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Are these two expressions equivalent: $a \neq \pm i\sqrt{2}b$ and $b \neq \pm i \frac{1}{\sqrt{2}}a$ ?

I would assume the follwing shows that they are:

$a \neq \pm i\sqrt{2}b \iff a \frac{1}{\sqrt{2}}\neq \pm ib \iff ia \frac{1}{\sqrt{2}} \neq \pm b \iff b \neq \pm i \frac{1}{\sqrt{2}}a$

But im unsure as to whether these biimplications hold - or if multiplying by $i$ and changing $\pm$ sing from one side to the other, makes these biimplications false.

Spink
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  • (Small reformatting of my first comment 7 min ago, but I don't think it is worth editing it as an answer) Yes they are. More precisely, $a=i\sqrt b\iff b=-ia/\sqrt2$ and $a=-i\sqrt b\iff b=ia/\sqrt2$, Writing $$a=\pm i\sqrt b\iff b=\color{red}\mp ia/\sqrt2$$ makes it a little explicit. Or equivalently$$a\ne\pm i\sqrt b\iff b\ne\mp ia/\sqrt2$$ – Anne Bauval Dec 08 '23 at 21:17

2 Answers2

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Your reasoning is fine. To see why, expand out the formulas of the form $s \neq \pm t$ to what they actually mean, i.e., $s \neq t \land s \neq -t$, which we can also write as $\lnot (s = t \lor s = -t)$. But if $c \neq 0$, then $s = t$ and $cs = ct$ are equivalent and likewise $s = -t$ and $cs = -ct$ are logically equivalent. Your argument reduces to a chain of equivalences of this kind and hence is valid.

Rob Arthan
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I think you would be better writing $$a \neq \pm {\mathrm i} \sqrt 2 b \iff b \neq \mp {\mathrm i}\frac{1}{\sqrt 2}a$$

Fly by Night
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