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I think I just need help thinking about this one. Hints without the answer would be great.

Prove with just the field and order axioms. You may assume that any indicated square root exists in the field.

If $0 and $b=1-\sqrt{1-a}$ prove $0

I have:

$b+-1=1-\sqrt{1-a}+-1$ by additive inverses axiom

$b=(1-1)-\sqrt{1-a}$ by communitivity

$b-1=\sqrt{1-a}$

$(b-1)^2=\sqrt{1-a}^2$ by an axiom? I'm confused at this part

$1-a=(b-1)^2$

$a=1-(b-1)^2$

So then we know that $0 => $0<1-(b-1)^2<1$

From this point I can get that statement down to $0 or $1 (the latter should be ignored because I'm assuming the square root is meant only to be positive.)

So from this point I can see the answer is close, but I'm discouraged because I have no idea if this is what using only field and order axioms means. Any help would be appreciated. Thanks!

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    It is *very bad form* to write $b+-1$; better write $b+(-1)$ or $b-1$. Also, it's called "commutativity", not "communitivity".2011-10-01

2 Answers 2

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HINT $\ $ On $\rm\ I = (0,1)\ $ holds $\rm\ 1-\sqrt{1-x}\ <\ x\ \iff\ 1-x\ < \sqrt{1-x}\ \iff\ x\ < \sqrt{x}$

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    This helped completely. Do you think I should prove x<\sqrt{x} in the interval or is it okay to just state it?2011-10-02
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Does it help if you write $ b = \frac{(1-\sqrt{1-a})(1+\sqrt{1-a})}{1+\sqrt{1-a}} = \dots $ instead?