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Here are my "proofs" for Spivak's Calculus Chapter 1 Problem 12. I am new to this level of rigour and I am attempting to intimate myself with more advanced topics of mathematics to prepare for next year. I apologize in advance, as these so-called "proofs" are not likely to be nearly as rigorous as they should be. Any assistance on how to write the proofs better or any critiques on faulty logic would be greatly appreciated.

12) i) Prove that |xy|=|x|$\cdot$|y|

My proof is:

$x,y\in\mathbb{R} \therefore$ $|x|>0$ and $|y|>0$ $\therefore$ $|x|\cdot|y|>0$

${(|xy|)}^2 = (xy)^2 = x^2y^2 = {|x|}^2{|y|}^2 = (|x|\cdot|y|)^2$

Therefore, because $|x|\cdot|y|>0$, $\;$ $|xy|=|x|\cdot|y|$

12) ii) Prove that $\left|\frac{1}{x}\right|=\frac{1}{\left|x\right|}$

My proof is:

$\left|\frac{1}{x}\right|=|1|\cdot\left|x^{-1}\right|$, therefore by (i), $|1\cdot(x^{-1})|= |1|\cdot|x^{-1}|$

$1>0$, therefore $|1|=1$ and $|1|\cdot|x^{-1}|=\frac{1}{|x|}$, which is what we wish to prove.

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  • $\begingroup$ I think you should post each item in a question of its own. Most people won't read such a long question. Do not post a lot of questions in a row, though. $\endgroup$ – fonini Apr 26 '16 at 18:06
  • $\begingroup$ If z is non-real, $|z|^2 \neq z^2$ $\endgroup$ – Evariste Apr 28 '16 at 13:14
  • $\begingroup$ Thank you for your input! I wasn't thinking about clarifying that. However; this is the first chapter and complex numbers have not yet been introduced. I shall edit this. $\endgroup$ – Greyson Cox Apr 28 '16 at 13:17
  • $\begingroup$ I think your proof for $i$ is fine, but it really depends exactly on what you are 'allowed' to use. I'm not sure what Spivak intends. Your argument for $ii$ is not correct. See the answer posted below. $\endgroup$ – jdods Apr 29 '16 at 14:36
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At this point, I'm not sure you've established that $a^2=b^2$ implies that $a=\pm b$. Spivak's intent is that you work with the absolute value definition in terms of cases. So, for the first problem, break it down to the cases where $x$ and $y$ are positive, zero, or negative. It's not quite as tedious as it seems.

For the second, recall that for $a\ne 0$, $1/a$ is defined to be the unique number whose product with $a$ is $1$. So you only need to show that $|1/x|\cdot |x| = 1$; again it takes two cases.

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  • $\begingroup$ Thank you. I'm still trying to learn how to write these correctly (and hopefully elegantly), so your critiques are extremely helpful. $\endgroup$ – Greyson Cox May 2 '16 at 13:55
  • $\begingroup$ Also, looking back I realized we have established that $a^2=b^2$ implies $a=\pm{b}$ in problem 6 (d) of the 3rd. Edition. $\endgroup$ – Greyson Cox May 2 '16 at 14:23
  • $\begingroup$ @Greyson, in that case, your argument for the first is fantastic. $\endgroup$ – Ted Shifrin May 3 '16 at 4:57

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