# absolute value proof

• September 7th 2010, 01:02 PM
mathhelp1
absolute value proof
does anybody know how to solve the following..

prove that, for real numbers x and y, |x| < |y| iff x^2 < y^2.
Is the result true if the absolute value signs are removed from the statement?
• September 7th 2010, 01:23 PM
Plato
Quote:

Originally Posted by mathhelp1
prove that, for real numbers x and y, |x| < |y| iff x^2 < y^2.

If $x^2 then $\sqrt{x^2}<\sqrt{y^2}$.
How do we write $\sqrt{a^2}~?$

If $|x|<|y|$ then $|x|^2<|x||y|~\&~|x||y|<|y|^2$
Recall that $|a|^2=a^2$.
• September 7th 2010, 01:24 PM
HallsofIvy
Quote:

Originally Posted by mathhelp1
does anybody know how to solve the following..

prove that, for real numbers x and y, |x| < |y| iff x^2 < y^2.
Is the result true if the absolute value signs are removed from the statement?

First, I think you are misunderstanding what "Linear and Abstract Algebra" means. This probably belongs in the "Pre-University Algebra" section.

I presume you know that an "iff" (if and only if) statement has to be proved both ways. So you must prove both "if |x|< |y| then $x^2< y^2$" and "if $x^2< y^2$ then |x|< |y|".

Now to prove the first, look at |x|< |y|. As is almost always the case with absolute value, break this into parts. If $x\ge 0$, |x|= x while if $x< 0$ |x|= -x. Since each of x and y can be "non-negative" or "negative", there will be four cases.

1) Suppose x and y are both non-negative. Then |x|< |y| means $0\le x< y$. Now what can you say?
2) Suppose x and y are both negative. Then |x|< |y| means $0< -x< -y< 0$. Now what can you say about their squares?
3) Suppose x is non-negative and y is negative. Then |x|< |y| means $0\le x< -y$. Now what can you say about their squares?
4) Suppose x is negative and y is non-negative. Then |x|< |y| means $0< -x< y$. Now what can you say about their squares?

I rather suspect that you can do the other way by looking at the same four cases.