# Simple Math Proofs

### Even and Odd Integers

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, if n is an integer, then

n is even \( \Leftrightarrow \exists \) an integer k such that \(n=2k\)

n is odd \( \Leftrightarrow \exists \) an integer k such that \(n=2k+1\)

### Prime Numbers

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 intger \(n\) is composite if, and only if \(n\)>1 and \(r=rs\) for some integers \(r\) and \(s\) with \(1<r<n\) and \(1<s<n\).

In symbols:

\(n\) is prime \( \Leftrightarrow \forall \) positive integers \(r\) and \(s\) if \(n=rs\) then either \(r=1\) and \(s=n\) or \(r=n\) and \(s=1\).

\(n\) is composite \( \Leftrightarrow \exists \) positive integers \(r\) and \(s\) such that \(n=rs\) and \(1<r<n\) and \(1<s<n\).

### The sum of any two even integers is even

Suppose \(m\) and \(n\) are even integers. By definition of even, \(m=2r\) and \(n=2s\) for some integers \(r\) and \(s\). Then

\(m+n=2r+2s\)

\( =2(r+s)\)

Let \(t=r+s\). Note that \(t\) is an integer because it is a sum of integers. hence \(m+n=2t\) where t is an integer.

It follows by definition of even that \(m+n\) is even. QED.

### Rational Number

A real number r is rational if, and only if, it can be expressed as a quotent 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

\(r\) is rational \( \Leftrightarrow \exists \) integers \(a and b\) such that \(r=\frac{a}{b} and b\neq0.\)

### The sum of any two rational numbers is rational

Suppose r and s are rational numbers. Then by definition of rational, \(r=\frac{a}{b} and \frac{c}{d}\) for some integers \(a,b,c, and d\) with \(b\neq0\) and \(d\neq0\). Thus

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

\( =\frac{ad+bc}{bd}\)

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 all integers. Also \(q\neq0\) by the zero product property. Thus

\(r+s=\frac{p}{q}\) where p and q are integers and \(q\neq0\).

Therefore, \(r+s\) is rational by definition of a rational number. QED.

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