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?

Printable View

- Sep 7th 2010, 01:02 PMmathhelp1absolute 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? - Sep 7th 2010, 01:23 PMPlato
- Sep 7th 2010, 01:24 PMHallsofIvy
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 $\displaystyle x^2< y^2$" and "if $\displaystyle x^2< y^2$ 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 $\displaystyle x\ge 0$, |x|= x while if $\displaystyle x< 0$ |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 $\displaystyle 0\le x< y$. Now what can you say?

2) Suppose x and y are both negative. Then |x|< |y| means $\displaystyle 0< -x< -y< 0$. Now what can you say about their squares?

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

4) Suppose x is negative and y is non-negative. Then |x|< |y| means $\displaystyle 0< -x< y$. 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.