11 KiB
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Assumptions
-
In this text we assume familiarity with the laws of basic algebra, which are listed in Appendix A.
-
We also use the three properties of equality: For all objects
A,B, andC, (1)A = A, (2) ifA = B, thenB = 1, and (3) ifA = BandB = C, thenA = C. -
And we use the principle of substitution: For all objects
AandB, ifA = B, then we may substituteBwhenever we haveA. -
In addition, we assume that there is no integer between
0and1and that the set of all integers is closed under addition, subtraction, and multiplication. This means that sums, differences, and products of integers are integers.
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Definitions
An integer n is even if, and only if, n equals twice some integer. An
integer n is odd if, and only if, n equals twice some integer plus 1.
Symbolically, for any integer n
n \text{ is even} \Leftrightarrow n = 2k \text{ for some integer } k n \text{ is odd} \Leftrightarrow n = 2k + 1 \text{ for some integer } k Page 186
Definition
An integer n is prime if, and only if, n > 1 and for all positive
integers r and s, if n = rs, then either r or s equals n. An integer
n is composite if, and only if, n > 1 and n = rs for some integers r
and s with 1 < r < n and 1 < s < n.
In symbols: For each integer n with n > 1,
n \text{ is prime} \Leftrightarrow \forall \text{ positive integers } r \text{ and } s, \text{ if } n = rs \text{ then either } r = 1 \text{ and } s = n \text{ or } r = n \text{ and } s = 1 n \text{ is composite} \Leftrightarrow \exists \text{ positive integers } r \text{ and } s \text{ such that } n = rs \text{ and } 1 < r < n \text{ and } 1 < s < n Page 188
Disproof by Counterexample
To disprove a statement of the form
"\forall x \in D, \text{ if } P(x) \text{ then } Q(x)," find a value of x in
D for which the hypothesis P(x) is true and the conclusion Q(x) is false.
Such an x is called a counterexample.
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Generalizing from the Generic Particular
To show that every element of a set satisfies a certain property, suppose x
is a particular but arbitrarily chosen element of the set, and show that x
satisfies the property.
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Existential Instantiation
If the existence of a certain kind of object is assumed or has been deduce, then it can be given a name, as long as that name is not currently being used to refer to something else in the same discussion.
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Theorem 4.1.1
The sum of any two even integers is even.
Proof: Suppose m and n are any [particular but arbitrarily chosen]
even integers. [We must show that m + n is even.] By definition of even,
m = 2r and n = 2s for some integers r and s. Then
m + n = 2r + 2s \quad \text{ by substitution} \quad = 2(r + s) \quad \text{ by factoring out a 2} Let t = r + s. Note that t is an integer because it is a sum of integers.
Hence
m + n = 2r \quad \text{where } t \text{ is an integer} It follows by definition of even that m + n is even. [This is what we needed
to show.]
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Personal Note: The entirety of 4.2 is extremely helpful in breaking down in exactly how to write proofs (for beginners). I'd advise revisiting this entire section frequently.
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Definition
A real number r is rational if, and only if, it can be expressed as a
quotient of two integers with a nonzero denominator. A real number that is not
rational is irrational. More formally, if r is a real number, then
r \text{ is rational } \Leftrightarrow \exists \text{ integers } a \text{ and } b \text{ such that } r = \frac{a}{b} \text{ and } b \neq 0 Page 207
Zero Product Property
If neither of two real numbers is zero, then their product is also not zero.
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Theorem 4.3.1
Every integer is a rational number.
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Theorem 4.3.2
The sum of any two rational numbers is rational.
Proof:
Suppose r and s are any rational numbers. [We must show that r + s is
rational.] Then, by definition of rational, 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. Thus
r + s = \frac{a}{b} + \frac{c}{d} \quad \text{ by substitution} \quad = \frac{ad + bc}{bd} \quad \text{ by basic algebra} Let p = ad + bc and q = bd. Then p and q are integers because products
and sums of integers are integers and because a, b, c, and d are
integers. Also q \neq 0 by the zero product property. Thus
r + s = \frac{p}{q} \text{ where } p \text{ and } q \text{ are integers and } q \neq 0 Therefore, r + s is rational by the definition of a rational number [as was
to be shown].
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Corollary 4.2.3
The double of a rational number is rational.
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Definition
If n and d are integers then
n is divisible by d if, and only if, n equals d times some integer
and d \neq 0.
Instead of "n is divisible by d," we can say that
n is a multiple of d, or
d is a factor of n, or
d is a divisor of n, or
d divides n.
The notation d \mid n is read "d divides n." Symbolically, if n and d
are integers:
d \mid n \Leftrightarrow \exists \text{ an integer, say } k, \text{ such that } n = dk \text{ and } d \neq 0 The notation d \nmid n is read "d does not divide n."
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Theorem 4.4.1 A Positive Divisor of a Positive Integer
For all integers a and b, if a and b are positive and a divides b
then a \leq b.
Proof:
Suppose a and b are positive integers such that a divides b. [We must
show that a \leq b.] By definition of divisibility, there exists an integer
k so that b = ak. By property T25 of Appendix A, k must be positive
because both a and b are positive. It follows that
1 \leq k because every positive integer is greater than or equal to 1. Multiplying both
sides by a gives
a \leq ka = b because multiplying both sides of an inequality by a positive number preserves
the inequality by property T20 of Appendix A. Thus a \leq b [as was to be
shown].
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Theorem 4.4.2 Divisors of 1
The only divisors of 1 are 1 and -1.
Proof:
Since 1 \cdot 1 = 1 and (-1)(-1) = 1, both 1 and -1 are divisors of 1.
Now suppose m is any integer that divides 1. Then there exists an integer
n such that 1 = mn. By Theorem T25 in Appendix A, either both m and n
are positive or both m and n are negative. If both m and n are positive,
then m is a positive integer divisor of 1. By Theorem 4.4.1, m \leq 1,
and, since the only positive integer that is less than or equal to 1 is 1
itself, it follows that m = 1. On the other hand, if both m and n are
negative, then by Theorem T12 in Appendix A, (-m)(-n) = mn = 1. In this case
-m is a positive integer divisor of 1, and so, by the same reasoning,
-m = 1 and thus m = -1. Therefore there are only two possibilities: either
m = 1 or m = -1. So the only divisors of 1 are 1 and -1.
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For all integers n and d, d \nmid n \Leftrightarrow \frac{n}{d} is not an
integer.
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Theorem 4.4.3 Transitivity of Divisibility
For all integers a, b, and c, if a divides b and b divides c, then
a divides c.
Proof:
Suppose a, b, and c are any [particular but arbitrarily chosen] integers
such that a divides b and b divides c. [We must show that a divides
c.] By definition of divisibility,
b = ar \text{ and } c = bs \quad \text{ for some integers } r \text{ and } s By substitution
c = bs \quad = (ar)s \quad = a(rs) \quad \text{ by basic algebra} Let k = rs. Then k is an integer since it is a product of integers, and
therefore
c = ak \quad \text{ where } k \text{ is an integer} Thus a divides c by definition of divisibility. [This is what was to be
shown.]
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Theorem 4.4.4 Divisibility by a Prime
Any integer n > 1 is divisible by a prime number.
Proof:
Suppose n is a _[particular but arbitrarily chosen] integer that is greater
than 1. [We must show that there is a prime number that divides n.] If n
is prime, then n is divisible by a prime number (namely itself), and we are
done. If n is not prime, the as discussed in Example 4.1.2b,
n = r_0s_0 where r_0 and s_0 are integers and 1 < r_0 < n and
1 < s_0 < n.
It follows by definition of divisibility that r_0 \mid n.
If r_0 is prime, then r_0 is a prime number that divides n, and we are
done. If r_0 is not prime, then
r_0 = r_1s_1 where r_1 and s_1 are integers and 1 < r_1 < r_0 and
1 < s_1 < r_0.
It follows by the definition of divisibility that r_q \mid r_0. But we already
know that r_0 \mid n. Consequently, by transitivity of divisibility,
r_1 \mid n.
If r_1 is prime, then r_1 is a prime number that divides n, and we are
done. If r_1 is not prime, then
r_1 = r_2s_2 where r_2 and s_2 are integers and 1 < r_2 < r_1 and
1 < s_2 < r_1.
It follows by the definition of divisibility that r_2 \mid r_1. But we already
know that r_1 \mid n. Consequently, by transitivity of divisibility,
r_2 \mid n.
If r_2 is prime, then r_2 is a prime number that divides n, and we are
done. If r_2 is not prime, then we may repeat the previous process by
factoring r_2 as r_3s_3.
We may continue in this way, factoring successive factors of n until we find a
prime factor. We must succeed in a finite number of steps because each new
factor is both less than the previous one (which is less than n) and greater
than 1, and there are fewer than n integers strictly between 1 and n.
Thus we obtain a sequence
r_0, r_1, r_2, \dots, r_k where k \geq 0, 1 < r_k < r_{k - 1} < \dots < r_2 < r_1 < r_0 < n, and
r_i \mid n for each i = 0, 1, 2, \dots, k. The condition for termination is
that r_k should be prime. Hence r_k is a prime number that divides n.
[This is what we were to show.]
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Proposed Divisibility Property: For all integers a and b, if a \mid b
and b \mid a then a = b.
Counterexample: Let a = 2 and b = -2. Then -2 = (-1) \cdot 2 and
2 = (-1) \cdot (-2), and thus
a \mid b \text{ and } b \mid a, \text{ but } a \neq b \text{ because } 2 \neq -2 Therefore, the statement is false.
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Theorem 4.4.5 Unique Factorization of Integers Theorem (Fundamental Theorem of Arithmetic)
Given any integer n > 1, there exist a positive integer k, distinct prime
numbers p_1, p_2, \dots, p_k, and positive integers e_1, e_2, \dots, e_k
such that
n = p_1^{e_1}p_2^{e^2}p_3^{e^3}\dots p_k^{e_k} and any other expression for n as a product of prime numbers is identical to
this except, perhaps, for the order in which the factors are written.
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Definition
Given any integer n > 1, the standard factored form of n is an
expression of the form
n = p_1^{e_1}p_2^{e^2}p_3^{e^3}\dots p_k^{e_k} where n is a positive integer, p_1,p_2,\dots , p_k are prime numbers,
e_1,e_2,\dots ,e_k are positive integers, and p_1 < p_2 < \dots < p_k.