Question
Prove for that for any real numbers x and y,  |xy| = |x||y|
(∀ x ∈ R)(∀ y ∈ R) [|xy| = |x|*|y|]

So I know this problem will have a few cases.
What I have so far is...
Let x and y be arbitrary.

Case 1: |xy| ≥ 0, |x| ≥ 0 and |y| ≥ 0
Then xy ≥ 0 and x ≥0 and y ≥ 0
so |xy| =|x||y|

Case 2: |xy| ≥ 0, |x| < 0 and |y| < 0
xy ≥ 0 and |-x| = x , |-y| = y so |-x||-y| = xy
Thus |xy| = |x||y|

Are those on the right track?
Also, Is it necessary to do all the possible outcomes (I think there are 8 of them)?

Thanks

There are two issues with how you are approaching this.

Issue #1

You have case 2 as |xy| ≥ 0, |x| < 0 and |y| < 0

That is never true. How could it be possible that |y| < 0 ????

There are really four cases:

x ≥ 0 , y ≥ 0

x ≥ 0 , y < 0

x < 0 , y ≥ 0

x < 0 , y < 0

Problem 2: Whether or not xy ≥ 0 or xy < 0 is not a third part of the cases (making 8) because that inequality is determined by the two above.

Here is an outline of the proof:

Case 1: x ≥ 0 , y ≥ 0, so |x| = x and |y| = y.

In this case xy ≥ 0. Thus |xy| = xy = x*y = |x|*|y|.

Case 2: x ≥ 0 and y < 0, so |x| = x and |y| = -y.

In this case, xy ≤ 0. Thus |xy| = -xy = x*(-y) = |x||y|.

Case 3: x < 0 and y ≥ 0. The same as case 2, by symmetry.

Case 4: x < 0 and y < 0, so |x| = -x and |y| = -y.

In this case, xy > 0. Thus |xy| = xy = (-x)*(-y) = |x||y|.

QED

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