1. ## Can someone check a proof involving cases, please?

I seem to think my proof of the following theorem has been made more complicated/lengthy than necessary. Any thoughts will be highly appreciated =)

Prove that for all real numbers a and b, |a|≤ b iff -b ≤ a ≤ b.

ADDENDUM! I went forward and completed part b) of this exercise before any responses were posted, so I figured I attach that as well (they are related).
Prove that for any real number x, -|x|≤ x ≤ |x|.

Proof:
Let a and b be arbitrary real numbers.
(-->) Suppose |a|≤ b.
Case I. a ≥ 0. Then |a|= a and we have 0 ≤ |a| = a ≤ b. This also means that 0 ≤ b, and so -b ≤ 0. Thus, -b ≤ 0 ≤ |a| = a ≤ b. In particular, then, -b ≤ a ≤ b.
Case II. a < 0. Then 0 < |a| = -a ≤ b. But since |a| = -a ≤ b, -|a| = a ≥ -b. Thus, -b ≤ a. Also, though, we have that 0 < -a and 0 < b. Adding these two inequalities yields 0 < b - a. Adding a to each side gives a < b. Clearly, then, a < b or a = b, so we may state that a ≤ b. Finally, together: -b ≤ a ≤ b.
Since these cases are exhaustive and each results in -b ≤ a ≤ b, it follows that if|a|≤ b, then -b ≤ a ≤ b.
(<--) Suppose -b ≤ a ≤ b. This means -b ≤ a and a ≤ b.
Case I. a ≥ 0. Then |a| = a ≤ b. So in particular, |a|≤ b.
Case II. a < 0. Then |a| = -a ≤ b. So in particular, |a| ≤ b.
Since these cases are exhaustive and each results in |a| ≤ b, it follows that if -b ≤ a ≤ b, then |a| ≤ b.
Lastly, since we have proven both directions of the biconditional for arbitrary real numbers a and b, we may conclude that for all real numbers a and b, |a|≤ b iff -b ≤ a ≤ b.

Let x be an arbitrary real number. We now break the proof into the following exhaustive cases, each of which results in the desired outcome.
Case I. x ≥ 0. Then |x| = x, so clearly |x| = x or |x| ≤ x. Thus, we have that |x| ≤ x. With this, if we apply our theorem above (letting a = b = x), we may conclude that -x ≤ x ≤ x, which, implementing the equality established above, is the same as saying -|x|≤ x ≤ |x|.
Case II. x < 0. Then |x| = -x. We may thus state that |x| = -x or |x| < -x, by which we rephrase to assert |x|≤ -x. Again applying the above theorem (this time with a = x and b = -x), we may conclude that -(-x) ≤ x ≤ -x. Applying the equality established above, we rexpress this as: -|x|≤ x ≤ |x|.
Since x was chosen arbitrarily, we have shown that for any real number x, -|x|≤ x ≤ |x|.

2. ## Re: Can someone check a proof involving cases, please?

And incase it interests anyone, the two prior theorems motivate a proof of the triangle inequality, which I also figure I'll include :

Theorem (triangle inequality)- For all real numbers x and y, |x + y| ≤ |x| + |y|.

Proof- Let x and y be arbitrary real numbers. Note that x + y is also a real number, as is |x| + |y|. Using these notions, along with our first theorem above (for all real numbers a and b, |a|≤ b iff -b ≤ a ≤ b), letting a = x + y and b = |x| + |y|), we have |x + y| ≤ (|x| + |y|) iff -(|x| + |y|) ≤ x + y ≤ (|x| + |y|). Now, by our second theorem above (for any real number c, -|c|≤ c ≤ |c|, applying it once for c = x and again for c = y), we have both -|x|≤ x ≤ |x| and -|y|≤ y ≤ |y|. This means -|x|≤ x, -|y|≤ y, x ≤ |x|, and y ≤ |y|. Adding -|x|≤ x to -|y|≤ y yields -|x|-|y| ≤ x + y. Adding x ≤ |x|, to y ≤ |y| yields x + y ≤ |x|+|y|. If we then, in turn, add these resulting inequalities, we are left with -|x|-|y| ≤ x + y ≤ |x|+|y|, or otherwise, -(|x|+|y|) ≤ x + y ≤ |x|+|y|, as desired.

3. ## Re: Can someone check a proof involving cases, please?

Originally Posted by Syrus
I seem to think my proof of the following theorem has been made more complicated/lengthy than necessary. Any thoughts will be highly appreciated =)

Prove that for all real numbers a and b, |a|≤ b iff -b ≤ a ≤ b.

ADDENDUM! I went forward and completed part b) of this exercise before any responses were posted, so I figured I attach that as well (they are related).
Prove that for any real number x, -|x|≤ x ≤ |x|.

Proof:
Let a and b be arbitrary real numbers.
(-->) Suppose |a|≤ b.
Case I. a ≥ 0. Then |a|= a and we have 0 ≤ |a| = a ≤ b. This also means that 0 ≤ b, and so -b ≤ 0. Thus, -b ≤ 0 ≤ |a| = a ≤ b. In particular, then, -b ≤ a ≤ b.
Case II. a < 0. Then 0 < |a| = -a ≤ b. But since |a| = -a ≤ b, -|a| = a ≥ -b. Thus, -b ≤ a. Also, though, we have that 0 < -a and 0 < b. Adding these two inequalities yields 0 < b - a. Adding a to each side gives a < b. Clearly, then, a < b or a = b, so we may state that a ≤ b. Finally, together: -b ≤ a ≤ b.
Since these cases are exhaustive and each results in -b ≤ a ≤ b, it follows that if|a|≤ b, then -b ≤ a ≤ b.
(<--) Suppose -b ≤ a ≤ b. This means -b ≤ a and a ≤ b.
Case I. a ≥ 0. Then |a| = a ≤ b. So in particular, |a|≤ b.
Case II. a < 0. Then |a| = -a ≤ b. So in particular, |a| ≤ b.
Since these cases are exhaustive and each results in |a| ≤ b, it follows that if -b ≤ a ≤ b, then |a| ≤ b.
Lastly, since we have proven both directions of the biconditional for arbitrary real numbers a and b, we may conclude that for all real numbers a and b, |a|≤ b iff -b ≤ a ≤ b.
i think that one's good as much as it gets,but the addendum proof can be shorter.namely,if you take the proof you just did,you can use it with a=x and b=|x|.since

|x|=|x| for any real x then

|x|≤|x| for any real x.

by the theorem you proved in the first part of your post we get that

-|x|≤x≤|x|

and that's it.no cases at all.

4. ## Re: Can someone check a proof involving cases, please?

Ah, very insightful anonimnystefy! Thank you for your response. Strange how I did not notice that shortcut.