discrete_mathematics_with_applications/chapter_4/notes.md

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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, and C, (1) A = A, (2) if A = B, then B = 1, and (3) if A = B and B = C, then A = C.

• And we use the principle of substitution: For all objects A and B, if A = B, then we may substitute B whenever we have A.

• In addition, we assume that there is no integer between 0 and 1 and 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 

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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 

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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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Page 189

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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Page 191

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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Page 192

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 

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Page 207

Zero Product Property

If neither of two real numbers is zero, then their product is also not zero.

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Page 208

Theorem 4.3.1

Every integer is a rational number.

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Page 209

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.