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Given the equation $$a^2=b^2,$$ solving for $a$ gives $$\sqrt {a^2} = \sqrt {b^2}\\ |a| = |b|\\ a = \pm |b|.$$ How to validly get rid of the absolute value sign on $b$ to get the correct answer $a = \pm b$?

Dividing $a = \pm |b|$ into cases

  • Case one: $a = |b|$

    $b = a$ or $b = -a$

    $b = \pm a$

  • Case two: $a = -|b|$

    $b = (-a)$ or $b = -(-a)$

    $b = \pm (-a)$

and making $a$ the subject (correct me if I'm wrong): $$a = \pm b \quad\text{or}\quad a = -(\pm b).$$ I'm aware that $\pm b \neq - (\pm b),$ so, how does the previous line imply that $a = \pm b$ ?

ryang
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Mohammad muazzam ali
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    $a = \pm |b|$ is the same as $a = \pm b$. –  Jan 25 '22 at 13:51
  • @RamanujanXV doesn't that only work iff $b \geq 0$? – Mohammad muazzam ali Jan 25 '22 at 13:52
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    No,it is always valid. –  Jan 25 '22 at 13:53
  • @RamanujanXV but if $b < 0$, then i believe it will be $a = \pm (-b)$ and not $a = \pm b$, right? – Mohammad muazzam ali Jan 25 '22 at 13:56
  • But the two solutions $a = \pm (-b)$ and $a = \mp (-b)$ are the same.Thus,it doesn't make any difference. –  Jan 25 '22 at 13:56
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    But consider that $\pm b$ means: "either $+b$ or $-b$" which is the same as "either $-b$ or $+b$"- Thus, $-(\pm b)=\pm (-b)=\pm b$ – Mauro ALLEGRANZA Jan 25 '22 at 13:57
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    $a = ±b$ if and only if $a = b$ or $a = -b$... And $a = ±(-b)$ if and only if $a = -b$ or $a = -(-b)$ if and only if $a = -b$ or $a = b$. – ZAF Jan 25 '22 at 13:58
  • @MauroALLEGRANZA so it's mathematically correct to say $\pm b = \pm (-b)$? I thought that's mathematically wrong since we would then be saying $b = -b$ – Mohammad muazzam ali Jan 25 '22 at 14:14
  • @RamanujanXV I'm sorry but I don't really understand when you say that they have the same solution. Also, I thought we were talking about $\pm b$ and $\pm (-b)$ and not $\pm (-b)$ and $\mp (-b)$ – Mohammad muazzam ali Jan 25 '22 at 14:16
  • @Mohammadmuazzamali Ah,yes that was a typo. What I meant to say is that $a=b,-b$ is the solution set and it doesn't matter in which way we represent. –  Jan 25 '22 at 14:18
  • @RamanujanXV so $\pm b$ and $\pm (-b)$ are the same? does that mean they are equal? I'm sorry, I understand that you said they have the same solution set, but does that mean they're equal? – Mohammad muazzam ali Jan 25 '22 at 14:21
  • @Mohammadmuazzamali Yes they are equal. –  Jan 25 '22 at 14:25
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    IMO it is correct, because what is $\pm (-b)$ ? It is "either $+(-b)$ or $-(-b)$" that is "either $-b$ or $+b$" that is again: $\pm b$. – Mauro ALLEGRANZA Jan 25 '22 at 14:26
  • @RamanujanXV if $\pm b = \pm (-b)$, then wouldn't that mean $b = -b$? which I believe won't be right. – Mohammad muazzam ali Jan 25 '22 at 14:26
  • @Mohammadmuazzamali Well,$\pm b $ means $b,-b$ and $ \pm (-b)$ means $-b,b$.Thus,there's is no difference.. –  Jan 25 '22 at 14:29
  • @RamanujanXV yeah i believe they mean the same thing. but aren't we saying $b = -b$, $-b = -(-b)$? I'm a bit confused about that – Mohammad muazzam ali Jan 25 '22 at 14:31
  • @MauroALLEGRANZA I thought that the order of $\pm$ or $\mp$ matters? so saying $\pm b = \pm (-b)$ will bring the meaning $b = -b$ or $-b = -(-b)$. That's what I know with my limited knowledge – Mohammad muazzam ali Jan 25 '22 at 14:42

2 Answers2

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$$a^2=b^2\iff a^2-b^2=(a-b)(a+b)=0.$$

So $$a-b=0\text{ or }a+b=0.$$

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$$a = \pm |b|$$ Dividing $a = \pm |b|$ into cases $a = |b|$ and $a = -|b|$ $$a = \pm b\quad\text{or}\quad a = -(\pm b).$$ I'm aware that $\pm b \neq - (\pm b),$ so, how does the previous line imply that $a = \pm b$ ?

No, it is in fact true that $\pm b=- (\pm b),$ because $$y=\pm x$$ isn't a genuine equation but notational shorthand for $$y\in\{x,-x\}.$$

Even though my instinct is to continue from your first chunk of working by simply writing \begin{gather}a = &\pm|b|\\=&{\pm}(\pm b)\tag#\\=&{\pm}b,\end{gather} there is a caveat.

The intermediate expression $$\pm (\pm b)$$ is arguably ambiguous as to whether the top and bottom signs are meant to separately/independently correspond, i.e., is the expression meant to be understood as \begin{align}&{\pm}(\pm b)\\=&{+}(+b)\:\:\text{or}\:\:{-}({-}b)\\=&b,\end{align} or whether \begin{align}&{\pm} (\pm b)\\=&{+}(+b)\:\:\text{or}\:\:{+}({-}b)\:\:\text{or}\:\:{-}(+b)\:\:\text{or}\:\:{-}({-}b)\\=&{\pm}b.\end{align}

In $(\#)$ and the final line of this answer, I mean the second interpretation, however, typically, the expression $$(\pm \,a\mp b)$$ is interpreted the first way, as $$\pm(a-b).$$

Summarising: an expression containing multiple occurrences of $\pm$ and/or $\mp$ may be disambiguated from its context.

user1015917's answer is the cleanest, as it sidesteps all occurrences of ±.

ryang
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