Demki wrote:MathDoofus wrote:Meteoric wrote:Yes! That is a valid argument, and a perfect example of proof by induction. You have the base step, the induction step, and you correctly justify your conclusion using known properties of even and odd numbers.

But my argument still contains an assumption that I'd need to build out (I think), which is that every positive integer is either cleanly divisible by two or not. I'm not sure how I'd deal with in a proof by induction.

So instead of using the definitions:

"Even" - a number cleanly divisible by 2

"Odd" - a number not divisible by 2

Use the definitions:

"Even" - a number cleanly divisible by 2

"Odd" - a number one greater than a number cleanly divisible by 2

This way you are tasked to prove that every integer that isn't even is odd.

The steps are basically the same as what you said earlier, just with a bit more justification as to why "an odd number + 1 is even"

I mean, clearly if we replace the 2s in these definitions with a 3, then the first set of definition still cover every integer, yet the second set doesn't (every integer of the form 3n+2 is missing)

And this is why it is important to state your definitions clearly.

Gotcha - thinking through this, however, I guess I haven't accounted for the case of 1, which is odd (I think) but doesn't get swept into the definition of "a number greater than a number cleanly divisible by 2," unless we count 0 as a number (but it's not a positive integer).