tomit4/discrete_mathematics_with_applications
1
0
Fork 0
Code Issues Pull requests Projects Releases Packages Wiki Activity Actions

🚧 Fin 4.8

Browse source
This commit is contained in:
tomit4 2026-06-14 14:43:57 -07:00
parent 36f5f17ac8
commit a687430ecd
2 changed files with 780 additions and 2 deletions
Download patch file Download diff file Expand all files Collapse all files

776
chapter_4/exercises.md
View file

@ -7807,8 +7807,6 @@ c. State the contrapositive of the result of part (b). The results of exercise
Omitted.
RESUME HERE.
**Test for Primality**
Given an integer $n > 1$, to test whether $n$ is prime check to see if it is
@ -7871,17 +7869,60 @@ Page 256
is a rational number, which contradicts Theorem 4.8.1. Explain the
discrepancy.
The reason for this discrepancy is due to mistaking
$\sqrt{2} \approx 1.41.4213562$ for $\sqrt{2} = 1.414213562$. The calculator
cannot display $\sqrt{2}$ finitely as it is an irrational number and its
non-repeating decimal goes on forever. Therefore, you cannot find equivalencies
for 1.414213562 and express $\sqrt{2}$ as a rational number as it is
"demonstrated" here.
2. Example 4.3.1(h) illustrates a technique for showing that any repeating
decimal number is rational. A calculator display shows the result of a
certain calculation as $40.72727272727$. Can you be sure that the result of
the calculation is a rational number? Explain.
Yes. The reason you can be sure that the result of the calculation is a rational
number is because repeating decimal places can always be expressed as a rational
number usually by subtracting the repeating decimal places from a larger number
with the same repeating decimal places. For the given example:
Let $x = 40.72727272727 \dots$, so $100x = 4072.727272727 \dots$. Then:
$$ 100x - x = 99x = 4072.727272727\dots - 40.72727272727\dots = 4032 $$
$$ 99x = 4032 $$
$$ x = \frac{4032}{99} $$
Which is an expression for 40.72727272727 in rational form.
3. Could there be a rational number whose first trillion digits are the same as
the first trillion digits of $\sqrt{2}$? Explain.
Yes, because the first trillion digits of $\sqrt{2}$ is potentially finite. In
that case it is rational. In another case where the first trillion digits are
then repeated, then we know by 4.3.1(h) that this form of a decimal is also
rational. Similarly, if even smaller portions of those first trillion digits are
then repeated, this same principle applies.
4. A calculator display shows that the result of a certain calculation is $0.2$.
Can you be sure that the result of the calculation is a rational number?
Yes. Since the decimal $0.2$ has a finite amount of decimal places, we can
simply express it as a fraction:
Let $x = 0.2$ and $10x = 2$.
Then:
$$ 10x = 2 $$
$$ x = \frac{2}{10} $$
$$ x = \frac{1}{5} $$
Where $1$ and $5$ are integers and $5 \neq 0$. This is a rational number.
5. Let $s$ be the statement: The cube root of every irrational number is
irrational. This statement is true, but the following "proof" is incorrect.
Explain the mistake.
@ -7895,58 +7936,431 @@ irrational. This is a contradiction, and hence it is not true that the cube root
of every irrational number is rational. Thus the statement to be proved is
true."
This incorrect proof has two problems. One is that in its supposition, the
wording suggests that the cube root of _every_ irrational number is rational,
when the negation of the given statement would be that "Suppose the cube root of
_some_ irrational number is rational."
The author of this incorrect proof then goes onto use a specific example of
$2\sqrt{2}$ for their proof. While this is fine for a disproof by
counterexample, a proof by contradiction should be more general. It should
instead read as:
**Proof by contradiction:**
Suppose not. Suppose the cube root of some irrational number $x$, is rational.
Since the cube root of $x$ is rational, this means that
$\sqrt[3]{x} = \dfrac{a}{b}$ for some integers $a$ and $b$ where $b \neq 0$.
Then, by laws of algebra:
$$ \sqrt[3]{x} = \frac{a}{b} $$
$$ x = \left(\frac{a}{b}\right)^3 $$
$$ x = \frac{a^3}{b^3} $$
Now, $a^3$ and $b^3$ are integers by the product of integers, where $b^3 \neq 0$
by the zero product property. Thus $x$ is a rational number and an irrational
number.
This is a contradiction.
Q.E.D.
Determine which statements in 6-16 are true and which are false. Prove those
that are true and disprove those that are false.
6. $6 - 7\sqrt{2}$ is irrational.
**Proof by contradiction:**
Suppose not. Suppose $6 - 7\sqrt{2}$ is rational. Then by definition of
rational,
$$ 6 - 7\sqrt{2} = \frac{a}{b} $$
For some integers $a$ and $b$ where $b \neq 0$.
It follows that:
$$ 6 - 7\sqrt{2} = \frac{a}{b} $$
$$ -7\sqrt{2} = \frac{a}{b} - 6 $$
$$ \sqrt{2} = \frac{6 - \dfrac{a}{b}}{7} $$
$$ \sqrt{2} = \frac{6b - a}{7b} $$
Now, since $6b - a$ and $7b$ are integers by the product and difference of
integers and $7b \neq 0$ by the zero product property. This means that
$\sqrt{2}$ is rational. The $\sqrt{2}$, however, is known to not be rational by
Theorem 4.8.1.
This is a contradiction.
Q.E.D.
7. $3\sqrt{2} - 7$ is irrational.
**Proof by contradiction:**
Suppose not. Suppose $\sqrt{2} - 7$ is rational.
Since $\sqrt{2} - 7$ is rational, $\sqrt{2} - 7 = \dfrac{a}{b}$ for some
integers $a$ and $b$ where $b \neq 0$.
Then, by laws of algebra:
$$ \sqrt{2} - 7 = \frac{a}{b} $$
$$ \sqrt{2} = \frac{a}{b} + 7 $$
$$ \sqrt{2} = \frac{a + 7b}{b} $$
Now, $a + 7b$ is an integer by the product and sum of integers. Thus $\sqrt{2}$
is rational. We know, however, by Theorem 4.8.1 that $\sqrt{2}$ is irrational.
This is a contradiction.
Q.E.D.
8. $\sqrt{4}$ is irrational.
This is false. $\sqrt{4} = 2 = \dfrac{2}{1}$, which is rational.
9. $\dfrac{\sqrt{2}}{6}$ is irrational.
**Proof by contradiction:**
Suppose not. Suppose $\dfrac{\sqrt{2}}{6}$ is rational. Then by the definition
of rational:
$$ \frac{\sqrt{2}}{6} = \frac{a}{b} $$
for some integers $a$ and $b$ where $b \neq 0$.
Then, by laws of algebra:
$$ \sqrt{2} = \frac{6a}{b} $$
Now, $6a$ is an integer by the product of integers. Thus $\sqrt{2}$ is a
rational number. We know by Theorem 4.8.1 that $\sqrt{2}$ is irrational.
This is a contradiction.
Q.E.D.
10. The sum of any two irrational numbers is irrational.
**Proof by counterexample:**
Let $a = \sqrt{2}$, and let $b = -\sqrt{2}$. Then, their sum is:
$$ a + b = \sqrt{2} + (-\sqrt{2}) = 0 = \frac{0}{1} $$
Which is rational. This statement is false.
Q.E.D.
11. The difference of any two irrational numbers is irrational.
**Proof by counterexample:**
Let $a = \sqrt{2}$, and let $b = \sqrt{2}$. Then, their sum is:
$$ a + b = \sqrt{2} - \sqrt{2}) = 0 = \frac{0}{1} $$
Which is rational. This statement is false.
Q.E.D.
12. The positive square root of a positive irrational number is irrational.
$$ \forall x \in \mathbb{R} (I(x) \to I(\sqrt{x})) $$
Contrapositive:
$$ \forall x \in \mathbb{R} (\neg I(\sqrt{x}) \to \neg I(x)) $$
$$ \forall x \in \mathbb{R} (R(\sqrt{x}) \to R(x)) $$
**Proof by contraposition:**
Suppose $r$ is any positive real number such that $\sqrt{r}$ is rational.
Since $\sqrt{r}$ is rational, $\sqrt{r} = \dfrac{a}{b}$ where $a$ and $b$ are
integers and $b \neq 0$.
Then:
$$ \sqrt{r} = \frac{a}{b} $$
$$ r = \left(\frac{a}{b}\right)^2 $$
$$ r = \frac{a^2}{b^2} $$
Now, $a^2$ and $b^2$ are both integers by the product of integers and
$b^2 \neq 0$ by the zero product property.
Therefore $r$ is a rational number.
Q.E.D.
13. If $r$ is any rational number and $s$ is any irrational number, then
$\dfrac{r}{s}$ is irrational.
**Proof by counterexample:**
Let $r = 0$ and $s = \sqrt{2}$, then
$\dfrac{r}{s} = \dfrac{0}{\sqrt{2}} = 0 = \dfrac{0}{1}$ which is rational.
Therefore, this statement is false.
Q.E.D.
14. The sum of any two positive irrational numbers is irrational.
**Proof by counterexample:**
Let $x = \sqrt{2}$ and $y = 2 - \sqrt{2}$, then:
$$ x + y = \sqrt{2} + (2 - \sqrt{2}) = 2 = \dfrac{2}{1} $$
Thus, $x + y$ is a rational number.
Therefore, this statement is false.
Q.E.D.
15. The product of two irrational numbers is irrational.
**Proof by counterexample:**
$$ \sqrt{2} \cdot \sqrt{2} = (\sqrt{2})^2 = 2 = \frac{2}{1} $$
Which is a rational number.
This statement is false.
Q.E.D.
16. If an integer greater than $1$ is a perfect square, then its cube root is
irrational.
**Proof by counterexample:**
Consider $64 = 8^2$, then $64 > 1$ and $64$ is a perfect square. Then consider
$\sqrt[3]{64} = 4 = \dfrac{4}{1}$. Thus $\sqrt[3]{64}$ is rational.
Thus there is at least one integer greater than $1$ that is a perfect square and
its cube root is rational.
This statement is false.
Q.E.D.
17. Consider the following sentence: If $x$ is rational then $\sqrt{x}$ is
irrational. Is this sentence always true, sometimes true and sometimes
false, or always false? Justify your answer.
This statement is sometimes true and sometimes false. Consider when $x = 2$,
then $\sqrt{2}$ is irrational. This is the case when the statement is true. Then
consider when $x = 9$ then $\sqrt{9} = 3 = \dfrac{3}{1}$ is rational.. This is a
case when the statement is false. Therefore this statement is sometimes true and
sometimes false.
18.
a. Prove that for every integer $a$, if $a^3$ is even then $a$ is even.
**Proof by contrapositive:**
Suppose $a$ is any integer such that $a$ is odd.
Since $a$ is odd, $a = 2k + 1$ for some integer $k$.
Then:
$$ a^3 = (2k + 1)^3 $$
$$ a^3 = 8k^3 + 12k^2 + 6k + 1 $$
$$ a^3 = 2(4k^3 + 6k^2 + 3k) + 1 $$
Now, $4k^3 + 6k^2 + 3k$ is an integer by the product and sum of integers.
Therefore $a^3$ is odd by definition of odd integers.
Q.E.D.
b. Prove that $\sqrt[3]{2}$ is irrational.
**Proof by contradiction:**
Suppose not. That is, suppose $\sqrt[3]{2}$ is rational. Then, by definition of
rational:
$$ \sqrt[3]{2} = \frac{a}{b} $$
for some integers $a$ and $b$ where $b \neq 0$ and $\dfrac{a}{b}$ is written in
lowest terms.
Then:
$$ \sqrt[3]{2} = \frac{a}{b} $$
$$ 2 = \left(\frac{a}{b}\right)^3 $$
$$ 2 = \frac{a^3}{b^3} $$
$$ 2b^3 = a^3 $$
Now, by the definition of even integers, we know that $a^3$ is even. We also
know by part (a), that if $a^3$ is even, then $a$ is even.
Since $a$ is even, $a = 2k$ for some integer $k$. Then:
$$ 2b^3 = (2k)^3 $$
$$ 2b^3 = 8k^3 $$
$$ 2b^3 = 8k^3 $$
$$ b^3 = 4k^3 $$
$$ b^3 = 2(2k^3) $$
Now, $2k^3$ is an integer by the product of integers. Additionally, $b^3$ is
even by the definition of even integers. Additionally, by part (a) $b$ is even
since $b^3$ is even.
Since both $a$ and $b$ and even, $\sqrt[3]{2}$ is not in lowest terms, so
$\sqrt[3]{2}$ is not rational, which contradicts the supposition.
Q.E.D.
19.
a. Use proof by contradiction to show that for any integer $n$, it is impossible
for $n$ to equal both $3q_1 + r_1$ and $3q_2 + r_2$, where $q_1$, $q_2$, $r_1$,
and $r_2$ are integers, $0 \leq r_1 < 3$, $0 \leq r_2 < 3$, and $r_1 \neq r_2$.
**Proof by contradiction:**
Suppose not. That is, suppose $n$ is some integer such
$n = 3q_1 + r_1 = 3q_2 + r_2$, where $q_1$, $q_2$, $r_1$, and $r_2$ are some
integers such that $0 \leq r_1 < 3$ and $0 \leq r_2 < 3$ and $r_1 \neq r_2$.
Then:
$$ 3q_1 + r_1 = 3q_2 + r_2 $$
$$ 3q_1 - 3q_2 = r_2 - r_1 $$
$$ 3(q_1 - q_2) = r_2 - r_1 $$
This means that $3 \mid (r_2 - r_1)$.
Since $0 \leq r_1,r_2 < 3$, it follows that:
$$ -2 \leq r_2 - r_1 \leq 2 $$
This means that $3 \cancel{\mid} (r_2 - r_1)$, which contradicts the earlier
finding that $3 \mid (r_2 - r_1)$.
Q.E.D.
b. Use proof by contradiction, the quotient-remainder theorem, division into
cases, and the result of part (a) to prove that for every integer $n$, if $n^2$
is divisible by $3$ then $n$ is divisible by $3$.
**Proof by contradiction:**
Suppose not. That is, suppose $n$ is some integer such that $n^2$ is divisible
by $3$ and $n$ is not divisible by $3$.
Since $n$ is not divisible by $3$, then $n = 3k + 1$ or $n = 3k + 2$ for some
integer $k$.
_Case where $n = 3k + 1$:_
$$ n^2 = (3k + 1)^2 $$
$$ n^2 = 9k^2 + 6k + 1 $$
$$ n^2 = 3(3k^2 + 2k) + 1 $$
By the quotient-remainder theorem, this means that $3 \cancel{\mid} n^2$. This
contradicts the supposition.
_Case where $n = 3k + 2$:_
$$ n^2 = (3k + 2)^2 $$
$$ n^2 = 9k^2 + 12k + 4 $$
$$ n^2 = 9k^2 + 12k + 3 + 1 $$
$$ n^2 = 3(3k^2 + 4k + 1) + 1 $$
By the quotient-remainder theorem, this means that $3 \cancel{\mid} n^2$. This
contradicts the supposition.
In both cases, the supposition is contradicted.
Q.E.D.
c. Prove that $\sqrt{3}$ is irrational.
**Proof by contradiction:**
Suppose not. Suppose that $\sqrt{3}$ is rational.
Since $\sqrt{3}$ is rational, $\sqrt{3} = \dfrac{a}{b}$ for some integers $a$
and $b$ where $b \neq 0$ and $\dfrac{a}{b}$ is in lowest terms.
Then:
$$ \sqrt{3} = \frac{a}{b} $$
$$ 3 = \left(\frac{a}{b}\right)^2 $$
$$ 3 = \frac{a^2}{b^2} $$
$$ 3b^2 = a^2 $$
This means that $3 \mid a^2$. By part (b), we then know that $3 \mid a$. This
means that $a = 3k$ for some integer $k$. Then:
$$ 3b^2 = (3k)^2 $$
$$ 3b^2 = 9k^2 $$
$$ b^2 = 3k^2 $$
This means that $3 \mid b^2$. Then, however, $\dfrac{a}{b}$ is not in lowest
terms since $3 \mid a$ and $3 \mid b$. This is a contradiction.
Q.E.D.
20. Give an example to show that if $d$ is not prime and $n^2$ is divisible by
$d$, then $n$ need not be divisible by $d$.
We need to find a non prime number $d$ and a real number $n$ such that:
$$ d \mid n^2 $$
$$ d \cancel{\mid} n $$
Let $d = 4$ and $n = 2$.
Then:
$$ 4 \mid 4 $$
$$ 4 \cancel{\mid} 2 $$
21. The quotient-remainder theorem says not only that there exist quotients and
remainders but also that the quotient and remainder of a division are
unique. Prove the uniqueness. That is, prove that if $a$ and $d$ are
@ -7963,30 +8377,326 @@ then
$$ q_1 = q_2 \quad \text{ and } r_1 = r_2 $$
**Proof:**
Suppose $a$, $d$ are any integers and $q_1$, $q_2$, $r_1$, and $r_2$ are some
integers such that $a = dq_1 + r_1$ where $0 \leq r_1 < d$ and $a = dq_2 + r_2$
where $0 \leq r_2 < d$.
Then:
$$ dq_1 + r_1 = dq_2 + r_2 $$
$$ dq_1 - dq_2 = r_2 - r_1 $$
$$ d(q_1 - q_2) = r_2 - r_1 $$
Since $0 \leq r_1,r_2 < d$, this means that:
$$ -(d - 1) \leq r_2 - r_1 \leq d - 1 $$
And this means:
$$ |r_2 - r_1| < d $$
Since $d(q_1 - q_2) = r_2 - r_1$ means $d \mid (r_2 - r_1)$, but we also know
that $|r_2 - r_1| < d$, then it follows that $r_2 - r_1 = 0$. Thus, it then
follows that $r_1 = r_2$
Then by substitution:
$$ d(q_1 - q_2) = r_2 - r_1 $$
$$ q_1 - q_2 = 0 $$
Thus it follows that $q_1 = q_2$.
Therefore, it has been shown that $q_1 = q_2$ and $r_1 = r_2$.
Q.E.D.
22. Prove that $\sqrt{5}$ is irrational.
**Lemma 22:**
If $5$ divides $a^2$, then $5$ divides $a$.
**Proof by contradiction:**
Suppose not, that is suppose $a$ is some integer such that $5 \mid a^2$ and
$5 \cancel{\mid} a$.
Since $5 \cancel{\mid} a$, this means that $a = 5q + r$ for some unique integers
$q$ and $r$, such that $1 \leq r < 5$.
Then, by substitution:
$$ a^2 = (5q + r)^2 $$
$$ a^2 = 25q^2 + 10qr + r^2 $$
$$ a^2 = 5(5q^2 + 2qr) + r^2 $$
So:
$$ a^2 \equiv r^2 (\mod 5) $$
Then by cases:
$$ r = 1 \to r^2 (\mod 5) = 1 $$
$$ r = 2 \to r^2 (\mod 5) = 4 $$
$$ r = 3 \to r^2 (\mod 5) = 4 $$
$$ r = 4 \to r^2 (\mod 5) = 1 $$
So in all cases $r^2 \cancel{\equiv} 0 (\mod 5)$
Therefore $5 \cancel{\mid} a^2$, which contradicts the supposition.$
Q.E.D.
**Proof by contradiction:**
Suppose not. Suppose that $\sqrt{5}$ is rational.
Since $\sqrt{5}$ is rational, $\sqrt{5} = \dfrac{a}{b}$ for some integers $a$
and $b$ where $b \neq 0$ and $\dfrac{a}{b}$ is in lowest terms.
Then:
$$ \sqrt{5} = \frac{a}{b} $$
$$ 5 = \left(\frac{a}{b}\right)^2 $$
$$ 5 = \frac{a^2}{b^2} $$
$$ 5b^2 = a^2 $$
Thus we know that $5 \mid a^2$, and by Lemma 22, we then know that $5 \mid a$.
So, since $5 \mid a$, $a = 5k$ for some integer $k$.
Then by substitution:
$$ 5b^2 = (5k)^2 $$
$$ 5b^2 = 25k^2 $$
$$ b^2 = 5k^2 $$
So then we know that $5 \mid b^2$, and by Lemma 22, we then know that
$5 \mid b$. Then, however, $\dfrac{a}{b}$ is not in lowest terms. Therefore
$\sqrt{5}$ is irrational, which contradicts the supposition.
Q.E.D.
23. Prove that for any integer $a$, $9 \cancel{\mid} (a^2 - 3)$.
_Hint:_ This statement is true. If $a^2 - 3 = 9b$, then
$a^2 = 9b + 3 = 3(3b + 1)$, and so $a^2$ is divisible by $3$. Hence, by exercise
19(b), $a$ is divisible by $3$. Thus $a^2 =(3c)^2$ for some integer $c$.
**Proof by contradiction:**
Suppose not. That is, suppose that for some integer $a$, $9 \mid (a^2 - 3)$.
Since $9 \mid (a^2 - 3)$, then $a^2 - 3 = 9b$ for some integer $b$. This then
becomes:
$$ a^2 - 3 = 9b $$
$$ a^2 = 9b + 3 $$
$$ a^2 = 3(3b + 1) $$
So $3 \mid a^2$, by 9(b), we then know that $3 \mid a$. Since $3 \mid a$,
$a = 3c$ for some integer $c$. By substitution:
$$ (3c)^2 = 3(3b + 1) $$
$$ 9c^2 = 3(3b + 1) $$
$$ 3c^2 = 3b + 1 $$
$$ 3c^2 - 1 = 3b $$
But since $ 3 \cancel{\mid} 3c^2 - 1$ and $3 \mid 3b$, this statement can never
hold for $b$ and $c$. This is a contradiction.
Q.E.D.
24. An alternative proof of the irrationality of $\sqrt{2}$ counts the number of
$2$'s on the two sides of the equation $2n^2 = m^2$ and uses the unique
factorization of integers theorem to deduce a contradiction. Write a proof
that uses this approach.
Statement: $\sqrt{2}$ is irrational.
**Proof by contradiction:**
Suppose not. That is, suppose $\sqrt{2}$ is rational. Then there are some
integers $m$ and $n$ where $n \neq 0$ such that:
$$ \sqrt{2} = \frac{m}{n} $$
where $\dfrac{m}{n}$ have no common factors. Then:
$$ \sqrt{2} = \frac{m}{n} $$
$$ 2 = \frac{m^2}{n^2} $$
$$ 2n^2 = m^2 $$
By the unique factorization theorem, we can express $n$ as the product of prime
numbers as:
$$ n = 2^a \cdot p_1^{e_1} \dots \cdot p_k^{e_k} $$
Then, we can express $n^2$ as:
$$ n^2 = 2^{2a} $$
Since the exponent is even, we know the number of $2s$ in $n^2$ is even. It then
follows that $2n^2$ adds one more $2$:
$$ 2n^2 = 2^{2a + 1} $$
So $2n^2$ has an odd number of $2$s.
We can then apply the same logic to $m^2$.
$$ m = 2^b \cdot p_1^{e_1} \dots \cdot p_q^{e_q} $$
$$ m^2 = 2^{2b} $$
So $m^2$ has an even number of $2$s. Thus $2n^2$ has an odd number of $2$s and
$m^2$ has an even number of $2$s, which contradicts the statement $2n^2 = m^2$.
Q.E.D.
25. Use the proof technique illustrated in exercise 24 to prove that if $n$ is
any positive integer that is not a perfect square, then $\sqrt{n}$ is
irrational.
**Proof by contradiction:**
Suppose not. That is, suppose $n$ is some positive integer such that $n$ is not
a perfect square and $\sqrt{n}$ is rational.
Since $\sqrt{n}$ is rational, $\sqrt{n} = \dfrac{a}{b}$ for some integers $a$
and $b$, where $b \neq 0$ and where $\dfrac{a}{b}$ is in lowest terms.
Then:
$$ \sqrt{n} = \frac{a}{b} $$
$$ n = \left(\frac{a}{b}\right)^2 $$
$$ n = \frac{a^2}{b^2} $$
$$ nb^2 = a^2 $$
By the unique factorization theorem, $a$ can be written as a product of prime
numbers:
$$ a = p_1^{e_1}p_2^{e_2} \cdot \dots \cdot p_k^{e_k} $$
And:
$$ a^2 = p_1^{2e_1}p_2^{2e_2} \cdot \dots \cdot p_k^{2e_k} $$
This means that all prime exponents in $a^2$ are even.
Similarly:
$$ b = q_1^{f_1}q_2^{f_2} \cdot \dots \cdot q_m^{f_m} $$
And:
$$ b^2 = q_1^{2e_1}q_2^{2e_2} \cdot \dots \cdot q_k^{2e_m} $$
This means that all prime exponents in $b^2$ are even.
Since $nb^2 = a^2$, all prime exponents in $nb^2$ are even since all the prime
exponents in $a^2$ are even. Therefore $n$ is a perfect square
This contradicts the supposition that $n$ is not a perfect square.
Q.E.D.
26. Prove that $\sqrt{2} + \sqrt{3}$ is irrational.
**Proof by contradiction:**
Suppose not. That is suppose that $\sqrt{2} + \sqrt{3}$ is rational.
Since $\sqrt{2} + \sqrt{3}$ is rational, then
$\sqrt{2} + \sqrt{3} = \dfrac{a}{b}$ where $a$ and $b$ are some integers and
$b \neq 0$ and $\dfrac{a}{b}$ has no common factors.
Then, by substitution:
$$ \sqrt{2} + \sqrt{3} = \frac{a}{b} $$
$$ \left(\sqrt{2} + \sqrt{3}\right)^2 = \left(\frac{a}{b}\right)^2 $$
$$ \left(\sqrt{2} + \sqrt{3}\right)\left(\sqrt{2} + \sqrt{3}\right) = \frac{a^2}{b^2} $$
$$ 2 + 2\sqrt{2}\sqrt{3} + 3 = \frac{a^2}{b^2} $$
$$ 2\sqrt{6} + 5 = \frac{a^2}{b^2} $$
$$ 2\sqrt{6} = \frac{a^2}{b^2} - 5 $$
$$ \sqrt{6} = \frac{\dfrac{a^2}{b^2} - 5}{2} $$
$$ \sqrt{6} = \frac{a^2 - 5b^2}{2b^2} $$
By the proof given in exercise 25, we know that $\sqrt{6}$ is irrational.
Now, $a^2 - 5b^2$ is an integer by the product and difference of integers. Also
$2b^2$ is an integer by the product of integers and $2b^2 \neq 0$ by the zero
product property.
Therefore $\sqrt{6}$ is a rational number which contradicts what we derived from
the proof given in exercise 25 that $\sqrt{6}$ is irrational.
Q.E.D.
27. Prove that $\log_5(2)$ is irrational. (_Hint:_ Use the unique factorization
of integers theorem.)
Omitted.
28. Let $N = 2 \cdot 3 \cdot 5 \cdot 7 + 1$. What remainder is obtained when $N$
is divided by $2$?$3$?$5$?$7$? Justify your answer.
$$ N = 2(3 \cdot 5 \cdot 7) + 1 $$
$$ N \mod 2 = 1 $$
$$ N = 3(2 \cdot 5 \cdot 7) + 1 $$
$$ N \mod 3 = 1 $$
$$ N = 5(2 \cdot 3 \cdot 7) + 1 $$
$$ N \mod 5 = 1 $$
$$ N = 7(2 \cdot 3 \cdot 5) + 1 $$
$$ N \mod 7 = 1 $$
29. Suppose $a$ is an integer and $p$ is a prime number such that $p \mid a$ and
$p \mid (a + 3)$. What can you deduce about $p$? Why?
Since $p \mid a$ and $p \mid (a + 3)$, then $p \mid ((a + 3) - a)$. It follows
then that $p \mid 3$, and the only prime number that divides $3$ is $3$ itself.
$$ p = 3 $$
30. Let $p_1, p_2, p_3, \dots$ be a list of all prime numbers in ascending
order. Here is a table of the first six:
@ -7998,10 +8708,32 @@ a. Let
$N_1 = p_1, N_2 = p_1 \cdot p_2, N_3 = p_1 \cdot p_2 \cdot p_3, \dots N_6 = p_1 \cdot p_2 \cdot p_3 \cdot p_4 \cdot p_5 \cdot p_6$.
Calculate $N_1$, $N_2$, $N_3$, $N_4$, $N_5$, and $N_6$.
| $N_1$ | $N_2$ | $N_3$ | $N_4$ | $N_5$ | $N_6$ |
| ----- | ----- | ----- | ----- | ------ | ------- |
| $2$ | $6$ | $30$ | $210$ | $2310$ | $30030$ |
b. For each $i = 1, 2, 3, 4, 5, 6$, find whether $N_i$ is itself prime or just
has a prime factor less than itself. (_Hint:_ Use the test for primality from
exercise 31 in Section 4.7 to determine your answers.)
**Test for Primality**
Given an integer $n > 1$, to test whether $n$ is prime check to see if it is
divisible by a prime number less than or equal to its square root. If it is not
divisible by any of these numbers, then it is prime.
$N_1$ is prime.
$N_2$ is not prime.
$N_3$ is not prime.
$N_4$ is not prime.
$N_5$ is not prime.
$N_6$ is not prime.
For exercises 31 and 32, use the fact that for ever integer $n$,
$$ n! = n(n - 1) \dots 3 \cdot 2 \cdot 1 $$
@ -8013,13 +8745,41 @@ $$ n! = n(n - 1) \dots 3 \cdot 2 \cdot 1 $$
largest. Call it $p$. Let $M = p! + 1$. We will show that there is a prime
number $q$ such that $q > p$. Complete this proof.
**Proof by contradiction:**
Suppose not. That is suppose there are only finitely many prime numbers. Then
one is the largest. Call it $p$. Let $M = p! + 1$.
Let $q$ be a prime number such that $q \leq p$.
Since $q \leq p$, $q \mid p!$. This means that:
$$ p! \equiv 0 (\mod q) $$
It then follows that:
$$ M = p! + 1 \equiv 1 (\mod q) $$
So, $q \cancel{\mid} M$.
Since $q$ is a prime number that cannot divide $M$, either $M$ is prime itself
or $M$ has a prime factor greater than $p$, which contradicts the supposition.
Q.E.D.
We will show that there is a prime number $q$ such that $q > p$.
32. Prove that for every integer $n$, if $n > 2$ then there is a prime number
$p$ such that $n < p < n!$.
Omitted.
33. Prove that if $p_1, p_2, \dots$, and $p_n$ are distinct prime numbers with
$p_1 = 2$ and $n > 1$, then $p_1, p_2, \dots, p_n + 1$ can be written in the
form $4k + 3$ for some integer $k$.
Omitted.
34.
a. Fermat's last theorem says that for every integer $n > 2$, the equation
@ -8029,11 +8789,15 @@ $p > 2$, $x^p + y^p = z^p$ has no positive integer solution, then for any
integer $n > 2$ that is not a power of $2$, $x^n + y^n = z^n$ has no positive
integer solution.
Omitted.
b. Fermat proved that there are no integers $x$, $y$, and $z$ such that
$x^4 + y^4 = z^4$. Use this result to remove the restriction in part (a) that
$n$ not be a power of $2$. That is, prove that if $n$ is a power of $2$ and
$n > 4$, then $x^n + y^n = z^n$ has no positive integer solution.
Omitted.
For exercises 35-38 note that to show there is a unique object with a certain
property, show that (1) there is an object with the property and (2) if objects
$A$ and $B$ have the property, then $A = B$.
@ -8041,13 +8805,21 @@ $A$ and $B$ have the property, then $A = B$.
35. Prove that there exists a unique prime number of the form $n^2 - 1$, where
$n$ is an integer that is greater than or equal to $2$.
Omitted.
36. Prove that there exists a unique prime number of the form $n^2 + 2n - 3$,
where $n$ is a positive integer.
Omitted.
37. Prove that there is at most one real number $a$ with the property that
$a + r = r$ for every real number $r$. (Such a number is called an _additive
identity_.)
Omitted.
38. Prove that there is at most one real number $b$ with the property that
$br = r$ for every real number $r$. (Such a number is called a
_multiplicative identity_.)
Omitted.

6
chapter_4/test_yourself.md
View file

@ -234,12 +234,18 @@ Page 256
length $1$, the ratio of the length of the hypotenuse to the length of one of
the legs is not equal to a ratio of ______.
two integers.
2. One way to prove that $\sqrt{2}$ is an irrational number is to assume that
$\sqrt{2} = \dfrac{m}{n}$ for some integers $m$ and $n$ that have no common
factor greater than $1$, use the lemma that says that if the square of an
integer is even then ______, and eventually show that $m$ and $n$ ______.
that integer is even; have a common factor greater than 1.
3. One way to prove that there are infinitely many prime numbers is to assume
that there is a largest prime number $p$, construct the number ______, and
then show that this number has to be divisible by a prime number that is
greater than ______.
$N = (2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot \dots \cdot p) + 1$; $p$