# [SOLVED] relation question

• Sep 8th 2009, 06:34 PM
seld
[SOLVED] relation question
I think this goes here.

$a \epsilon \mathbb{R} \text{ and }0 0<1-\sqrt{1-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: ${a^2}<{1^2}$
• Sep 8th 2009, 06:47 PM
Bruno J.
If $0 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 = $\epsilon$)
• Sep 8th 2009, 07:06 PM
seld
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

$a

$a0 \Rightarrow ac$

So I guess in essence I need to prove for $0 x$
• Sep 8th 2009, 07:10 PM
Bruno J.
You have $x<1$. Multiply by $x$ on both sides and you have $x^2. Take the square root on both sides and you have $x<\sqrt x$.
• Sep 8th 2009, 07:31 PM
seld
thank you for both help with this and the latex