Proof by Contradiction Involving Inequalities

Question

I am attempting to prove a result (does not have to do with inequalities) using proof by contradiction. We know that by assuming the proposition P and for some quantity $x$ related to P, $4 \le x \le 10$. However, if I assume not P, I arrive at $6\le x \le 9$. Is this considered a valid contradiction to prove P?

Answer 1

This does not result in a contradiction. For values of $x$ between $4$ and $10$, it's entirely possible for all $x$ to be contained between $6$ and $9$. To put it a different way a little more concretely, let's just say that assumping your proposition $P$, that the only values $x$ can take on are $6,7,$ or $8$. This is entirely valid for assuming not $P$, as you've ONLY claimed $4\leq x\leq 10$, nothing about $x$ between $4$ and $6$ or between $9$ and $10$.

Answer 2

Not really. $P\implies Q$ does not really tell us much about what happens when $\lnot P$. It could be the case that $\lnot P\implies Q$ or $\lnot P\implies \lnot Q$. If you say "If I go to the restaurant, I'll have lunch" you're not saying what will happen if you don't go to the restaurant: it is possible that you have lunch at home or that you'll go hungry.