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Math
> Number theory
> Even and odd numbers Proof:
Even and odd numbers
Definitions and hypotheses regarding even and odd numbers.Sibling topics:
Contents:
Definition of even and odd numbers
For `x in ZZ`:
- "`x` is even" means: `(EE t in Z)(x=2t)` -- there exists an integer `t` such that `x=2t`. More concisely, `x` is even if `2 div x`.
- "`x` is odd" means: `(EE y in Z)(x=2y+1)` -- there exists an integer `y` such that `x=2t+1`. More concisely, `x` is odd if `2 !div x`.
Hypothesis: Sums of odd and even numbers
- If `a` is even and `b` is even, then `(a+b)` and `(a-b)` are even.
- If `a` is even and `b` is odd, then `(a+b)` and `(a-b)` are odd.
- If `a` is odd and `b` is odd, then `(a+b)` and `(a-b)` are even.
Theorem: Products of odd and even numbers
- If `a` is even or `b` is even, then `ab` is even.
- If `a` is odd and `b` is odd, then `ab` is odd.
- To prove (1) above, assume `a` is even, and `b` is any integer. In the special case where `a=0`, `ab=0`, which is even. If `a != 0`, then we have `2 div a and a div ab`. By Divisors are transitive, we have `2 | ab`, which means that `ab` is even. By a parallel argument, `ab` is even if `b` is even and `a` is any integer.
- To prove (2) above, let `a` and `b` be any two odd numbers. Then by definition there exist integers `m` and `n` such that `a=2m+1` and `b=2n+1`. Then `ab = (2m+1)(2n+1) = (4mn+2m+2n+1) = 2(2mn+m+n)+1`. Since there exists an integer `z = 2mn+m+n` where `2z+1 = ab`, then `ab` must be odd, by definition.