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Where r $ \epsilon \Bbb { R,}$ determine the range of r.

$$\left\lvert 1-2 \sqrt{-r}\right\rvert < 1$$

$$-1< (1-2 \sqrt{-r}) < 1$$

$$-2< (-2\sqrt -r) < 0 $$

$$1>(\sqrt -r) > 0$$

The answer is: $-1<r<0$

So why do I not believe it was found using the following?

continuing from above at $\quad1>(\sqrt -r)> 0 $ $$(1)^2>(\sqrt -r)^2>(0)^2 $$ $$1> -r > 0$$ $$-1<r<0$$

This is where I began to wonder about $\;(\sqrt -r)^2$ and why not...$\;(-{\sqrt -r})^2$ too?

Because $\quad(-{\sqrt -r})*(-{\sqrt -r})\neq \sqrt -r$ or does it? I must keep firm that r is not complex, r is in the reals. I know I am missing something but am not sure what that is. Can anyone please tell me what I am not understanding about these statements?

Thanks.

user451
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  • Let $r'=-r$ and do the computation with $r'$ instead. It may be way less confusing this way. – Suzet Jul 25 '18 at 01:31
  • "Because $(-\sqrt{-r})(-\sqrt{-r})\neq \sqrt{-r}$ or does it?*" You surely have a typo there somewhere, the expression on the left is of cardinality $|r|$ while the expression on the right is only of cardinality $|\sqrt{r}|$. Either way, you seem to be reaching for something along the lines of how $\sqrt{(-1)\times(-1)}\neq (\sqrt{-1})^2$. – JMoravitz Jul 25 '18 at 01:38
  • "I began to wonder about $(\sqrt{-r})^2$ and why not ... $(-\sqrt{-r})^2$ too". $(\sqrt{-r})^2= (-\sqrt{-r})^2$. "Because $(-\sqrt{-r})(-\sqrt{-r})\neq \sqrt{-r}$ or does it". Well, no, but $(-\sqrt{-r})(-\sqrt{-r})=(\sqrt{-r})*(\sqrt{-r})=-r$. – fleablood Jul 25 '18 at 02:40

1 Answers1

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$$\left\lvert 1-2 \sqrt{-r}\right\rvert < 1$$

Let r=-x, then we have $$\left\lvert 1-2 \sqrt{x}\right\rvert < 1$$

$$-1<1-2 \sqrt{x}<1$$

$$-2<-2 \sqrt{x}<0$$

$$0<2 \sqrt{x}<2$$ $$0<\sqrt{x}<1$$ $$0<x<1$$ $$0<-r<1$$ $$-1<r<0$$

Mohammad Riazi-Kermani
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