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Even and odd numbersEven and odd numbers
Definitions and hypotheses regarding even and odd numbers.
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Definition of even and odd numbers
[tags:
even odd]
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`.
Note that according to this definition, the number zero is considered even.
Hypothesis: Sums of odd and even numbers
[tags:
even odd]  If `a` is even and `b` is even, then `(a+b)` and `(ab)` are even.
 If `a` is even and `b` is odd, then `(a+b)` and `(ab)` are odd.
 If `a` is odd and `b` is odd, then `(a+b)` and `(ab)` are even.
Theorem: Products of odd and even numbers
[tags:
even odd]  If `a` is even or `b` is even, then `ab` is even.
 If `a` is odd and `b` is odd, then `ab` is odd.
Proof:  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.