Suppose A and B are real numbers(adsbygoogle = window.adsbygoogle || []).push({});

If [tex] 0 < A < 1 [/tex] and [tex] B = 1 - \sqrt{1-A} [/tex] then [tex]0 < B < A [/tex]

I tried proving this by assuming the "If" and then manipulating it to no avail.

I then tried proving it by contradiction and I was hoping someone could check my work.

First I negate the question so it is now(If A then B becomes: A and not B):

[tex] 0 < A < 1 [/tex] and [tex] B = 1 - \sqrt{1-A} [/tex] and [tex] A < B < 0 [/tex]

Then:

[tex] A < 1 - \sqrt{1-A} < 0 [/tex]

Looking at the right inequality:

[tex]1 - \sqrt{1-A} < 0 [/tex]

[tex]1 < \sqrt{1-A} [/tex]

[tex]1 < 1 - A [/tex]

[tex]0 < -A [/tex]

[tex]0 > A [/tex]

But [tex] 0 < A < 1 [/tex], so that is a contradiction and so by theorem 0 < B < A.

I'm not very good at writing proofs down so if anyone has any tips for me or advice on what I can improve on that would be nice. Also I've never used this site before, so if anyone could give me advice on the etiquette of my first post that would be nice. Thanks :)

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# Inequality proof

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