Prove if x,y are reals that |xy| = |x||y|

[Of Mike and Men]

Homework Statement

This is my second proofs course, but I've always felt uncomfortable with absolute values and inequalities because I feel like my proofs are circular or too simple. In this case, I'm not sure if I'm showing enough in my steps.

I would just like to know if this is a way to prove this. I've seen other results online, but I want to be sure I understand this.

Homework Equations

Prove if x,y are reals that |xy| = |x||y|

The Attempt at a Solution

Let x,y belong to the set of real numbers. Since |xy|=|x||y|=0 if either x or y is 0, we can assume that x,y is non-zero.

Case 1: x,y > 0
|xy| = xy = |x||y|

Case 2: x,y < 0
|xy| = (-x) (-y) = |x||y|

Case 3: Either x or y < 0, but not both
WLOG let x < 0 and y > 0 then
|xy| = (-x)(y) = |x||y|

.:. |xy| = |x||y| for x,y in the set of real numbers.

 

[fresh_42]

Yes, this is correct. If one wants to be very petty, then an argument, why we may assume $x<0$ and $y>0$ without loss of generality should be added, like symmetry or the possible swap to $-x,-y$. Good authors add this reason to help their readers to figure out what to do in an opposite case. Of course the length of the w.l.o.g. argument is longer than the fourth case would have been in this case.

 

[Of Mike and Men]

Yes, this is correct. If one wants to be very petty, then an argument, why we may assume $x<0$ and $y>0$ without loss of generality should be added, like symmetry or the possible swap to $-x,-y$. Good authors add this reason to help their readers to figure out what to do in an opposite case. Of course the length of the w.l.o.g. argument is longer than the fourth case would have been in this case.

Thanks. To be sure, I don't need to show an intermittent step on some of the cases? E.G.
Case 2:
x,y < 0
|(-x)(-y)| = |xy| = xy = |x||y|

It's stuff like this where I'm not sure my approach is sufficient in the cases which gives me the feeling of uncertainty.

 

[fresh_42]

Thanks. To be sure, I don't need to show an intermittent step on some of the cases? E.G.
Case 2:
x,y < 0
|(-x)(-y)| = |xy| = xy = |x||y|

It's stuff like this where I'm not sure my approach is sufficient in the cases which gives me the feeling of uncertainty.

Formally you would have to. But you left out a few steps: $|xy|=|(-x)(-y)|\stackrel{(1)}{=}|(-x)|\cdot |(-y)|\stackrel{D}{=}|x| \cdot |y|$ by case (1), which we already know is true, and property $D$ of the absolute value function. And in case you don't have $|x|=|-x|$ you have to insert these steps, too, like you did with the detour to $|x| = -x$ for negative $x$.

All these subtle steps are needed, if we want to figure out which part of the definition or which axioms are actually used. So for the same reason it would have to be $|x\cdot 0|=|0 \cdot y|=|0|= 0 = |x| \cdot 0 = 0 \cdot |y| = |x| \cdot |0| = |0| \cdot |y|$ where you also could label each equation with the property used. However, in this example every proof between "obvious" to "a complete list with labeled equations" can be seen as correct.

 

[member 587159]

Another proof:

$|xy| = \sqrt{(xy)^2} = \sqrt{x^2}\sqrt{y^2} = |x||y|$