Feedback on my proof of showing if $|a - b| \leq 1$, then $|a| \leq |b| + 1$ i'm taking a non-credit real analysis course online and this is a proof i wrote for one of my answers in my first problem set.
my solution is different from the instructor's, but he said in the video that there's more than one valid way to solve this problem, so i'd like to know if i did that. i'm not sure if what i did makes any sense because instead of proving if A then B, i proved if B then A, and it's not always the case that converses of true propositions are also true.
Any feedback? thanks in advance.
prove: if $|a - b| \leq 1$, then $|a| \leq |b| + 1$
proof:
$|a| \leq |b| + 1$
$|a| - |b| \leq 1$
$(|a| - |b|)^2 \leq 1$
$(|a| - |b|)(|a| - |b|) \leq 1$
$|a|^2 - 2|a||b| + |b|^2 \leq 1$
note: $|a| \geq a, |b| \geq b,$ so
$a^2 - 2ab + b^2 \leq |a|^2 - 2|a||b| + |b|^2 \leq 1$
$(a - b)^2 = |a - b|^2 \leq 1$
 A: No, this is wrong. The main problem is that you're trying to prove the converse of the statement, which as you point out yourself, isn't the same thing as proving the statement.
In this case, the converse is false. Try $a=0, b=2$, for example, and see where the logic falls apart.
A: First, there's the major mistake you mentioned, that you're attempting to prove the converse instead of the statement in the question, so even if your proof were correct, it wouldn't be proving the right implication.
And note I said "even if your proof were correct", from which you might correctly guess it's not correct.  The easiest way to see this is that $|a| \leq |b| + 1$ is true for $a=-1, b=3$, but then $|a-b| \leq 1$ is false, so the converse isn't even true.
There are two steps you made an error in:
$$|a|-|b| \leq 1 \implies (|a|-|b|)^2 \leq 1 \\ |a|\geq a, |b|\geq b \implies a^2 - 2ab+b^2 \leq |a|^2 - 2|a||b| + |b|^2$$
Both of these are invalid steps.  Can you see why?
A: You're right in that you've proved the converse of the statement (incorrectly, as other responses have pointed out), not the statement itself. Generally, one way you can check whether your proof still holds is to reverse the order of your proof, and see if it still works. There may be a step that no longer works, in which case the proof is invalid. But, if it still works in the reverse order, then you're good to go!
So, for this proof, you would start with:
$$ |a-b|^2 \le 1 $$
$$ \Rightarrow (a-b)^2 \le 1 ,$$
and continue back up your proof until you find a mistake or get back to the other end.
Another common way to approach something like this is to prove the contrapositive. In this case, you could do the following.
We want to prove that $|a-b| \le 1 \Rightarrow |a| \le |b| - 1 $. The contrapositive of this statement is:
$$ \neg (|a-b| \le 1) \Leftarrow \neg (|a| \le |b| - 1) $$
$$ \Leftrightarrow (|a| > |b| - 1 \Rightarrow |a-b| > 1) . $$
So, if you can prove that $|a| > |b| - 1$ implies $|a-b| > 1$, then you've also proved that $|a-b| \le 1$ implies $|a| \le |b| - 1$.
