$$\begin{align*} |x|=x &\text{if }x\geq 0\\ |x|=-x &\text{if }x\lt 0. \end{align*}$$ Show that $|xy|=|x||y|$.

I try to prove it as follows:

$|xy|=xy$ if $xy\geq 0$, but $xy\geq 0$ if and only if $x\geq 0$ and $y\geq 0$. Since $x\geq 0$ and $y\geq 0$, then $x=|x|$ and $y=|y|$ hence from $|xy|=xy$ we have $|xy|=|x||y|$.

Am I correct?

How can I show the following two cases?

i. Show that $|x+y|\leq |x|+|y|$.

ii. If $y\gt 0$ and $-y\leq x \leq y$, then $|x|\lt y$

  • 4
    $\begingroup$ Please make your title more descriptive and finish your question. $\endgroup$ Mar 28, 2011 at 14:46
  • 6
    $\begingroup$ That is a rather bad, undescriptive title for your question, don't you think? $\endgroup$ Mar 28, 2011 at 14:46

1 Answer 1

  1. No, you are not correct. It is false that $xy\geq 0$ if and only if $x\geq 0$ and $y\geq 0$. For example, if $x=-1$ and $y=-1$, then $x\lt 0$, $y\lt 0$, and $xy = (-1)(-1) = 1\geq 0$. So your claim is incorrect, and thus your argument is incorrect. There are other possibilities, and you need to consider them too.

  2. Even if your argument were correct for that part, you would not be done yet. You only considered the case in which $xy\geq 0$. What if $xy\lt 0$? You also need to do that case.

  3. The last statement is incorrect. For example, if $x=y$, then $-y\leq x\leq y$ is true, but $|x|\lt y$ is false.

  4. To show (i), consider the different cases. What happens if $x$ and $y$ are both positive? If $x$ and $y$ are both negative? If $x$ is positive and $y$ negative and $x+y\geq 0$? If $x$ is positive, $y$ is negative, and $x+y\lt 0$? If $x$ is negative and $y$ positive?

  5. To show (a corrected version of) (ii), consider the cases. What happens if $0\leq x\leq y$? What happens if $-y\lt x\lt 0$?


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