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if-part

  1. $ab\geq 0$
  2. $a,b\in P\cup\{0\}$
  3. $\lvert a+b\rvert=a+b$, since if $a,b\in P\cup\{0\}$, then $a+b\in P\cup\{0\}$
  4. $a,b\geq0\rightarrow a=\lvert a\rvert,b=\lvert b\rvert\rightarrow a+b=\lvert a\rvert+\lvert b\rvert$
  5. $\lvert a+b\rvert=\lvert a\rvert+\lvert b\rvert$

only-if part

  1. $\lvert a+b\rvert=\lvert a\rvert+\lvert b\rvert$
  2. To be honest I don't know how to start with this side, so the question is how to start from point 1. Therefore, the following would be prove by contraposition.
  3. WLOG, $a\lt 0,b\geq 0$
  4. $\lvert a\rvert=-a,\lvert b\rvert=b$
  5. $b-a\neq a+b \neq -a-b$

Any suggestions to refine the only-if proof? The book I use is Introduction to real analysis by Robert G. Bartle and Donald R. Sherbert

Andes Lam
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1 Answers1

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Note that $|a+b|=|a|+|b|\iff a^2+b^2+2ab=a^2+b^2+2|a||b|\iff ab=|ab|\iff ab\ge0$.

J.G.
  • 115,835
  • Is square root included in the $a^2+b^2+2ab=a^2+b^2+2\lvert a\rvert\lvert b\rvert\rightarrow \lvert a+b\rvert=\lvert a\rvert+\lvert b\rvert$ relation since the book has yet established square root at this stage? – Andes Lam Feb 25 '20 at 12:20
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    @AndersLam I'm using $|x|^2=x^2$ (which you can separately prove for $x\ge0$ and $x<0$) and $x^2=y^2\land x,,y\ge0\implies x=y$ (which follows from $x^2-y^2=(x+y)(x-y)$). No knowledge of square roots is needed. – J.G. Feb 25 '20 at 12:22