Hello all I am wondering if anyone has the correct proof that I should use for Spivak calculus ( chapter 1, question 12 ) that says

$$|xy|=|x| \cdot |y|$$

from past times I know it is true , but I am not sure the best way to prove it, and I need this property to use in the rest of the proofs as well.

Should I write something like

$|xy|= xy$ , for $xy \ge 0$ and $|xy|= -(xy)$ if $xy \le 0 ?$

Or should I write $$|xy|=\sqrt{(xy)^{2}}$$ and expand from that or what?

Thanks a lot everyone, I just want to make sure I have covered it all correctly.

Preferably the answer will be from someone that is familiar with Spivak , because I would like to be able to prove it the way he would have wanted, i.e., only using what we had learnt up to this point.



The safest and easiest approach is to consider all four cases using the definition:

  • If $x\ge0$ and $y\ge0$, then $xy\ge0$ and so $|xy|=xy=|x|\,|y|$

  • If $x\ge0$ and $y\le0$, then $xy\le0$ and so $|xy|=-(xy)=(x)(-y)=|x|\,|y|$

  • If $x\le0$ and $y\ge0$, then $xy\le0$ and so $|xy|=-(xy)=(-x)(y)=|x|\,|y|$

  • If $x\le0$ and $y\le0$, then $xy\ge0$ and so $|xy|=xy=(-x)(-y)=|x|\,|y|$

The bottom half of page 7 is directly relevant here because it contains proofs that $(-a)b=-(ab)$ and $(-a)(-b)=ab$.


A bit shorter: Since $|x| = |-x|$ (more or less by definition), changing the sign of $x$ (or $y$) leaves both sides of the tentative equality unchanged. Hence, we may as well assume that $x$ and $y$ are both positive, in which case the equality is obvious.


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