Prove that

0<x<1

I know of 2 proofs. One, which is mine, and anouther that is better. Try to find both!

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- February 16th 2011, 02:06 PMChris11Inequality puzzle
Prove that

0<x<1

I know of 2 proofs. One, which is mine, and anouther that is better. Try to find both! - February 17th 2011, 01:29 AMOpalg
- February 17th 2011, 09:11 AMChris11
That's actualy neither! I like it though. My proof expaned the darn polynomial via induction, showing that the coefficents were all one. Then, I subtracted that expansion from the geometric series for . The slicker proof is just a multiplication by (1-x) ! You didn't have to multiply by that product. Of course, you can do a lot to the exponent and maintain that inequality!

- February 17th 2011, 09:58 AMmelese
Write . (Only later did I notice that it's actually and not , but I wanted to post this anyway.)

If we expand the infinite product , a typical term is , where is a sum of powers of two.

The following is the**Basis Representation Theorem**for base two.

*Every*positive integer can be represented*uniquely*as the sum of powers of two. Example: .

Since the exponents of the terms in the infinite series consist of*all possible combinations*of powers of two, it follows from the theorem that the exponents range over*all*the positive integers.

Therefore, .

We have , i.e. .

Here, .

So, a third one?(Rock) - February 17th 2011, 11:28 AMChris11