Spivak’s explanation Ch 1, Q 12.1

absolute valuealgebra-precalculusinequality

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

There have been many answers, (1,2,3,4,)

But, I would like to ask about Spivak's answer, as it most likely gives an insight that would be helpful.

$(|xy|)^{2} = (xy)^{2} = x^{2}y^{2} = |x|^{2}|y|^{2} = (|x|\cdot|y|)^{2}$ Sińce $|xy|$ and $|x|\cdot|y|$ arę both $>0$, this proves that $|xy|=|x|\cdot|y|$

Just about all the answers referenced make use of individual cases as a proof. Spivak, in this proof, seems to rely heavily on the fact that $(-a)^{2} = (a)^{2}$ unless I am missing something. So, it is understandable that the first part of his proof is correct as any value $x,y <0$ is positive after squaring. What I am not understanding is how the last statement

Sińce $|xy|$ and $|x|\cdot|y|$ arę both $>0$, this proves that
$|xy|=|x|\cdot|y|$

follows the first, and justifies the proof he states.

EDIT

In response to one of the comments, here is what I would like clarified.

(a) Spivak has shown If $a < 0$ and $b < 0$ then $ab>0$, which allows me to conclude that $(xy)^2 > 0$. and thus, $(|xy|)^2 >0$ as Spivak shows in the first part of the proof. But, why is this even necessary, as surely any absolute value, for example, $|xy|$ is by definition $>0$. I would really like to understand his rationale for this.

(b) Why does the last part of his proof ( the second block quote above) even need to be justified by part a? It is almost as if it could stand alone.

I hope this helps clarify the question.

Best Answer

$(|xy|)^{2} = (xy)^{2} = x^{2}y^{2} = |x|^{2}|y|^{2} = (|x|\cdot|y|)^{2}$. Since $|xy|$ and $|x|\cdot|y|$ arę both $>0$, this proves that $|xy|=|x|\cdot|y|$.

It seems that you want to understand what Spivak motivated to give the above proof and which ingredients are possibly dispensable. In fact he uses exactly two ingredients:

  1. $u^2 = \lvert u \rvert^2$. It is applied for $u = xy$ and $u = x, y$.

  2. If $u, v \ge 0$, then $u^2 = v^2$ implies $u = v$. It is applied for $u = \lvert xy \rvert$, $v = \lvert x \rvert \cdot \lvert y \rvert$.

None of these "lemmas" can be dropped in Spivak's approach. However, both have to be proved formally (which is very easy), but I couldn't find an explicit proof in Spivak's book.

Therefore I agree to you that it would be more transparent to prove directly that $|xy|=|x|\cdot|y|$. To do so, you have to consider the four cases $x \ge 0, y \ge 0$ / $x \ge 0, y \le 0$ etc. But 1. requires two cases and 2. requires 6.(d) which is by no means easier or more elegant than the direct proof.

Related Question