write a convincing argument that if the sum of two integers is odd?

Question

then exactly one of them if odd

Answer 1

If an integer is odd then it can be written in the form 2n + 1 where n is an integer.
If an integer is even then it can be written in the form 2n, where n is an integer.
Suppose both of the integers to be added are odd.
Then you could write them as 2a + 1 and 2b + 1 where a and b are integers.
But then their sum would be 2a + 2b + 1 + 1 = 2(a + b+ 1) which is an even number.
Suppose neither of them is odd
Then you could write them as 2a and 2b, and their sum would be 2(a+b) which is again an even integer.
If one is even and the other odd then you could write the even one as 2a and the odd one as 2b+1 and their sum would be 2(a+b) + 1, an odd integer
This is the only way you could get an odd integer, as the other two possibilities give even integers.
This is not a formal proof, just a convincing arguement. A formal proof would be along these lines but it would be a tighter.

Answer 2

A number can either be even or odd. Even implies it is divisible by 2.
Real integer numbers alternate between even and odd numbers.

even + odd = odd
odd + odd = even
even + even = even

This is a 'hand-waving' proof...

Because the only way to get an odd number is by adding an even number to an odd number (or vice versa), then exactly one number is even, and exactly one other is odd. /QED

Answer 3

let's say that the two integers are a & b

1. if a+b were odd, then a+b=2X+1 ....odd
2. if a & b were both even, then a=2c, b=2d
a+b=2c+2d=2(c+d)=2X
if a&b were both odd, then a=2c+1, b=2d+1
a+b=(2c+1)+(2d+1)=2(c+d+1)=2X

3. 2. contradicts 1. a+b=2X+1 so exactly one of them has to be odd