Proof by contradiction

The 3 lines:

let p = 2m+1 and q = 2s+1 (both now odd)
pq = 2(2ms + m + s) +1
pq is odd


prove that if p and q are odd then pq is odd.

It's not clear whether you understand/accept this or not - please elaborate.

~snip~ See how much I don't get?

Without wanting to get snarky, I asked you a question about whether you understood 3 particular lines in the proof, and I can't see that anything in the paragraph you posted actually answers my question.

I think I do. If you multiply in integer by 2 then it’s even and by adding 1 you make it odd
Multiplying pand q gives 4ms + 2m + 2s +. 1 and that can be factorised as shown. The first term is even because an integer has been multiplied by 2 so Pi is odd.

As an aside
https://nrich.maths.org/4717
Is pretty good and emphasises that some things are more natural/easier to prove by contradiction. For problems like the OP, there are several ways to prove it, all fairly straightforward.

As you've already gone "meta": I didn't want to detail things too much, but I'm really not liking " If pq is even then neither p nor q is even" as an implied negation of "If pq is even then at least one of p or q is even" (you wouldn't want to claim the negation of "if p is prime then p is even" is "if p is prime then p is odd"), but I fear it would only confuse things further at this point.

OK, so we've shown that if p, q are both odd, then pq is odd (*)
Then the argument goes:

Claim: If pq is even then at least one of p or q is even.
Assume (for contradiction) that pq is even but neither p or q is even (**)
Then p and q must both be odd.
And then by (*) pq must be odd.
But (**) says that pq is even, so we have a contradiction.
Therefore (**) can't ever be true.

If you still don't agree with the conclusion, try to decide exactly which line in the above argument you disagree with.

I agree with up to and including But(**
However, all that's been shown is that if p, q are odd then pq is not even but we haven't necessarily shown that if p and/or q is even then pq is - maybe two evens multiply to give an odd (BTW I know they don't). That extra bit hasn't been shown so I don't think the conclusion is perfect.

The original question claims that: "If pq is even then at least one of p or q is even."
It does NOT claim that "pq is even if p and/or q is".

You are complaining that something hasn't proved that was never attempted to be proved.

Again, more generally:

The question is of the form: "Show that: if A is true, then B is true".
You are complaining that the proof does not show: "If B is true, then A is true".

OK. I get that. That has been the missing link.
Many thanks.

I find this topic really tricky and sometimes I can't fully believe the "Therefore" part.

Here's an example

Prove: If pq is even then at least one of p or q is even.
Assume: If pq is even then neither p nor q is even

let p = 2m+1 and q = 2s+1 (both now odd)
pq = 2(2ms + m + s) +1
pq is odd

This contradicts the assumption
Therefore if pq is even then at least one of p or q is even

BUT does it prove that? Yes in as much as we know that and even times and odd gives an even as does an even times an even. But if we don't already know that an odd times an odd gives an odd then why can we make the final statement without proof that and even times and even or an odd times an even gives an even by some other means?

Thanks

NO BECAUSE WHY WAS THIS THE ACTUAL QUESTION ON PAPER 1 pure maths edexcel 2022 😐😐