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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.

Thanks!

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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 $(-a)(-b)=ab$.

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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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