Proof of Inequality 0<B<A

[kidmode01]

Suppose A and B are real numbers
If $$0 < A < 1$$ and $$B = 1 - \sqrt{1-A}$$ then $$0 < B < A$$

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):

$$0 < A < 1$$ and $$B = 1 - \sqrt{1-A}$$ and $$A < B < 0$$

Then:

$$A < 1 - \sqrt{1-A} < 0$$

Looking at the right inequality:

$$1 - \sqrt{1-A} < 0$$
$$1 < \sqrt{1-A}$$
$$1 < 1 - A$$
$$0 < -A$$
$$0 > A$$

But $$0 < A < 1$$, 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 :)

 

[tiny-tim]

Welcome to PF!

Hi kidmode01! Welcome to PF!

(have a square-root: √ )

Sorry, but you started completely wrong …

since A and B have to be positive, the negation is 0 < A < B …

you can't suppose A < 0.

Hint: You have to prove that 1 - √(1 - A) < A.

Start again, rewriting it as 1 - A < √(1 - A)

 

[MrSnoopy]

Hi

You started correctly only you have to finished it.
$$0 < B < A$$ for $$B=1-\sqrt{1-A}$$. You can proof for $$A\in C(\re)$$.
$$0 < 1-\sqrt{1-A} < A$$
First: $$0 <1-\sqrt{1-A}$$
$$\sqrt{1-A} < 1 \;\; /^2$$
$$1-A < |1|^{2}$$
$$1-|1|^{2} < A$$
$$1-1 < A$$
$$0 < A$$

and second:
$$1-\sqrt{1-A} < A$$
$$1-A < \sqrt{1-A}\;\; /^2$$
$$|1-A|^{2} < 1 - A$$
and if $$A \in C$$
$$|1-A|^{2}=(1-A)(1-\overline{A}) = 1 - \overline{A} - A + A\overline{A}$$
$$1 - \overline{A} - A + A\overline{A} < 1 - A$$
$$A\overline{A} - \overline{A} < 0$$
$$A\overline{A} < \overline{A}$$
$$A < 1$$

This shows: $$0 < A < 1$$.
This proof shows that A can be defined in complex space and will have same define area as in real.
I think this is all you want if not send me email. I hope I helped you.

With respect
MrSnoopy

 

[MrSnoopy]

Sorry for bad writing - some problems with Physics server.

 

[kidmode01]

I was wondering if someone could further Mr Snoopy's explanation on why his proof is valid. I don't understand why proving it for A as an element of the complex numbers proves it for A as an element of the real numbers.

Thank you,
kidmode01

 

[JG89]

Since 0 < A < 1 then we have:

A < SqRoot(A)
A^2 < A
A^2 - A < 0

Adding 1 to both sides and rearranging the terms:

1 - A + A^2 < 1

Subtracting A from both sides:

1 - 2A + A^2 < 1 - A
(1 - A)^2 < 1 - A

Taking the square root of both sides:

1 - A < SqRoot(1-A)

1 < SqRoot(1-A) + A
1 - SqRoot(1-A) < A

And of course since B = 1 - SqRoot(1-A) then:

B < A

What's left to prove now is that 0 < B. Use the fact that 0 < A < 1
And so 0 < 1 - A < 1 Which means that 0 < SqRoot(1 - A) < 1

 

[kidmode01]

Thanks for your post JG89 but I know how to prove it for real numbers. I'm questioning Mr Snoopy's thought process on proving it in complex space.

 

[soumyashant]

Suppose A and B are real numbers
If $$0 < A < 1$$ and $$B = 1 - \sqrt{1-A}$$ then $$0 < B < A$$

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):

Before negating, let's look at the statement logically..
Decompose the compound statement into simple ones..
p: 0<B<A <==>(0<B) AND (B<A)
The equivalence of both the statements can be easily verified...

By De Morgan's Rule, $$p^{c}$$ is $$(0<B)^{c}$$ OR $$(B<A)^{c}$$..
So the negation would look like: (B<0) OR (A<B) where we are using the inclusive OR...

Now about the solution to the problem.. we can directly do it..

Substitute $$A=sin^{2} \theta=4 \cdot \sin^{2} \frac{\theta}{2} \cdot \cos^{2} \frac{\theta}{2}$$ where $$\theta \in (0,\frac{\pi}{2})$$...

So $$B=1-\cos \theta=2 \cdot \sin^{2} \frac{\theta}{2}$$

Obviously B>0.

B<A
$$\Leftrightarrow 2 \cdot \sin^{2} \frac{\theta}{2} < 4 \cdot \sin^{2} \frac{\theta}{2} \cdot \cos^{2} \frac{\theta}{2}$$
$$\Leftrightarrow \frac{1}{\sqrt{2}} < \cos \frac{\theta}{2}$$
which is obviously true as $$\frac{\theta}{2} \in (0, \frac{\pi}{4})$$
Working backwards we get our desired result...
So 0<B<A