[SOLVED] relation question

**seld** · Sep 8th 2009, 06:34 PM

I think this goes here.

$a \epsilon \mathbb{R} \text{ and }0<a<1 \Rightarrow<br /> 0<1-\sqrt{1-a}<a$

I'm not sure where to start with this proof, rather i'm not sure where to start with this proof if I do not use the fact that I can square both sides.

because I think: ${a^2}<{1^2}$

**Bruno J.** · Sep 8th 2009, 06:47 PM

If $0<x<1$ then $\sqrt x > x$ .

Now $0<1-a<1$ , hence $\sqrt{1-a}>1-a$ and you have it.

(Note that in LaTeX, \in = $\in$ and \epsilon = $\epsilon$ )

**seld** · Sep 8th 2009, 07:06 PM

It took me a while to figure out the latex thing i've never used it before.

Though I'm not sure I'm allowed this fact
$\sqrt x > x$ for |x|<1

I know I'm allowed to use:

$a<b \text{ and } b<c \Rightarrow a<c$

$a<b \text{ and } c\in \mathbb{R} \Rightarrow a+c < b+c$

$a<b \text{ and } c>0 \Rightarrow ac<bc<br />$

So I guess in essence I need to prove for $0<x<1\text{ }\sqrt x > x$