Thread: [SOLVED] relation question

I think this goes here.

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

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: $\displaystyle {a^2}<{1^2}$

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

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

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

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
$\displaystyle \sqrt x > x$ for |x|<1

I know I'm allowed to use:

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

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

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

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

You have $\displaystyle x<1$. Multiply by $\displaystyle x$ on both sides and you have $\displaystyle x^2<x$. Take the square root on both sides and you have $\displaystyle x<\sqrt x$.

thank you for both help with this and the latex