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.
$$\left\lvert 1-2 \sqrt{-r}\right\rvert < 1$$
Let r=-x, then we have $$\left\lvert 1-2 \sqrt{x}\right\rvert < 1$$
$$-1<1-2 \sqrt{x}<1$$
$$-2<-2 \sqrt{x}<0$$
$$0<2 \sqrt{x}<2$$ $$0<\sqrt{x}<1$$ $$0<x<1$$ $$0<-r<1$$ $$-1<r<0$$