Proof by cases(4)

If a and b  are real numbers.

Case 1: Assume that a >= 0, b >= 0. Then |a| = a, |b| = b, and |a - b| = a - b. So in this case
||a| - |b|| = |a - b| =< |a - b|.

Case 2: Assume that a >= 0, and b < 0. Then |a| = a and |b| = -b.

-Sub-case 1: Assume that |a| >= |b|, then 0=< a + b =< a =< a - b.
So |a + b| < |a - b| giving us

||a| - |b|| = |a + b| < |a - b|.

-Sub-case 2: Assume that |b| > |a|, then we have both 0 < -b =< a - b, and b < a + b < 0. So it follows that |a + b| < |a - b| giving us

||a| - |b|| = |a + b| < |a - b|

Case 3: Assume that a < 0, and b >= 0. Then |a| = - a and |b| = b.
-Sub-case 1: Assume that |a| >= |b|, then 0 < - a =< b - a = |a - b|, and a=< a + b =< 0. So that |a + b| =< |a - b|. It follows that

||a| - |b|| = |- a - b| = |a + b| =< |a - b|.

-Sub-case 2: Assume that |b| > |a|, then a - b =< - a - b< 0. This gives us |- a - b| =< |a - b|. So again
||a| - |b|| = |- a - b| =< |a - b|.

Case 4: Assume that a < 0, and b < 0. Then |a| = - a and |b| = - b.
Then in this case
||a| - |b|| = |- a + b| = |a - b| =< |a - b|.

So in all cases ||a| - |b|| =< |a - b|

QED

Proofs by Case (3)

(3) If n is an even integer, then n = 4j or n = 4j - 2 for some integer j.

Proof. Assume that n is an even integer. Then n = 2k for some integer k.
Case 1: Assume that k is even. Then k = 2p for some integer p. So in this case

n = 2k = 2(2p) = 4p

and the results holds.

Case 2: Assume that k is odd. Then k = 2p + 1 for some integer p. So in this case

n = 2k = 2(2p + 1) = 4p + 2 + 4 - 4 = 4(p - 1) - 2 = 4j - 2

where j = p - 1 is an integer, and therefore the result holds.

So in all cases n = 4j or n = 4j - 2 for some integer j.

QED