac{1}{-2} + \dfrac{1}{2} = 0, and 0 is an integer.

7. There are real numbers a and b such that

 \sqrt{a + b} = \sqrt{a} + \sqrt{b} 

For example, let a = 0 and b = 9, then \sqrt{0 + 9} = \sqrt{9} = 3 = 0 + 3 = \sqrt{0} + \sqrt{9}.

8. There is an integer n > 5 such that 2^n - 1 is prime.

For example, let n = 7, then 2^7 - 1 = 127, and 127 is prime.

9. There is a real number x such that x > 1 and 2^x > x^{10}.

For example, let x = \dfrac{1}{2}, then $2^{\frac{1}{2}} \approx 1.414213562 > 0.0009765625 = \left(\frac{1}{2}\right)^{10}$

Definition: An integer n is called a perfect square if, and only if, n = k^2 for some integer k.

10. There is a perfect square that can be written as a sum of two other perfect squares.

Let n = 4 and m = 3, and let l = k^2 be the sum of their squares:

 l = k^2 = n^2 + m^2 = (4)^2 + (3)^2 = 16 + 9 = 25 

 l = k^2 = 25 

So l = 25 can be written as n^2 + m^2 where n = 4 and m = 3, and since both n and m are integers, we can say that l is a perfect square by definition of a perfect square and l can be written as the sum of two other perfect squares.

11. There is an integer n such that 2n^2 - 5n + 2 is prime.

For example let n = 3, then 2n^2 - 5n + 2 = 2(3)^2 - 5(3) + 2 = 2(9) - 15 + 2 = 18 - 15 + 2 = 3 + 2 = 5, and 5 is prime.

In 12-13, (a) write a negation for the given statement, and (b) use a counterexample to disprove the given statement. Explain how the counterexample actually shows that the given statement is false.

12. For all real numbers a and b, if a < b the a^2 < b^2.

(a)

Original:

 \forall a \in \mathbb{R} \forall b \in \mathbb{R}((a < b) \to (a^2 < b^2)) 

Negation:

 \exists a \in \mathbb{R} \exists b \in \mathbb{R}((a < b) \wedge (a^2 \geq b^2)) 

There exist real numbers a and b such that a < b and a^2 \geq b^2.

(b)

Counterexample:

Let a = -2 and let b = -1. The hypothesis a < b holds as -2 < -1 is true, but the conclusion of the original statement a^2 < b^2 is false as (-2)^2 = 4 \cancel{<} 1 = (-1)^2.

Since the original statement claims that the implication holds true for all real numbers a and b, a single counterexample is sufficient to show that the statement is false.

13. For every integer n, if n is odd then \dfrac{n - 1}{2} is odd.

(a)

Original:

Let P(n) = n \text{ is odd}

Let Q(m) = m \text{ is odd}

 \forall n \in \mathbb{Z}(P(n) \to Q\left(\frac{n - 1}{2}\right)) 

Negation:

 \exists n \in \mathbb{Z}(P(n) \wedge \neg Q\left(\frac{n - 1}{2}\right)) 

There exists some integer n such that n is odd and \dfrac{n - 1}{2} is not odd.

(b)

Counterexample:

Let n = 1. n is odd as 1 can be expressed as n = 1 = 2(k) + 1, where k = 0. This means that 1 is odd by the definition of an odd integer, and the hypothesis of the original statement is true. The conclusion of the original statement, however, is false, as \dfrac{n - 1}{2} = \dfrac{1 - 1}{2} = \dfrac{0}{2} = 0, and 0 is not odd.

Since the original statement claims that the implication holds true for all integers n, a single counterexample is sufficient to show that the statement is false.

Disprove each of the statements in 14-16 by giving a counterexample. In each case explain how the counterexample actually disproves the statement.

14. For all integers m and n, if 2m + n is odd then m and n are both odd.

Let m = 2 and let n = 1, the hypothesis 2m + n is odd is true as 2(2) + 1 = 5, and 5 is odd, but the conclusion that both m and n are odd is false, as m is even.

15. For every integer p, if p is prime then p^2 - 1 is even.

Let p = 2. The hypothesis holds true as 2 is prime, but the conclusion "p^2 - 1 is even" is false for this p as (2)^2 - 1 = 4 - 1 = 3, and 3 is not even.

16. For every integer n, if n is even then n^2 + 1 is prime.

Let n = 0, the hypothesis "n is even" holds true for this n as 0 is even. The conclusion "n^2 + 1 is prime" fails for this n as 0^2 + 1 = 0 + 1 = 1, and 1 is not prime.

In 17-20, determine whether the property is true for all integers, true for no integers, or true for some integers and false for other integers. Justify your answers.

17. (a + b)^2 = a^2 + b^2

This property is true for some integers and not others.

For example where it is true, consider a = 0 and b = 1, then (0 + 1)^2 = 1^2 = 1 = 0 + 1 = (0)^2 + (1)^2 holds true for at least two integers.

For example where it is false, consider a = 1 and b = 1, then (1 + 1)^2 = 2^2 = 4 \neq 2 = 1 + 2 = (1)^2 + (1)^2. Since this provides a counterexample, this property cannot hold true for all integers.

Therefore, this property holds true for some integers and not others.

18. \dfrac{a}{b} + \dfrac{c}{d} = \dfrac{a + c}{b + d}

This is true for a = c = 0 and $b = d = 1$as:

 \frac{0}{1} + \frac{0}{1} = 0 + 0 = 0 = \frac{0}{2} = \frac{0 + 0}{1 + 1} 

This is false for a = b = c = d = 1, as:

 \frac{1}{1} + \frac{1}{1} = 1 + 1 =  2 \neq 1 = \frac{2}{2} = \frac{1 + 1}{1 + 1} 

Therefore, this property holds true for some integers and not others.

19. -a^n = (-a)^n

This is true for a = -1 and n = 1.

 -(-1)^1 = (-(-1))^1 

 -(-1) = (-(-1)) 

 1 = 1 

This is false for a = -1 and n = 2.

 -(-1)^2 = (-(-1))^2 

 -(1) = (1)^2 

 -1 \neq 1 

Therefore, this property holds true for some integers and not others.

20. The average of any two odd integers is odd.

Let m and n be odd integers. Let m = 2k + 1 and n = 2p + 1 where k and p are any integers.

We are asserting that \dfrac{m + n}{2} is odd. By substitution, we can express this as:

 \frac{m + n}{2} = \frac{(2k + 1) + (2p + 1)}{2} = \frac{2k + 2p + 2}{2} = \frac{2(k + p + 1)}{2} = k + p + 1 

In order to prove that k + p + 1 is odd, we need to be able to express it in the form of 2(\text{some integer}) + 1 by the definition of an odd integer.

An example where this is true is if k = 2 and p = 4, then k + p + 1 = 2 + 4 + 1 = 7, and 7 is an odd integer.

A counterexample where this is false is if k = 3 and p = 4, then k + p + 1 = 3 + 4 + 1 = 8, and 8 is not an odd integer.

Therefore, this property holds true for some integers and not others.

Prove the statement in 21 and 22 by the method of exhaustion.

21. Every positive even integer less than 26 can be expressed as a sum of three or fewer perfect squares. (For instance, 10 = 1^2 + 3^2 and 16 = 4^2.)

Let's first establish all positive even integers less than 26:

 \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24\} 

 2 = 1^2 + 1^2 

 4 = 2^2 

 6 = 2^2 + 1^2 + 1^2 

 8 = 2^2 + 2^2 

 10 = 3^2 + 1^2 

 12 = 2^2 + 2^2 + 2^2 

 14 = 3^2 + 2^2 + 1^2 

 16 = 4^2 

 18 = 4^2 + 1^2 + 1^2 

 20 = 4^2 + 2^2 

 22 = 3^2 + 3^2 + 2^2 

 24 = 4^2 + 2^2 + 2^2 

22. For each integer n with 1 \leq n \leq 10, n^2 - n + 11 is a prime number.

Let's establish all possible values for n:

 \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} 

n = 1:

 (1)^2 - (1) + 11 = 1 - 1 + 11 = 11 \text{ is prime} 

n = 2:

 (2)^2 - (2) + 11 = 4 - 2 + 11 = 13 \text{ is prime} 

n = 3:

 (3)^2 - (3) + 11 = 9 - 3 + 11 = 17 \text{ is prime} 

n = 4:

 (4)^2 - (4) + 11 = 16 - 4 + 11 = 23 \text{ is prime} 

n = 5:

 (5)^2 - (5) + 11 = 25 - 5 + 11 = 31 \text{ is prime} 

n = 6:

 (6)^2 - (6) + 11 = 36 - 6 + 11 = 41 \text{ is prime} 

n = 7:

 (7)^2 - (7) + 11 = 49 - 7 + 11 = 53 \text{ is prime} 

n = 8:

 (8)^2 - (8) + 11 = 64 - 8 + 11 = 67 \text{ is prime} 

n = 9:

 (9)^2 - (9) + 11 = 81 - 9 + 11 = 83 \text{ is prime} 

n = 10:

 (10)^2 - (10) + 11 = 100 - 10 + 11 = 101 \text{ is prime} 

Each of the statements in 23-26 is true. For each, (a) rewrite the statement with the quantification implicit as If _____, then _____, and (b) write the first sentence of a proof (the "starting point") and the last sentence of a proof (the "conclusion to be shown"). (Note that you do not need to understand the statements in order to be able to do these exercises.)

23. For every integer m, if m > 1 then 0 < \dfrac{1}{m} < 1.

(a) If an integer is greater than 1, then its reciprocal is between 0 and 1.

(b)

Starting Point: Suppose m is any integer such that m > 1.

To Show: 0 < \dfrac{1}{m} < 1

24. For every real number x, if x > 1 then x^2 > x.

(a) If a real number is greater than 1, then it's square is greater than itself.

(b)

Starting Point: Suppose x is any real number such that x > 1.

To Show: x^2 > x.

25. For all integers m and n, if mn = 1 then m = n = 1 or m = n = -1.

(a) If the product of any two integers is equal to 1, then both integers either equal 1 or -1.

(b)

Starting Point: Suppose m and n are any integers such that mn = 1.

To Show: m = n = 1 or m = n = -1.

26. For every real number x, if 0 < x < 1 then x^2 < x.

(a) If a real number is between 0 and 1, then its square is less than itself.

(b)

Starting Point: Suppose x is any real number such that 0 < x < 1.

To Show: x^2 < x.

27. Fill in the blanks in the following proof.

Theorem: For every odd integer n, n^2 is odd.

Proof: Suppose n is any ___ (a) ___. By definition of odd, n = 2k + 1 for some integer k. Then

 n^2 = \left(___(b)____\right)^2  \quad \text{ by substitution} 

 \quad = 4k^2 + 4k + 1 \quad \text{ by multiplying out} 

 \quad = 2(2k^2 + 2k) + 1 \quad \text{ by factoring out a 2} 

Now 2k^2 + 2k is an integer because it is a sum of products of integers. Therefore n^2 equals 2 \cdot (\text{an integer}) + 1, and so ___ (c) ___ is odd by definition of odd.

Because we have not assumed anything about n except that it is an odd integer, it follows from the principle of ___ (d) ___ that for every odd integer n, n^2 is odd.

a. odd integer.

b. 2k + 1

c. n^2

d. universal generalization

In each of 28-31:

a. Rewrite the theorem in three different ways:

as \forall _____, if _____ then _____, as \forall _____, _____ (without using the words if or then),

and as If _____, then _____ (without using an explicit universal quantifier).

b. Fill in the blanks in the proof of the theorem.

Theorem: the sum of any two odd integers is even.

Proof: Suppose m and n are any [particular but arbitrarily chosen] odd integers.

[We must show that m + n is even.]

By __ (a) __, m = 2r + 1 and n = 2s + 1 for some integers r and s.

Then

 m + n = (2r + 1) + (2s + 1) \quad \text{k by \_\_ (b) \_\_} 

 \quad = 2r + 2s + 2 

 \quad = 2(r + s + 1) \quad \text{ by algebra} 

Let u = r + s + 1. Then u is an integer because r, s, and 1 are integers and because __ c __.

Hence m + n = 2u, where u is an integer, and so, by __ (d) __, m + n is even [as was to be shown].

a.

Theorem: the sum of any two odd integers is even.

\forall integers m and n, if m and n are odd, then m + n is even.

\forall odd integers m and n, m + n is even.

If any two integers are odd, then their sum is even.

b.

(a) the definition of an odd integer

(b) substitution

(\c) any sum of integers is an integer

(d)$ the definition of an even integer

Theorem: The negative of any integer is even.

Proof: Suppose n is any [particular but arbitrarily chosen] even integer.

[We must show that -n is even.]

By __ (a) __, n = 2k for some integer k.

Then

 -n = -(2k)  \quad \text{ by \_\_ (b) \_\_} 

 \quad = 2(-k) \quad \text{ by algebra} 

Let r = -k. Then r is an integer because (-1) and k are integers and __ c __.

Hence -n = 2r, where r is an integer, and so -n is even by __ (d) __ [as was to be shown].

a.

Theorem: The negative of any integer is even.

\forall integers n, if n is negative, then n is even.

\forall negative integers n, n is even.

If an integer is negative, then it is even.

b.

(a) the definition of an even integer

(b) substitution

c the product of any two integers is an integer

(d) the definition of an even integer

Theorem 4.1.2: The sum of any even integer and any odd integer is odd.

Proof: Suppose m 8s any even integer and n is __ (a) __. By definition of even, m = 2 for some __ (b) __, and by definition of odd, n = 2s + 1 for some integer s. By substitution and algebra,

 m + n = \text{\_\_ (c) \_\_} = 2(r + s) + 1 

Since r and s are both integers, so is their sum r + s. Hence m + n has the form twice some integer plus one, and so __ (d) __ by definition of odd.

a.

Theorem 4.1.2: The sum of any even integer and any odd integer is odd.

\forall integers m and n, if m is an even integer and n is an odd integer, then m + n is odd.

\forall even integers m and odd integers n, m + n is odd.

If m is an even integer and n is any odd integer, then m + n is odd.

b.

(a) any odd integer

(b) integer r

c 2r + (2s + 1)

(d) m + n is odd

Theorem: Whenever n is an odd integer, 5n^2 + 7 is even.

Proof: Suppose n is any [particular but arbitrarily chosen] odd integer.

[We must show that 5n^2 + 7 is even.]

By definition of odd, n = __ (a) __ for some integer k.

Then

 5n^2 + 7 = \text{\_\_ (b) \_\_}  \quad \text{ by substitution} 

 \quad = 5(4k^2 + 4k + 1) + 7 

 \quad = 20k^2 + 20k + 12 

 \quad = 2(10k^2 + 10k + 6) \quad \text{ by algebra} 

Let t = __ c __. Then t is an integer because products and sums of integers are integers.

Hence 5n^2 + 7 = 2t, where t is an integer, and thus __ (d) __ by definition of even [as was to be shown].

a.

Theorem: Whenever n is an odd integer, 5n^2 + 7 is even.

\forall integers n, if n is an odd integer, then 5n^2 + 7 is even.

\forall odd integers n, 5n^2 + 7 is even.

If n is an odd integer, then 5n^2 + 7 is even.

b.

(a) 2k + 1

(b) 5(2k + 1)^2 + 7

c 10k^2 + 10k + 6

(d) 5n^2 + 7 is even

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Exercise Set 4.2

Page 204

Prove the statements in 1-11. In each case use only the definitions of the terms and the Assumptions listed on page 161, not any previously established properties of odd and even integers. Follow the directions given in this section for writing proofs of universal statements.

1. For every integer n, if n is odd then 3n + 5 is even.

Theorem: Suppose n is any odd integer.

Proof:

Since n is odd, n = 2k + 1 for some integer k.

Then

 3n + 5 = 3(2k + 1) + 5 \quad \text{ by substitution} 

 \quad = 6k + 3 + 5 

 \quad = 6k + 8 

 \quad = 2(3k + 4)  \quad \text{ by algebra} 

Let t = 3k + 4.

Then 3n + 5 = 2(3k + 4) = 2t, where t is an integer because products and sums of integers are integers.

Therefore 3n + 5 is even by the definition of even integers.

Q.E.D.

2. For ever integer m, if m is even then 3m + 5 is odd.

Theorem: Suppose m is any even integer.

Proof:

Since m is even, m = 2k for some integer k.

Then:

 3m + 5 = 3(2k) + 5 \quad \text{ by substitution} 

 \quad = 6k + 5 

 \quad = 6k + 4 + 1 

 \quad = 2(3k + 2) + 1 \quad \text{ by algebra} 

Let t = 3k + 2.

Then 3m + 5 = 2(3k + 2) + 1 = 2t + 1 where t is an integer because the product and sum of integers are integers.

Therefore 3m + 5 is odd by the definition of odd integers.

Q.E.D.

3. For every integer n, 2n - 1 is odd.

Theorem:

Suppose n is any integer.

Proof:

Then:

 2n - 1 = 2n - 2 + 1 \quad \text{ by algebra} 

 \quad = 2(n - 1) + 1 \quad \text{ by factoring} 

Let t = n - 1.

Then 2n - 1 = 2(n - 1) + 1 = 2t + 1 where t is an integer because the difference of integers is an integer.

Therefore 2n - 1 is odd by the definition of an odd integer.

Q.E.D.

4. Theorem 4.2.2: The difference of any even integer minus any odd integer is odd.

Theorem: Suppose m is any even integer and n is any odd integer.

Proof:

Since m is even and n is odd, m = 2k and n = 2s + 1 where k is some integer and s is some integer.

Then

 m - n = 2k - (2s + 1)  \quad \text{ by substitution} 

 \quad = 2k - 2s - 1 

 \quad = 2k - 2s - 2 + 1 

 \quad = 2(k - s - 1) + 1 

Let t = k - s - 1.

Then m - n = 2(k - s - 1) + 1 = 2t + 1 where t is an integer because the difference of integers is an integer.

Therefore m - n is odd by the definition of odd integers.

Q.E.D.

5. If a and b are any odd integers, then a^2 + b^2 is even.

Theorem: Suppose a is any odd integer and b is any odd integer.

Proof:

Since a is an odd integer and b is an odd integer, a = 2k + 1 and b = 2s + 1 where k is some integer and s is some integer.

Then:

 a^2 + b^2 = (2k + 1)^2 + (2s + 1)^2 \quad \text{ by substitution} 

 \quad = (2k + 1)(2k + 1) + (2s + 1)(2s + 1) \quad \text{ by exponentiation} 

 \quad = (4k^2 + 4k + 1) + (4s^2 + 4s + 1) 

 \quad = 4k^2 + 4k + 1 + 4s^2 + 4s + 1 

 \quad = 4k^2 + 4k + 4s^2 + 4s + 2 

 \quad = 2(2k^2 + 2k + 2s^2 + 2s + 1) 

Let t = 2k^2 + 2k + 2s^2 + 2s + 1.

Then a^2 + b^2 = 2(2k^2 + 2k + 2s^2 + 2s + 1) = 2t where t is an integer because the product and sum of integers is an integer.

Therefore a^2 + b^2 is even by the definition of even integers.

Q.E.D.

6. If k is any odd integer and m is any even integer, then k^2 + m^2 is odd.

Theorem:

Suppose k is any odd integer and m is any even integer.

Proof:

Since k is an odd integer and m is an even integer, k = 2a + 1 and m = 2b where a is some integer and b is some integer.

Then:

 k^2 + m^2 = (2a + 1)^2 + (2b)^2 \quad \text{ by substitution} 

 \quad = (2a + 1)(2a + 1) + (2b)(2b) \quad \text{ by exponentiation} 

 \quad = (4a^2 + 4a + 1) + (4b^2) 

 \quad = 4a^2 + 4a + 4b^2 + 1 

 \quad = 2(2a^2 + 2a + 2b^2) + 1 

Let t = 2a^2 + 2a + 2b^2.

Then k^2 + m^2 = 2(2a^2 + 2a + 2b^2) + 1 = 2t + 1 where t is an integer because the product and sum of integers is an integer.

Therefore k^2 + m^2 is odd by the definition of an odd integer.

Q.E.D.

7. The difference between the squares of any two consecutive integers is odd.

Theorem:

Suppose n is any integer.

Proof:

Since n is an integer, n + 1 is a consecutive integer of n.

Then:

 n^2 - (n + 1)^2 = n^2 - (n + 1)(n + 1) \quad \text{ by exponentiation} 

 \quad = n^2 - (n^2 + 2n + 1) 

 \quad = n^2 - n^2 - 2n - 1 \quad \text{ by distribution} 

 \quad = -2n - 1 \quad \text{ by distribution} 

 \quad = -2n - 2 + 1 

 \quad = 2(-n - 1) + 1 

Let t = -n - 1.

Then n^2 - (n + 1)^2 = 2(-n - 1) + 1 = 2t + 1 where t is an integer because the product and difference of integers is an integer.

Therefore n^2 - (n + 1)^2 is odd by the definition of an odd integer.

Q.E.D.

8. For any integers m and n, if m is even and n is odd then 5m + 3n is odd.

Theorem:

Suppose m is any even integer and n is any odd integer.

Proof:

Since m is an even integer and n is an odd integer, m = 2k and n = 2s + 1 where k is some integer and s is some integer.

Then:

 5m + 3n = 5(2k) + 3(2s + 1) \quad \text{ by substitution} 

 \quad = 10k + 6s + 3 

 \quad = 10k + 6s + 2 + 1 

 \quad = 2(5k + 3s + 1) + 1 

Let t = 5k + 3s + 1.

Then 5m + 3n = 2(5k + 3s + 1) + 1 = 2t + 1 where t is an integer because the products and sums of integers is an integer.

Therefore 5m + 3n is odd by the definition of an odd integer.

Q.E.D.

9. If an integer greater than 4 is a perfect square, then the immediately preceding integer is not prime.

Theorem:

Suppose n is any integer where n > 4 and n is a perfect square.

Proof:

Since n is a perfect square and n > 4, then n = k^2 for some integer k where k > 2 or k < -2.

Then:

 n - 1 = k^2 - 1 \quad \text{ by substitution} 

 \quad = (k + 1)(k - 1) \quad \text{ by algebra} 

In order for n - 1 to be prime, either k + 1 or k - 1 must be equal to 1.

If k > 2, then both k + 1 > 1 and k - 1 > 1 are true.

If k < -2, then both k + 1 < 1 and k - 1 < 1 are true.

Therefore neither k + 1 nor k - 1 can ever be equal to 1.

Therefore n - 1 is not prime by the definition of a prime number.

Q.E.D.

10. If n is any even integer, then (-1)^n = 1.

Theorem:

Suppose n is any even integer.

Proof:

Since n is an even integer, then n = 2k where k is some integer.

Then:

 (-1)^n = (-1)^{2k} \quad \text{ by substitution} 

 \quad = (-1)^{2 \cdot k} 

 \quad = ((-1)^2)^k 

 \quad = 1^k 

 \quad = 1 \quad \text{ by the laws of exponents} 

Therefore (-1)^n = 1.

Q.E.D.

11. If n is any odd integer, then (-1)^n = -1.

Theorem:

Suppose n is any odd integer.

Proof:

Since n is an odd integer, then n = 2k + 1 where k is some integer.

Then:

 (-1)^n = (-1)^{2k + 1} \quad \text{ by substitution} 

 (-1)^n = (-1)^{2 \cdot k} \cdot (-1)^1 

 (-1)^n = ((-1)^2)^k \cdot (-1)^1 

 (-1)^n = 1^k \cdot -1 

 (-1)^n = 1 \cdot -1 

 (-1)^n = -1 \quad \text{ by the laws of exponents} 

Therefore (-1)^n = -1.

Q.E.D.

Prove that the statements in 12-14 are false.

12. There exists an integer m \geq 3 such that m^2 - 1 is prime.

Take the negation first:

For all integers m \geq 3, m^2 - 1 is not prime.

Theorem:

There is no integer m \geq 3 such that m^2 - 1 is prime.

Proof:

By algebra, we know that:

 m^2 - 1 = (m + 1)(m - 1) 

We also know that for m^2 - 1 to be prime, either m + 1 or m - 1 must be equal to 1.

Since m \geq 3, we know that both m + 1 \geq 4 and m - 1 \geq 2 are both true. Thus both factors are greater than 1.

Therefore m^2 - 1 is a product of two integers greater than 1, so it is not prime.

Therefore m^2 - 1 is not prime by the definition of prime numbers.

Q.E.D.

13. There exists an integer n such that 6n^2 + 27 is prime.

Take the negation first:

For all integers n, 6n^2 + 27 is not prime.

Theorem:

There is no integer n such that 6n^2 + 27 is prime.

Proof:

By algebra we know that:

 6n^2 + 27 = 3(2n^2 + 9) 

Since n^2 is always positive or 0, by the laws of exponentiation and by algebra, we can conclude that 2n^2 + 9 \geq 9 is true.

Since 3 > 1 and 2n^2 + 9 > 1, we then know that 6n^2 + 27 is a product of two integers greater than 1, so it is not prime.

6n^2 + 27 is not prime by the definition of prime numbers.

Q.E.D.

14. There exists an integer k \geq 4 such that 2k^2 - 5k + 2 is prime.

Take the negation first:

For all integers k \geq 4, 2k^2 - 5k + 2 is not prime.

Theorem:

There is no integer k \geq 4 such that 2k^2 - 5k + 2 is prime.

Proof:

By algebra we know:

 2k^2 - 5k + 2 = (k - 2)(2k - 1) 

Since we know that k \geq 4, we know that k - 2 \geq 2 and 2k - 1 \geq 7.

Since k - 2 > 1 and 2k - 1 > 1, we then know that 2k^2 - 5k + 2 is a product of two integers greater than 1, so it is not prime.

Therefore 2k^2 - 5k + 2 is not prime by definition of prime numbers.

Q.E.D.

Find the mistakes in the "proofs" shown in 15-19.

Theorem: For every integer k, if k > 0 then k^2 + 2k + 1 is composite.

"Proof: For k = 2, k > 0 and k^2 + 2k + 1 = 2^2 + 2 \cdot 2 + 1 = 9. And since 9 = 3 \cdot 3, then 9 is composite. Hence the theorem is true."

Answer: This proof just shows that the theorem is true for a single case, k = 2, in order to prove a universal claim as the theorem presents, the proof must prove the conclusion true for every integer k where k > 0.

Theorem: The difference between any odd integer and any even integer is odd.

"Proof: Suppose n is any odd integer, and m is any even integer. By definition of odd, n = 2k + 1 where k is an integer, and by definition of even, m = 2k where k is an integer. Then

 n - m = (2k + 1) - 2k = 1 

Answer: This proof makes the mistake of using k to represent two different quantities. By setting n = 2k + 1 and m = 2k, the proof implies that n = m + 1, and thus deduces the conclusion for only this situation. This proof falsely then "proves" that the difference between any even and odd integer will always equal 1, but taking most examples of even and odd integers as cases for this would show that this is false. In essence, this proof makes the mistake of assigning the same variable name to represent two different integers, and then by algebra comes to a false conclusion.

Theorem: For every integer k, if k > 0, then k^2 + 2k + 1 is composite.

"Proof: Suppose k is any integer such that k > 0. If k^2 + 2k + 1 is composite, then k^2 + 2k + 1 = rs for some integers r and s such that

 1 < r < k^2 + 2k + 1 

and

 1 < s < k^2 + 2k + 1 

Since

 k^2 + 2k + 1 = rs 

and both r and s are strictly between 1 and k^2 + 2k + 1, then k^2 + 2k + 1 is not prime. Hence k^2 + 2k + 1 is composite as was to be shown."

Answer: This proof makes the mistake of assuming what is to be proved. Instead of proving that k^2 + 2k + 1 is composite, it assumes the definition of composite numbers applies to the expression and then extrapolates logic about r and s that cannot be known because it has not yet been proven that k^2 + 2k +1 is composite. This starts at the line starting with "Since", which cannot be asserted as that is an assertion of the conclusion, not the hypothesis.

Teacher's answer: This incorrect proof assumes what is to be proved. The word since in the third sentence is completely unjustified. The second sentence tells only what happens if k^2 + 2k + 1 is composite. But at that point in the proof, it has not been established that k^2 + 2k + 1 is composite. In fact, that is exactly what is to be proved.

Theorem: The product of any even integer and any odd integer is even.

"Proof: Suppose m is any even integer and n is any odd integer. If m \cdot n is even, then by definition of even there exists an integer r such that m \cdot n = 2r. Also since m is even, there exists an integer p such that m = 2p, and since n is odd there exists an integer q such that n = 2q + 1. Thus

 mn = (2p)(2q + 1) = 2r 

where r is an integer. By definition of even, then, m \cdot n is even, as was to be shown."

Answer: This incorrect proof exhibits confusion between what is known and what is still to be shown. The writer correctly uses the definitions of even and odd integers to express m and n as 2p and 2q + 1, but assumes the conclusion that mn must be an expression of 2r, which is exactly what is to be shown, but has not yet been proven. In essence, they have jumped to the conclusion.

Theorem: The sum of any two even integers equals 4k for some integer k.

"Proof: Suppose m and n are any two even integers. By definition of even, m = 2k for some integer k and n = 2k for some integer k. By substitution,

 m + n = 2k + 2k = 4k 

That is what was to be shown."

Answer: This incorrect proof suffers from multiple problems. One is that it uses the same variable name k to represent two potentially different integers when expressing both m and n as even integers. The writer then incorrectly sums them to 4k and concludes they have proven the conclusion, but the form of 4k does not explicitly show that m + n is even by the definition of even integers.

In 20-38 determine whether the statement is true or false. Justify your answer with a proof or a counterexample, as appropriate. In each case use only the definitions of the terms and the Assumptions listed on page 161, not any previously established properties.

20. The product of any two odd integers is odd.

Theorem:

Suppose n is any odd integer and m is any odd integer.

Proof:

Since n and m are odd integers, n = 2k + 1 and m = 2s + 1 where k is some integer and s is some integer.

Then:

 n \cdot m = (2k + 1)(2s + 1) \quad \text{ by substitution} 

 \quad = 4ks + 2s + 2k + 1 

 \quad = 2(2ks + s + k) + 1  \quad \text{ by algebra} 

Let t = 2ks + s + k.

Then n \cdot m = 2(2ks + s + k) + 1 = 2t + 1 where t is an integer because the products and sums of integers is an integer.

Therefore n \cdot m is odd by the definition of odd integers.

Q.E.D.

21. The negative of any odd integer is odd.

Theorem:

Suppose n is any odd integer.

Proof:

Since n is odd, n = 2k + 1 where k is some integer.

Then:

 -n = -(2k + 1) \quad \text{ by substitution} 

 \quad = -2k - 1 

 \quad = -2k - 2 + 1 

 \quad = 2(-k - 1) + 1 

Let t = -k - 1.

Then -n = 2(-k - 1) + 1 = 2t + 1 where t is an integer because the products and differences of integers is an integer.

Therefore -n is odd by definition of an odd integer.

Q.E.D.

22. For all integers a and b, 4a + 5b + 3 is even.

False. Intuition says if a = b = 0 then 4a + 5b + 3 = 3 which is not even. Let's prove this more formally.

Take the negation:

There exists some integer a and some integer b such that 4a + 5b + 3 is not even.

Theorem: There is some integer a and some integer b such that 4a + 5b + 3 is not even.

Let a = 0 and let b = 0

Then:

 4a + 5b + 3 = 4(0) + 5(0) + 3 \quad \text{ by substitution} 

 \quad = 0 + 0 + 3 

 \quad = 3 

Since 3 is not even, 4a + 5b + 3 is not even for the given a and b.

Therefore there exists integers a and b such that 4a + 5b + 3 is not even and the original given statement is false.

Q.E.D.

23. The product of any even integer and any integer is even.

Theorem:

Suppose n is any even integer and m is any integer.

Proof:

Since n is even, n = 2k for some integer k.

Then:

 n \cdot m = (2k)(m) \quad \text{ by substitution} 

 \quad = 2km 

 \quad = 2(km) 

Let t = km.

Then n \cdot m = 2(km) = 2t where t is an integer because the product of integers is an integer.

Therefore n \cdot m is even by definition of even integers.

Q.E.D.

24. If a sum of two integers is even, then one of the summands is even. (In the expression a + b, a and b are called summands.)

This is false, quickly consider 1 + 3 = 4 where both the summands are odd, but the sum is even.

Take the negation first:

There exists two integers whose sum is even but neither integer is even.

Claim:

There is some integer a and there is some integer b such that a + b is even and neither a nor b is even.

Proof:

Let a = 1 and b = 3.

Then:

 a + b = 4 

 \quad = 2(2) 

Then a + b is even by the definition of even integers.

Then:

 a = 1 

 \quad = 2(0) + 1 

And:

 b = 3 

 \quad = 2(1) + 1 

Then both a and b are odd by the definition of odd integers.

Therefore a + b is even, but a and b are not even for the given a and b, therefore the statement is false.

Q.E.D.

25. The difference of any two even integers is even.

Theorem:

Suppose m is an even integer and n is an even integer.

Proof:

Since m and n are even integers, m = 2k and n = 2s where k is some integer and s is some integer.

Then:

 n - m = (2s) - (2k) \quad \text{ by substitution} 

 \quad = 2s - 2k 

 \quad = 2(s - k) \quad \text{ by algebra} 

Let t = s - k.

Then n -m = 2(s - k) = 2t where t is an integer because the difference of integers is an integer.

Therefore n - m is even by the definition of even integers.

Q.E.D.

26. For all integers a, b, and c, if a, b, and c are consecutive, then a + b + c is even.

This is false. Take the negation for the claim.

Claim:

There exists some integer a, some integer b, and some integer c such that a, b, and c are consecutive and a + b + c is not even.

Proof:

Let a = 2, b = 3, c = 4.

Then:

 a + b + c = 2 + 3 + 4 \quad \text{ by substitution} 

 \quad = 9 

 \quad = 8 + 1 

 \quad = 2(4) + 1 

Therefore for the given a, b, and c, a + b + c is not even, by the definition of an odd number.

Therefore the given a, b, and c are consecutive numbers, but their sum is not even. The statement is false.

Q.E.D.

27. The difference of any two odd integers is even.

Theorem:

Suppose n is any odd integer and m is any odd integer.

Proof:

Since n and m are odd integers, n = 2k + 1 and m = 2s + 1 where k is some integer and s is some integer.

Then:

 n - m = (2k + 1) - (2s + 1) \quad \text{ by substitution} 

 \quad = 2k + 1 - 2s - 1 

 \quad = 2k - 2s 

 \quad = 2(k - s) 

Let t = k - s.

Then n - m = 2(k - s) = 2t where t is an integer because the difference of integers is an integer.

Therefore n - m is even by definition of an even integer.

Q.E.D.

28. For all integers n and m, if n - m is even then n^3 - m^3 is even.

Theorem:

Suppose n is any integer and m is any integer and n - m is even.

Proof:

Since we know that n - m is even, n - m = 2k where k is some integer.

Then:

 n^3 - m^3 = (n - m)(n^2 + nm + m^2) \quad \text{ by factoring} 

 \quad = 2k(n^2 + nm + m^2) \quad \text{ by substitution} 

 \quad = 2[k(n^2 + nm + m^2)] 

Let t = k(n^2 + nm + m^2).

Then n^3 - m^3 = 2t where t is an integer because products and sums of integers is an integer.

Therefore n^3 - m^3 is even by the definition of even integers.

Q.E.D.

29. For every integer n, if n is prime then (-1)^n = -1.

This is false when n = 2. Let's prove our claim.

Claim:

There exists some integer n such that n is prime and (-1)^n \neq -1.

Proof:

Let n = 2.

Then:

 (-1)^n = (-1)^2 \quad \text{ by substitution} 

 \quad = 1 

 1 \neq -1 

Therefore since there is a prime number for n such that (-1)^n \neq -1, the given statement is false.

Q.E.D.

30. For every integer m, if m > 2 then m^2 - 4 is composite.

This is false. If m = 3, then m^2 - 4 = 9 - 4 = 5 which is not composite. Let's prove our claim.

Claim:

There exists some integer m such that m > 2 and m^2 - 4 is not composite.

Proof:

Let m = 3.

Then:

 m^2 - 4 = (3)^2 - 4 \quad \text{ by substitution} 

 \quad = 9 - 4 

 \quad = 5 

 \quad = (1)(5) 

Since m^2 - 4 cannot be written as the product of 2 factors where both factors are greater than 1, m^2 - 4 is not composite.

Therefore since there is some integer m such that m > 2 and m^2 - 4 is not composite, this statement is false.

Q.E.D.

31. For every integer n, n^2 - n + 11 is a prime number.

This is false for when n = 11, let's formalize our claim.

Claim:

There exists some integer n such that n^2 - n + 11 is not a prime number.

Proof:

Let n = 11.

Then:

 n^2 - n + 11 = (11)^2 - (11) + 11 \quad \text{ by substitution} 

 \quad = 121 - 11 + 11 

 \quad = 121 

 \quad = (11)(11) 

Therefore n^2 - n + 11 is not a prime number since it is divisible by a number other than 1 and itself for this given n.

Thus there exists some integer n such that n^2 - n + 11 is not a prime number, and therefore the given statement is false.

Q.E.D.

32. For every integer n, 4(n^2 + n + 1) - 3n^2 is a perfect square.

Theorem:

Suppose n is any integer.

Proof:

Then:

 4(n^2 + n + 1) - 3n^2 = 4n^2 + 4n + 4 - 3n^2 \quad \text{ by distribution} 

 \quad = n^2 + 4n + 4 

 \quad = (n + 2)(n + 2) 

 \quad = (n + 2)^2 

Let t = n + 2.

Then 4(n^2 + n + 1) - 3n^2 = t^2 where t is an integer because the sum of integers is an integer.

Therefore 4(n^2 + n + 1) - 3n^2 is a perfect square by the definition of perfect squares.

Q.E.D.

33. Every positive integer can be expressed as a sum of three or fewer perfect squares.

This is false.

Claim:

There exists some positive integer x such that x cannot be expressed as the sum of three or fewer perfect squares.

Proof:

Let x = 7. We check all sums of three nonnegative perfect squares a^2 + b^2 + c^2, where a, b, c \in \{0, 1, 2\} because 3^2 = 9 > 7.

Possible squares: 0^2 = 0, 1^2 = 1, 2^2 = 4.

Now we check all sums

1. Using only 0 and 1:

• 0 + 0 + 1 = 1, 0 + 1 + 1 = 2, 1 + 1 + 1 = 3

All of these are too small and do not add up to 7.

2. Using a 4 (2^2) with 0 and 1:

• 4 + 0 + 0 = 4, 4 + 0 + 1 = 5, 4 + 1 + 1 = 6, 4 + 4 + 0 = 8

All of these do not equal 7.

No combination sums to 7.

Therefore, since all possible combinations from the given set of numbers that could potentially sum to 7 when each individual number is squared have been exhausted, it can be concluded that x = 7 cannot be expressed as the sum of three or fewer perfect squares.

Therefore there exists at least one integer x such that x cannot be expressed as a sum of three or fewer perfect squares, and this statement is false.

Q.E.D.

34. (Two integers are consecutive if, and only if, one is one more than the other.) Any product of four consecutive integers is one less than a perfect square.

Theorem:

Suppose n is any integer.

Proof:

Then:

 (n)(n + 1)(n + 2)(n + 3) = (n(n + 3))(n + 1)(n + 2) 

 \quad = (n^2 + 3n)(n^2 + 3n + 2)

 \quad = (n^2 + 3n)((n^2 + 3n) + 2)

Let x = n^2 + 3n.

Then:

 \quad = (x)((x) + 2) 

 \quad = x^2 + 2x 

 \quad = x^2 + 2x + 1 - 1 

 \quad = (x^2 + 2x + 1) - 1 

 \quad = (x + 1)(x + 1) - 1 

 \quad = (x + 1)^2 - 1 

Then remove the substitution:

 \quad = ((n^2 + 3n) + 1)^2 - 1 

 \quad = (n^2 + 3n + 1)^2 - 1 

Since n^2 + 3n + 1 is an integer because the products and sum of integers is an integer, this means that (n^2 + 3n + 1)^2 is a perfect square and (n^2 + 3n + 1)^2 - 1 is one less than a perfect square.

Therefore the product of any four consecutive integers is one less than a perfect square.

Q.E.D.

35. If m and n are any positive integers and mn is a perfect square, then m and n are perfect squares.

This is false.

Claim:

There is a positive integer m and there is a positive integer n such that mn is a perfect square and m and n are not perfect squares.

Proof:

Let m = 2 and n = 8.

Then:

 mn = (2)(8) \quad \text{ by substitution} 

 \quad = 16 

 \quad = 4^2 

Then mn is a perfect square, but m and n are not perfect squares.

Therefore there exists some m and there exists some n such that mn is a perfect square and m and n are not perfect squares, proving the statement false.

Q.E.D.

36. The difference of the squares of any two consecutive integers is odd.

Theorem:

Suppose n is any integer.

Proof:

Then:

 (n + 1)^2 - n^2 = (n + 1)(n + 1) - n^2 \quad \text{ by substitution} 

 \quad = n^2 + 2n + 1 - n^2 

 \quad = 2n + 1 

Therefore (n + 1)^2 - n^2 is odd by the definition of an odd integer.

Q.E.D.

37. For all nonnegative real numbers a and b, \sqrt{ab} = \sqrt{a}\sqrt{b}. (Note that if x is a nonnegative real number, then there is a unique nonnegative real number y, denoted \sqrt{x}, such that y^2 = x.)

Theorem:

Suppose a is any nonnegative real number and b is any nonnegative real number.

Proof:

Since a \geq 0 and b \geq 0, we know that \sqrt{a} and \sqrt{b} are defined nonnegative real numbers such that:

 (\sqrt{a})^2 = a 

and

 (\sqrt{b})^2 = b 

Then:

 (\sqrt{a}\sqrt{b})^2 = (\sqrt{a})^2(\sqrt{b})^2 = ab \quad \text{ by the product of powers property} 

We then know that \sqrt{a}\sqrt{b} \geq 0 because both factors are nonnegative.

So \sqrt{a}\sqrt{b} is a nonnegative real number whose square is ab.

Therefore \sqrt{ab} = \sqrt{a}\sqrt{b} by the definition of square root (uniqueness of the nonnegative number whose square is ab).

Q.E.D.

38. For all nonnegative real numbers a and b,

 \sqrt{a + b} = \sqrt{a} + \sqrt{b} 

This is false.

Claim:

There is some nonnegative real number a and some nonnegative real number b such that \sqrt{a + b} \neq \sqrt{a} + \sqrt{b}.

Proof:

Let a = 9 and b = 16.

Then:

 \sqrt{a + b} = \sqrt{9 + 16} \quad \text{ by substitution} 

 \quad = \sqrt{25} 

 \quad = 5 

Then:

 5 \stackrel{?}{=} \sqrt{a} + \sqrt{b} 

 5 \stackrel{?}{=} \sqrt{9} + \sqrt{16} \quad \text{ by substitution} 

 5 \stackrel{?}{=} 3 + 4 

 5 \neq 7 

Therefore for the given a and b, we have shown that \sqrt{a + b} \neq \sqrt{a} + \sqrt{b}, thus proving the statement false.

Q.E.D.

39. Suppose that integers m and n are perfect squares. Then m + n + 2\sqrt{mn} is also a perfect square. Why?

Theorem:

Suppose m is any perfect square and n is any perfect square.

Proof:

Since m and n are perfect squares, m = k^2 and n = s^2 for some integer k and some integer s.

Then:

 m + n + 2\sqrt{mn} = (k^2) + (s^2) + 2\sqrt{(k^2)(s^2)} \quad \text{ by substitution} 

 \quad = k^2 + s^2 + 2\sqrt{k^2}\sqrt{s^2} \quad \text{ by the product of powers property} 

 \quad = k^2 + s^2 + 2ks 

 \quad = k^2 + 2ks + s^2 \quad \text{ by the commutative property} 

 \quad = (k + s)(k + s) 

 \quad = (k + s)^2 

Therefore m + n + 2\sqrt{mn} is a perfect square by the definition of a perfect square.

Q.E.D.

40. If p is a prime number, must 2^p - 1 also be prime? Prove or give a counterexample.

False.

Claim:

There is some prime number p such that 2^p - 1 is not prime.

Let p = 11.

Then:

 2^p - 1 = 2^{11} - 1 \quad \text{ by substitution} 

 \quad = 2048 - 1 

 \quad = 2047 

 \quad = (23)(89) 

Thus there is a case where p is a prime number and 2^p - 1 is not prime, and therefore the given statement is false.

Q.E.D.

41. If n is a nonnegative integer, must 2^{2n} + 1 be prime? Prove or give a counterexample.

False.

Claim:

There exists a nonnegative integer n such that 2^{2n} + 1 is not prime.

Proof:

Let n = 5.

Then:

 2^{2n} + 1 = 2^{2 \cdot 5} + 1 \quad \text{ by substitution} 

 \quad = 2^{10} + 1 

 \quad = 1024 + 1 

 \quad = 1025 

 \quad = (25)(41) 

Thus there exists an nonnegative integer n such that 2^{2n} + 1 is not prime, and therefore the given statement is false.

Q.E.D.

---

Exercise Set 4.3

Page 210

The numbers in 1-7 are all rational. Write each number as a ratio of two integers.

1. -\dfrac{35}{6}

 -\frac{35}{6} = -\frac{35}{6} 

2. 4.6037

 4.6037 = \frac{46037}{10000} 

3. \dfrac{4}{5} + \dfrac{2}{9}

 \frac{4}{5} + \frac{2}{9} = \frac{9(4) + 5(2)}{45} = \frac{46}{45} 

4. 0.37373737\dots

Let x = 0.37373737\dots, then 100x = 37.373737\dots, so 100x - x = 99x = 37. Therefore:

 x = 0.37373737\dots = \frac{37}{99} 

5. 0.56565656\dots

Let x = 0.56565656\dots, then 100x = 56.565656\dots, so 100x - x = 99x = 56. Therefore:

 x = 0.56565656\dots = \frac{56}{99} 

6. 320.5492492492\dots

 x = 320.5492492492\dots 

 10000x = 3205492.492492492\dots 

 10x = 3205.492492492\dots 

 10000x - 10x = 9990x = 3202287 

 x = \frac{3202287}{9990} 

7. 52.4672167216721\dots

 x = 52.4672167216721\dots 

 100000x = 5246721.672167216721\dots 

 10x = 524.672167216721\dots

 100000x - 10x = 99990x  = 5246197 

 x = 52.4672167216721\dots = \frac{5246197}{99990} 

8. The zero product property, says that if a product of two real numbers is 0, then one of the numbers must be 0.

a. Write this property formally using quantifiers and variables.

Let P(x) be "x = 0."

Let Q(y) be "y = 0."

Let R(x, y) be "(x)(y) = 0."

 \forall x \in \mathbb{R} \forall y \in \mathbb{R} (R(x, y) \to (P(x) \vee Q(y))) 

b. Write the contrapositive of your answer to part (a).

 \forall x \in \mathbb{R} \forall y \in \mathbb{R} ((\neg P(x) \wedge \neg Q(y)) \to \neg R(x, y)) 

c. Write an informal version (without quantifier symbols or variables) for your part to part (b).

If any two real numbers do not equal zero, then their product does not equal zero.

9. Assume that a and b are both integers and that a \neq 0 and b \neq 0. Explain why \dfrac{(b - a)}{(ab^2)} must be a rational number.

A rational number is a ratio of integers with a nonzero denominator. The given fraction

 \frac{(b - q)}{(ab^2)} 

is rational, the numerator is an integer as the difference of integers are integers, and the denominator is an integer because the product of integers are integers, also the assumption states that both a and b are not 0, so the denominator cannot be 0 by the zero product property. Hence the given fraction is a rational number.

10. Assume that m and n are both integers and that n \neq 0. Explain why \dfrac{(5m - 12n)}{(4n)} must be a rational number.

Given that m and n are both integers, in the given fraction

 \frac{(5m -12n)}{4n} 

The numerator 5m - 12n is an integer because the difference of integers are integers. The denominator 4n is an integer because the product of integers are integers. Also, the since n \neq 0, 4n \neq 0 by the zero product property. Hence the given fraction is a rational number.

11. Prove that every integer is a rational number.

Theorem:

Suppose x is any integer.

Proof:

Then:

 x = x \cdot 1  

 \dfrac{x}{1} = x 

Then x is an integer and 1 is an integer where 1 \neq 0. Hence x can be expressed as a quotient of integers with a nonzero denominator and therefore x is a rational number by definition of a rational number.

Q.E.D.

12. Let S be the statement "The square of any rational number is rational." A formal version of S is "For every rational number r, r^2 is rational." Fill in the blanks in the proof for S.

Proof:

Suppose that r is __ (a) __. By definition of rational, r = \dfrac{a}{b} for some __ (b) __ with b \neq 0. By substitution,

 r^2 = \text{\_\_ (c) \_\_} = \frac{a^2}{b^2} 

Since a and b are both integers, so are the products a^2 and __ (d) __. Also b^2 \neq 0 by the __ (e) __. Hence r^2 is a ratio of two integers with a non-zero denominator,n and so __ (f) __ by definition of rational.

a. a rational number

b. integers a and b

c. \left(\frac{a}{b}\right)^2

d. b^2

e. zero product property

f. r^2 is a rational number

13. Consider the following statement: The negative of any rational number is rational.

a. Write the statement formally using a quantifier and a variable.

 \forall q \in \mathbb{Q} (-q \in \mathbb{Q}) 

Alternatively:

 \forall q ((q \in \mathbb{Q}) \in (-q \in \mathbb{Q})) 

b. Determine whether the statement is true or false and justify your answer.

Theorem:

Suppose q is any rational number.

Proof:

Since q is a rational number, q can be expressed as \dfrac{a}{b} where a and b are integers and b \neq 0.

Then:

 -q = -\left(\frac{a}{b}\right) \quad \text{ by substitution} 

 -q = \frac{-a}{b} 

Then the numerator -a is an integer because the product of integers are integers. The denominator b is an integer and b \neq 0 by assumption of q as a rational number. Hence -q can be expressed as the ratio of two integers with a nonzero denominator, and therefore -q is a rational number by definition of rational numbers.

Q.E.D.

14. Consider the statement: The cube of any rational number is a rational number.

a. Write the statement formally using a quantifier and a variable.

 \forall q ((q \in \mathbb{Q}) \to (q^3 \in \mathbb{Q})) 

b. Determine whether the statement is true or false and justify your answer.

Theorem:

Suppose q is any rational number.

Proof:

Since q is a rational number, q = \dfrac{a}{b} where a and b are integers and b \neq 0.

Then:

 q^3 = \left(\frac{a}{b}\right)^3 \quad \text{ by substitution} 

 q^3 = \frac{a^3}{b^3} \quad \text{ by power of a quotient} 

Then the numerator a^3 is an integer because the products of integers are integers. Also the denominator b^3 is an integer because the product of integers are integers and b^3 \neq 0 by the zero product property.

Thus q^3 can be expressed as a ratio of two integers with a nonzero denominator and therefore q^3 is a rational number by definition of a rational number.

Q.E.D.

Determine which of the statements in 15-19 are true and which are false. Prove each true statement directly from the definitions, and give a counterexample for each false statement. For a statement that is false, determine whether a small change would make it true. If so, make the change and prove the new statement. Follow the directions for writing proofs on page 173.

15. The product of any two rational numbers is a rational number.

Theorem:

Suppose q and r are rational numbers.

Proof:

Since q and r are rational numbers, then q = \dfrac{a}{b} and r = \dfrac{c}{d} where a, b, c, and d are some integers and b \neq 0 and d \neq 0.

Then:

 qr = \left(\frac{a}{b}\right)\left(\frac{c}{d}\right) \quad \text{ by substitution} 

 qr = \frac{ac}{bd} 

Then the numerator ac is an integer because the product of integers are integers. The denominator bd is an integer because the product of integers are integers, and bd \neq 0 by the of the zero product property.

Thus qr can be expressed as a ratio of two integers with a nonzero denominator and therefore qr is a rational number by the definition of a rational number.

Q.E.D.

16. The quotient of any two rational numbers is a rational number.

This is false.

Claim:

There exists some rational number q and some rational number r such that \dfrac{q}{r} is not a rational number.

Proof:

Let q = \dfrac{1}{2} and r = \dfrac{0}{1}.

Then:

 \frac{q}{r} = \frac{\dfrac{1}{2}}{\dfrac{0}{1}} \quad \text{ by substitution} 

 \quad = \frac{1}{2 \cdot 0} 

 \quad = \text{ undefined} 

So the numerator of the given \dfrac{q}{r} is 1 which is an integer, but the denominator is 0, which means \dfrac{q}{r} is not any number, and therefore not a rational number.

Thus there exists two rational numbers whose quotients are not a rational number, therefore the statement is false.

Q.E.D.

A small change that would make this true were if the statement were reworded as:

For any two rational numbers, the quotient of those two numbers is a rational number as long as the rational number in the divisor doesn't equal 0.

17. The difference of any two rational numbers is a rational number.

Theorem:

Suppose that q and r are any rational numbers.

Proof:

Since q and r are rational numbers, q = \dfrac{a}{b} and r = \dfrac{c}{d} where a, b, c, and d are some integers and b \neq 0 and d \neq 0.

Then:

 q - r = \frac{a}{b} - \frac{c}{d} \text{ by substitution} 

 \quad = \frac{ad - cb}{bd} 

Then the numerator ad - cb is an integer because the difference and products of integers are integers. The denominator bd is an nonzero integer because the products of integers are integers and because of the zero product property.

Thus q - r can be expressed as a ratio of two integers with a nonzero denominator, and therefore q - r is a rational number by the definition of a rational number.

Q.E.D.

18. If r and s are any two rational numbers, then \dfrac{r + s}{2} is rational.

Theorem:

Suppose r and s are any two rational numbers.

Proof:

Since r and s are rational numbers, then r = \dfrac{a}{b} and s = \dfrac{c}{d} where a, b, c, and d are integers and b \neq 0 and d \neq 0.

Then:

By substitution:

 \frac{r + s}{2} = \frac{\dfrac{a}{b} + \dfrac{c}{d}}{2} 

 \quad = \frac{1}{2}\left(\frac{a}{b} + \frac{c}{d}\right) 

 \quad = \frac{a}{2b} + \frac{c}{2d} 

 \quad = \frac{ad + bc}{2bd} 

Then the numerator ad + bc is an integer because the products and sums of integers are integers. The denominator 2bd is a nonzero integer because the products of integers are integers and because of the zero product property.

Thus \dfrac{r + s}{2} can be expressed as the ratio of two integers with a nonzero denominator, and therefore \dfrac{r + s}{2} is a rational number by the definition of a rational number.

Q.E.D.

19. For all real numbers a and b, if a < b then a < \dfrac{a + b}{2} < b.

    (You may use the properties of inequalities in T17-T27 of Appendix A.)

Theorem:

Suppose a and b are any real numbers and that a < b.

Proof:

Then:

By T19:

 a + a < a + b 

 2a < a + b 

By T20:

 a < \frac{a + b}{2} 

And:

By T19:

 a + b < b + b 

 a + b < 2b 

By T20:

 \frac{a + b}{2} < b 

Therefore a < \dfrac{a + b}{2} < b.

Q.E.D.

20. Use the results of exercises 18 and 19 to prove that given any two rational numbers r and s with r < s, there is another rational number between r and s. An important consequence is that there are infinitely many rational numbers in between any two distinct rational numbers. See Section 7.4.

Theorem:

Suppose r is any rational number and s is any rational number where r < s.

Proof:

By 18, we know that \dfrac{r + s}{2} is a rational number.

By 19, we know that if r < s, then r < \dfrac{r + s}{2} < s.

Therefore there exists some rational number \dfrac{r + s}{2} that is between r and s, [as was to be shown].

Q.E.D.

Use the properties of even and odd integers that are listed in Example 4.3.3 to do exercises 21-23. Indicate which properties you use to justify your reasoning.

21. True or false? If m is any even integer and n is any odd integer, then m^2 + 3n is odd. Explain.

Theorem:

Suppose m is any even integer and n is any odd integer.

Proof:

By 3, the product of any two odd integers is odd, then:

 3n \text{ is odd} 

Since m is even, m = 2k for some integer k.

Then:

By substitution:

 m^2 = (2k)^2 

 \quad = 4k^2 

 \quad = 2(2k^2) 

Then m^2 is even by the definition of an even integer.

 m^2 \text{ is even} 

By 5, the sum of any odd integer and any even integer is odd.

Thus m^2 + 3n is odd, therefore the statement is true.

Q.E.D.

22. True or false? If a is any odd integer, then a^2 + a is even. Explain.

Theorem:

Suppose a is any odd integer.

Proof:

Then:

 a^2 = a \cdot a 

By 3, the product of any two odd integers is odd, then:

 a^2 \text{ is odd} 

By 2, the sum and difference of any two odd integers are even, then:

 a^2 + a \text{ is even} 

Therefore the statement is true.

Q.E.D.

23. True or false? If k is any even integer and m is any odd integer, then (k + 2)^2 - (m - 1)^2 is even. Explain.

Theorem:

Suppose k is any even integer and m is any odd integer.

Proof:

By 1, the sum of any two even integers is even, then:

 k + 2 \text{ is even} 

By 1, the product of any two even integers is even, then:

 (k + 2)^2 = (k + 2)(k + 2) 

 (k + 2)^2 \text{ is even} 

By 2 the difference of any two odd integers is even, then:

 m - 1 \text{ is even} 

By 1 the product of any two even integers is even, then:

 (m - 1)^2 = (m - 1)(m - 1) 

 (m - 1)^2 \text{ is even} 

By 1 the difference of any two even integers is even, then:

 (k + 2)^2 - (m - 1)^2 \text{ is even} 

Therefore the statement is true.

Q.E.D.

Derive the statements in 24-26 as corollaries of Theorems 4.3.1, 4.3.2, and the results of exercises 12, 13, 14, 15, and 17.

24. For any rational numbers r and s, 2r + 3s is rational.

Theorem:

By 15, the product of any two rational numbers is a rational number, then:

 2r \text{ is rational} 

and

 3s \text{ is rational} 

By Theorem 4.3.2, the sum of any two rational numbers is rational, then:

 2r + 3s \text{ is rational} 

Therefore the statement is true.

Q.E.D.

Proof:

25. If r is any rational number, then 3r^2 - 2r + 4 is rational.

By 15, the product of any two rational numbers is a rational number, then:

 r^2 = r \cdot r 

 r^2 \text{ is rational} 

 3r^2 = 3 \cdot r^2 

 3r^2 \text{ is rational} 

and

 2r = 2 \cdot r 

 2r \text{ is rational} 

By 17, the difference of any two rational numbers is a rational number, then:

 3r^2 - 2r \text{ is rational} 

By Theorem 4.3.2, the sum of any two rational numbers is rational, then:

 (3r^2 - 2r) + 4 \text{ is rational} 

Therefore the statement is true.

Q.E.D.

26. For any rational number s, 5s^3 + 8s^2 - 7 is rational.

Theorem:

Suppose s is any rational number.

Proof:

By 15, the product of any two rational numbers is a rational number, then:

 s^2 = s \cdot s 

 s^2 \text{ is rational} 

 s^3 = s^2 \cdot s 

 s^3 \text{ is rational} 

 5s^3 = 5 \cdot s^3 

 5s^3 \text{ is rational} 

and

 8s^2 = 8 \cdot s^2 

 8s^2 \text{ is rational} 

By 17, the difference of any two rational numbers is a rational number, then:

 8s^2 - 7 \text{ is rational} 

By Theorem 4.3.2, the sum of any two rational numbers is rational, then:

 5s^3 + (8s^2 - 7) \text{ is rational} 

Therefore the statement is true.

Q.E.D.

27. It is a fact that if n is any nonnegative integer, then

 1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots + \frac{1}{2^n} = \frac{1 - \left(\dfrac{1}{2^{n + 1}}\right)}{1 - \left(\dfrac{1}{2}\right)}  

(A more general form of this statement is proved in Section 5.2.) Is the right-hand side of this equation rational? If so, express it as a ratio of two integers.

Theorem:

Suppose n is any nonnegative integer.

Proof:

Consider:

 \frac{1 - \left(\dfrac{1}{2^{n + 1}}\right)}{1 - \left(\dfrac{1}{2}\right)} 

The denominator can be simpilifed as:

 1 - \frac{1}{2} = \frac{1}{2}  

Then:

 \frac{1 - \left(\dfrac{1}{2^{n + 1}}\right)}{\frac{1}{2}} 

 \quad = 2\left(1 - \frac{1}{2^{n + 1}}\right)

 \quad = 2 - \frac{2}{2^{n + 1}} 

 \quad = 2 - \frac{1}{2^n} 

 \quad = \frac{2 \cdot 2^n - 1}{2^n}

 \quad = \frac{2^{n + 1} - 1}{2^n}

Since n is a nonnegative integer, the numerator 2^{n + 1} - 1 is an integer because the products and differences of integers are integers. And the denominator 2^n is a positive integer because of the products of integers and because n is a nonnegative integer.

Therefore right-hand side of the given equation is a rational number by the definition of rational numbers.

Q.E.D.

28. Suppose a, b, c, and d are integers and a \neq c. Suppose also that x is a real number that satisfies the equation

 \frac{ax + b}{cx + d} = 1 

Must x be rational? If so, express x as a ratio of two integers.

Theorem:

Suppose a, b, c, and d are any integers and suppose x is a real number that satisfies the equation:

 \frac{ax + b}{cx + d} = 1 

Proof:*

Consider:

 \frac{ax + b}{cx + d} = 1 

 ax + b = cx + d 

 ax - cx = d - b 

 x(a - c) = d - b 

 x = \frac{d - b}{a - c} 

Then the numerator d - b is an integer because the difference of integers are integers. The denominator a - c must be a nonzero integer because the difference of integers are integers and because a \neq c.

Therefore x must be a rational number by the definition of rational numbers.

Q.E.D.

29. Suppose a, b, and c are integers and x, y, and z are nonzero real numbers that satisfy the following equations:

 \frac{xy}{x + y} = a \quad \text{ and } \quad \frac{xz}{x + z} = b \quad \text{ and } \quad \frac{yz}{y + z} = c 

Is x rational? If so, express it as ratio of two integers.

Omitted.

30. Prove that if one solution for a quadratic equation of the form x^2 + bx + c = 0 is rational (where b and c are rational), then the other solution is also rational. (Use the fact that if the solutions of the equation are r and s, then x^2 + bx + c = (x - r)(x - s).)

Theorem:

Suppose there is any rational number r that is a solution to a quadratic equation of the form:

 x^2 + bx + c = 0 

Where b and c are rational.

And suppose s is the other solution to the given equation.

Proof:

Given that both r and s are solutions to the given equation, then:

 x^2 + bx + c = (x - r)(x - s) = x^2 - rx - sx + rs 

This means that:

 bx = (-r - s)x 

 b = -1(r + s) 

And:

 c = rs 

Let's analyze b and isolate s.

 b = -1(r + s) 

 -b = r + s 

 -b - r = s 

Since both b and r are rational numbers, then b = \dfrac{g}{h} and r = \dfrac{i}{j} where g, h, i, and j are some integers and h \neq 0 and j \neq 0.

Then:

 s = -b - r = -\left(\frac{g}{h}\right) - \left(\frac{i}{j}\right) \text{ by substitution} 

 \quad = \frac{-1(gj + hi)}{hj} 

The numerator -1(gj + hi) is an integer because the sum and products of integers are integers. The denominator is a nonzero integer because the products of integers are integers and because of the zero product property.

Therefore s can be expressed as a ratio of two integers where the denominator is nonzero. Thus s is a rational number by the definition of rational numbers, therefore the other solution is rational.

Q.E.D.

31. Prove that if a real number c satisfies a polynomial equation of the form

 r_3x^3 + r_2x^2 + r_1x + r_0 = 0 

where r_0, r_1, r_2, and r_3 are rational numbers, then c satisfies an equation of the form

 n_3x^3 + n_2x^2 + n_1x + n_0 = 0 

where n_0, n_1, n_2, and n_3 are integers.

Definition: A number c is called a root of a polynomial p(x) if, and only if, p(c) = 0.

Theorem:

Suppose c is any real number that satisfies a polynomial equation of the form

 r_3x^3 + r_2x^2 + r_1x + r_0 = 0 

where r_0, r_1, r_2, and r_3 are rational numbers.

Proof:

Since c is a real number that satisfies the given equation, then:

 r_3c^3 + r_2c^2 + r_1c + r_0 = 0 

Since r_3, r_2, r_1, and r_0 are rational numbers, then r_3 = \dfrac{a_3}{b_3}, r_2 = \dfrac{a_2}{b_2}, r_1 = \dfrac{a_1}{b_1}, and r_0 = \dfrac{a_0}{b_0} where a_3, a_2, a_1, a_0 are some integers and b_3, b_2, b_1, b_0 are some nonzero integers.

Then, by substitution:

 r_3c^3 + r_2c^2 + r_1c + r_0 = \left(\frac{a_3}{b_3}\right)c^3 + \left(\frac{a_2}{b_2}\right)c^2 + \left(\frac{a_1}{b_1}\right)c + \frac{a_0}{b_0} = 0 

 \quad = \frac{a_3b_2b_1b_0c^3 + a_2b_3b_1b_0c^2 + a_1b_3b_2b_0c + a_0b_3b_2b_1}{b_3b_2b_1b_0} = 0 

 \quad = a_3b_2b_1b_0c^3 + a_2b_3b_1b_0c^2 + a_1b_3b_2b_0c + a_0b_3b_2b_1 = 0 

Let n_3 = a_3b_2b_1b_0, and n_2 = a_2b_3b_1b_0, and n_1 = a_1b_3b_2b_0, and n_0 = a_0b_3b_2b_1.

Then n_3, n_2, n_1, and n_0 are integers because of the product of integers.

Thus we can write the given equation as:

 n_3c^3 + n_2c^2 + n_1c + n_0 = 0 

Where c is a real number that satisfies the equation:

 n_3x^3 + n_2x^2 + n_1x + n_0 = 0 

Q.E.D.

32. Prove that for every real number c, if c is a root of a polynomial with rational coefficients, then c is a root of a polynomial with integer coefficients.

Theorem:

Suppose c is a root of a polynomial with rational coefficients.

Proof:

Then

 r_nx^n + r_{n - 1}x^{n - 1} + \dots + r_1x + r_0 = 0 

where each r_i is rational.

Then each r_i can be written as a ratio of integers with nonzero denominators. Let D be a common multiple of all denominators of the r_i. Multiplying the equation by D gives

 s_nx^n + s_{n - 1}x^{n - 1} + \dots + s_1x + s_0 = 0 

where each s_i is an integer.

Thus c is a root of a polynomial with integer coefficients.

Q.E.D.

Use the properties of even and odd integers that are listed in Example 4.3.3 to do exercises 33 and 34.

33. When expressions of the form (x - r)(x - s) are multiplied out, a quadratic polynomial is obtained. For instance, (x - 2)(x - (-7)) = (x - 2)(x + 7) = x^2 + 5x - 14.

a. What can be said about the coefficients of the polynomial obtained by multiplying out (x - r)(x - s) when both r and s are odd integers? When both r and s are even integers? When one of r and s is even and the other odd?

Case when both r and s are odd integers:

Theorem:

Suppose r and s are odd integers.

Conclusion:

Let x be some real number.

Then:

 (x - r)(x - s) = x^2 - rx - sx - rs = x^2 + (-1)(r + s)x + (-1)rs 

We know the coefficient of x^2 is 1.

By 2, we know the sum of any two odd integers are even, then:

We know the coefficient of -1(r + s) is even.

By 3, we know the product of any two odd integers is odd, then:

We know the coefficient of rs is odd.

Case when both r and s are even integers:

Theorem:

Suppose r and s are even integers.

Conclusion:

Let x be some real number.

Then:

 (x - r)(x - s) = x^2 - rx - sx - rs = x^2 + (-1)(r + s)x + (-1)rs 

We know the coefficient of x^2 is 1.

By 1 we know the sum of any two even integers is even, then:

We know the coefficient of (-1)(r + s)x is even.

By 1 we know the product of any two even integers is even, then:

We know the coefficient of (-1)rs is even.

Case where r and s is even and the other odd:

Theorem:

Suppose r and s are any integers where one is even and the other is odd.

Conclusion:

Let x be some real number.

Then:

 (x - r)(x - s) = x^2 - rx - sx - rs = x^2 + (-1)(r + s)x + (-1)rs 

We know the coefficient of x^2 is 1.

By 5, we know the sum of any odd integer and any even integer is odd, then:

We know the coefficient of (-1)(r +s)x is odd.

By 4, we know the product of any even integer and any odd integer is even, then:

We know the coefficient of (-1)rs is even.

b. It follows from part (a) that x^2 - 1253x + 255 cannot be written as a product of two polynomials with integer coefficients. Explain why this is so.

Because in all cases from part (a), the middle coefficient and the third coefficient were always either even and odd or odd and even. Since both 1253 and 255 are odd, this expression cannot be expressed as the product of two polynomials with integer coefficients.

34. Observe that

 (x - r)(x -s)(x - t) = x^3 - (r + s + t)x^2 + (rs + rs + st)x - rst 

a. Derive a result for cubic polynomials similar to the result in part (a) of exercise 33 for quadratic polynomials.

Case where r and s and t are all even:

The coefficient of x^3 is 1.

By 1, the sum of any two even integers is even, then:

 r + s \text{ is even} 

 (r + s) + t \text{ is even} 

The coefficient of (r + s + t)x^2 is even.

By 1, the sum and product of any two even integers is even, then:

 rs \text{ is even} 

 st \text{ is even} 

 (rs + rs + rt) \text{ is even} 

The coefficient of (rs + rs + st)x is even.

By 1, the product of any two even integers is even, then:

 rs \text{ is even} 

 rst \text{ is even} 

The coefficient of rst is even.

Case where r is odd and s and t are even:

The coefficient of x^3 is 1.

By 1 the sum of any two even integers is even, then:

 s + t \text{ is even} 

By 5 the sum of any odd integer and any even integer is odd.

 r + (s + t) \text{ is odd} 

The coefficient of (r + s + t)x^2 is odd.

By 4, the product of any even integer and any odd integer is even, then:

 rs \text{ is even} 

By 1, the sum and product of any two even integers is even.

 st \text{ is even} 

 rs + st \text{ is even} 

 rs + (rs + st) \text{ is even} 

The coefficient of (rs + rs + st)x is even.

 (rs)t \text{ is even} 

The coefficient of rst is even.

Case where r and s are odd and t is even:

The coefficient of x^3 is 1.

By 2, the sum of any two odd integers is even.

 r + s \text{ is even} 

By 1, the sum of any two even integers is even.

 (r + s) + t \text{ is even} 

The coefficient of (r + s + t)x^2 is even.

By 3, the product of any two odd integers is odd.

 rs \text{ is odd} 

By 4, the product of any even integer and any odd integer is even.

 st \text{ is even} 

By 2, the sum of any two odd integers is even.

 rs + rs \text{ is even} 

By 1, the sum of any two even integers is even.

 (rs + rs) + st \text{ is even} 

The coefficient of (rs + rs + st)x is even

By 4, the product of any even integer and any odd integer is even.

 (rs)t \text{ is even} 

The coefficient of rst is even.

Case where r and s and t are all odd:

The coefficient of x^3 is 1.

By 2, the sum of any two odd integers is even.

 r + s \text{ even} 

By 5, the sum of any odd integer and any even integer is odd.

 (r + s) + t \text{ is odd} 

The coefficient of (r + s + t)x^2 is odd.

By 3, The product of any two odd integers is odd.

 rs \text{ is odd} 

 st \text{ is odd} 

By 2, the sum of any two odd integers is even.

 rs + rs \text{ is even} 

By 5, the sum of any odd integer and any even integer is odd.

 (rs + rs) + st \text{ is odd} 

The coefficient of (rs + rs + st)x is odd.

 (x - r)(x -s)(x - t) = x^3 - (r + s + t)x^2 + (rs + rs + st)x - rst 

By 3, The product of any two odd integers is odd.

 (rs)t \text{ is odd.} 

The coefficient of rst is odd.

b. Can x^3 + 7x^2 - 8x - 27 be written as a product of three polynomials with integer coefficients? Explain.

In all cases, the order of the second through fourth terms are never: "odd, even, odd". Therefore the given polynomial x^3 + 7x^2 - 8x - 27 can be written as a product of three polynomials with integer coefficients.

In 35-39 find the mistakes in the "proofs" that the sum of any two rational numbers is a rational number.

"Proof: Any two rational numbers produce a rational number when added together. So if r and s are particular but arbitrarily chosen rational numbers, then r + s is rational."

This proof assumes what is to be proved.

"Proof: Let rational numbers r = \dfrac{1}{4} and s = \dfrac{1}{2} be given. Then r + s = \dfrac{1}{4} + \dfrac{1}{2} = \dfrac{3}{4}, which is a rational number. This is what was to be shown."

This proof is arguing from examples.

"Proof: Suppose r and s are rational numbers. By definition of rational, r = \dfrac{a}{b} for some integers a and b with b \neq 0, and s = \dfrac{a}{b} for some integers a and b with b \neq 0. Then

 r + s = \frac{a}{b} + \frac{a}{b} = \frac{2a}{b} 

Let p = 2a. Then p is an integer since it is a product of integers. Hence r + s = \dfrac{p}{b}, where p and b are integers and b \neq 0. Thus r + s is a rational number by definition of rational. This is what was to be shown."

This incorrect proof uses the same letter to mean two different things.

"Proof: Suppose r and s are rational numbers. Then r = \dfrac{a}{b} and s = \dfrac{c}{d} for some integers a, b, c, and d with b \neq 0 and d \neq 0 (by definition of rational.) Then

 r + s = \frac{a}{b} + \frac{c}{d} 

But this is a sum of two fractions, which is a fraction. So r - s is a rational number since a rational number is a fraction."

This incorrect proof exhibits confusion between what is known and what is still to be shown. Additionally, they simply abandon what is to be shown since what is to be shown is r + s is rational, not r - s is rational.

"Proof: Suppose r and s are rational numbers. If r + s is rational, then by definition of rational r + s = \dfrac{a}{b} for some integers a and b with b \neq 0. Also since r and s are rational, r = \dfrac{i}{j} and s = \dfrac{m}{n} for some integers i, j, m, and n with j \neq 0 and n \neq 0. It follows that

 r + s = \frac{i}{j} + \frac{m}{n} = \frac{a}{b} 

which is a quotient of two integers with a nonzero denominator. Hence it is a rational number. This is what is to be shown.

This incorrect prove is assuming what is to be proved.

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Exercise Set 4.4

Page 220

Give a reason for your answer in each of 1-13. Assume that all variables represent integers.

1. is 52 divisible by 13?

2. Does 7 \mid 56?

3. Does 5 \mid 0?

4. Does 3 divide (3k + 1)(3k + 2)(3k + 3)?

5. Is 6m(2m + 10) divisible by 4?

6. Is 29 a multiple of 3?

7. Is -3 a factor of 66?

8. Is 6a(a + b) a multiple of 3a?

9. Is 4 a factor of 2a \cdot 34b?

10. Does 7 \mid 34?

11. Does 13 \mid 73?

12. If n = 4k + 1, does 8 divide n^2 - 1?

13. If n = 4k + 3, does 8 divide n^2 - 1?

14. Fill in the blanks in the following proof that for all integers a and b, if a \mid b then a \mid (-b).

Proof:

Suppose a and b are integers such that __ (a) __. By definition of divisibility, there exists an integer r such that __ (b) __. By substitution,

 -b = -(ar) = a(-r) 

Let t = __ (c) __. Then t is an integer because t = (-1) \cdot r, and both -1 and r are integers. Thus, by substitution, -b = at, where t is an integer, and by the definition of divisibility, __ (d) __, as was to be shown.

Prove statements 15 and 16 directly from the definition of divisibility.

15. For all integers a, b, and c, if a \mid b and a \mid c then a \mid (b + c).

16. For all integers a, b, and c, if a \mid b then a \mid c then a \mid (b - c).

17. For all integers a, b, c, and d, if a \mid c and b \mid d then ab \mid cd.

18. Consider the following statement: The negative of any multiple of 3 is a multiple of 3.

a. Write the statement formally using a quantifier and a variable.

b. Determine whether the statement is true or false and justify your answer.

19. Show that the following statement is false: For all integers a and b, if 3 \mid (a + b) then 3 \mid (a - b).

For each statement in 20-32, determine whether the statement is true or false. Prove the statement directly from the definitions if it is true, and give a counterexample if it is false.

20. The sum of any three consecutive integers is divisible by 3.

21. The product of any two even integers is a multiple of 4.

22. A necessary condition for an integer to be divisible by 6 is that it be divisible by 2.

23. A sufficient condition for an integer to be divisible by 8 is that it be divisible by 16.

24. For all integers a, b, and c, if a \mid b and a \mid c then a \mid (2b - 3c).

25. For all integers a, b, and c, if a is a factor of c and b is a factor of c then ab is a factor of c.

26. For all integers a, b, and c, if ab \mid c then a \mid c and b \mid c.

27. For all integers a, b, and c, if a \mid (b + c) then a \mid b or a \mid c.

28. For all integers a, b, and c, if a \mid bc then a \mid b or a \mid c.

29. For all integers a and b, if a \mid b then a^2 \mid b^2.

30. For all integers a and n, if a \mid n^2 and a \leq n then a \mid n.

31. For all integers a and b, if a \mid 10b then a \mid 10 or a \mid b.

32. A fast-food chain has a contest in which a card with numbers on it is given to each customer who makes a purchase. If some of the numbers on the card add up to 100, then the customer wins $100. A certain customer receives a card containing the numbers

72, 21, 15, 36, 69, 81, 9, 27, 42, and 63.

Will the customer win $100? Why or why not?

33. Is it possible to have a combination of nickels, dimes, and quarters that add up to $4.72? Explain.

34. Consider a string consisting of a's, b's, and c's where the number of b's is three times the number of a's and the number of c's is five times the number of a's. Prove that the length of the string is divisible by 3.

35. Two athletes run a circular track at a steady pace so that the first completes one round in 8 minutes and the second in 10 minutes. If they both start from the same spot at 4 P.M., when will be the first time they return to the start?

36. It can be shown (see exercises 44-48) that an integer is divisible by 3 if, and only if, the sum of its digits is divisible by 3; an integer is divisible by 9 if, and only if, the sum of its digits is divisible by 9; an integer is divisible by 5 if, and only if, its right-most digit is a 5 or a 0; and an integer is divisible by 4 if, and only if, the number formed by its right-most two digits is divisible by 4. Check the following integers for divisibility by 3, 4, 5, and 9.

a. 637,425,403,705,125

b. 12,858,306,120,312

c. 517,924,440,926,512

d. 14,328,083,360,232

37. Use the unique factorization theorem to write the following integers in standard factored form.

a. 1,176

b. 5,733

c. 3,675

38. Let n = 8,424.

a. Write the prime factorization for n.

b. Write the prime factorization for n^5.

c. Is n^5 divisible by 20? Explain.

d. What is the least positive integer m so that 8,424 \cdot m is a perfect square?

39. Suppose that in standard factored form a = p_1^{e_1}p_2^{e_2} \dots p_k^{e^k}, where k is a positive integer; p_1, p_2, \dots, p_k are prime numbers; and e_1, e_2, \dots, e_k are positive integers.

a. What is the standard factored form for a^3?

b. Find the least positive integer k such that 2^4 \cdot 3^5 \cdot 7 11^2 \cdot k is a perfect cube (that is, it equals an integer to the third power). Write the resulting product as a perfect cube.

a. If a and b are integers and 12a = 25b, does 12 \mid b? does 25 \mid a? Explain.

b. If x and y are integers and 10x = 9y, does 10 \mid y? does 9 \mid x? Explain.

41. How many zeros are at the end of 45^8 \cdot 88^5? Explain how you can answer this question without actually computing the number. (Hint: 10 = 2 \cdot 5.)

42. If n is an integer and n > 1, then n! is the product of n and every other positive integer that is less than n. For example, 5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1.

a. Write 6! in standard factored form.

b. Write 20! in standard factored form.

c. Without computing the value of (20!)^2 determine how many zeros are at the end of this number when it is written in decimal form. Justify your answer.

43. At a certain university 2/3 of the mathematics students and 3/5 of the computer science students have taken a discrete mathematics course. The number of mathematics students who have taken the course equals the number of computer science students who have taken the course. If there are at least 100 mathematics students at the university, what are the least possible number of mathematics students and the least possible number of computer science students at the university?

Definition: Given any nonnegative integer n, the decimal representation of n is an expression of the form

 d_kd_{k + 1} \dots d_2d_1d_0 

where $kr is a nonnegative integer, d_0, d_1, d_2, \dots, d_k (called the decimal digits of n) are integers from 0 to 9 inclusive, dk \neq 0 unless n = 0 and k = 0, and

 n = d_k \cdot 10^k + d_{k + 1} \cdot 10^{k + 1} + \dots + d_2 \cdot 10^2 + d_1 \cdot 10 + d_0 

(For example, 2,503 = 2 \cdot 10^3 + 5 \cdot 10^2 + 0 \cdot 10 + 3.)

44. Prove that if n is any nonnegative integer whose decimal representation ends in 0, then 5 \mid n. (Hint: If the decimal representation of a nonnegative integer n ends in d_0, then n = 10m + d_0 for some integer m.)

45. Prove that if n is any nonnegative integer whose decimal representation ends in 5, then 5 \mid n.

46. Prove that if the decimal representation of a nonnegative integer n ends in d_1d_0 and if 4 \mid (10d_1 + d_0), then 4 \mid n. (Hint: If the decimal representation of a nonnegative integer n ends in d_1d_0, then there is an integer s such that n = 100s + 10d_1 + d_0.)

47. Observe that

7,524 = 7 \cdot 1,000 + 5 \cdot 100 + 2 \cdot 10 + 4 \ \quad = 7(999 + 1) + 5(99 + 1) + 2(9 + 1) + 4 \ \quad = (7 \cdot 99 + 7) + (5 \cdot 99 + 5) + (2 \cdot 9 + 2) + 4 \ \quad = (7 \cdot 999 + 5 \cdot 99 2 \cdot 9) + (7 + 5 + 2 + 4) \ \quad = (7 \cdot 111 \cdot 9 + 5 \cdot 11 \cdot 9 + 2 \cdot 9) + (7 + 5 + 2 + 4) \ \quad = (7 \cdot 111 + 5 \cdot 11 + 2) \cdot 9 + (7 + 5 + 2 + 4) \ \quad = (\text{an integer divisible by 9})i + (\text{the sum of the digits of } 7,524)

Since the sum of the digits of 7,524 is divisible by 9, 7,524 can be written as a sum of two integers each of which is divisible by 9. It follows from exercise 15 that 7,524 is divisible by 9.

Generalize the argument given in this example to any nonnegative integer n. In other words, prove that for any nonnegative integer n, if the sum of the digits of n is divisible by $9r, then n is divisible by 9.

48. Prove that for any nonnegative integer n, if the sum of the digits of n is divisible by 3, then n is divisible by 3.

49. Given a positive integer n written in decimal form, the alternating sum of the digits of n is obtained by starting with the right-most digit, subtracting the digit immediately to its left, adding the next digit to the left, subtracting the next digit, and so forth. For example, the alternating sum of the digits of 180,928 is 8 - 2 + 9 - 0 + 8 - 1 = 22. Justify the fact that for any nonnegative integer n, if the alternating sum of the digits of n is divisible by 11, then n is divisible by 11.

50. The integer 123,123 has the form abc,abc, where a, b, and c are integers from 0 through 9. Consider all six-digit integers of this form. Which prime numbers divide every one of these integers? Prove your answer.