# Proof Related to the Binomial Theorem

1. Nov 30, 2011

### icestone111

1. The problem statement, all variables and given/known data
Use the above to prove that given a rational number a > 1 and A any other rational number, there exists b ε N such that ab > A.

2. Relevant equations
The above refers to the proving, by use of both induction and binomial theorem, that (1+a)n ≥ 1+na.

Binomial Theorem: (i=0 to n)Ʃ(n choose i)ai

3. The attempt at a solution

So I tried using the binomial theorem to get the value aN.
I get that aN must be greater than (i=0 to n-1)Ʃ(n choose i)ai
So how do I choose an N so that this holds?
Could you just let N > (i=0 to n-1)Ʃ(n choose i)ai?

2. Nov 30, 2011

### Office_Shredder

Staff Emeritus
Let's keep it simple and use this result:
(1+a)n ≥ 1+na

The a in your problem and the a in the result they tell you to use are not going to be the same number. What should the relationship between them be?

3. Nov 30, 2011

### icestone111

So, would the a in my problem be equivalent to (1+a), and then you would choose b = 1?
Then you would get (1+a)b ≥ 1+ba,
And using the expansion formula, they are equivalent when b = 1, so as a result b ≥ 1?

4. Nov 30, 2011

### Office_Shredder

Staff Emeritus
No, there's a difference between A and a.

The objective now is to show that given any a>0, A>0, there exists some b such that (1+a)b>A. Use the fact that (1+a)b[/sub]>1+ab at this point

5. Nov 30, 2011

### icestone111

Sorry for the slowness...
It looks awfully similar to a Cauchy sequence proof.

But I do realize (?) that I am trying to solve for b in terms of the given/fixed values a and A.

When I look at the "hint" that was given, I keep seeing that (i=2 to b)Ʃ(b choose i)ai ≥ 0

Does this train of thought lead me anywhere?

Last edited: Nov 30, 2011
6. Nov 30, 2011

### icestone111

Oh!
Would it work if we just choose b > (A-1)/a?

If it is I feel bad for missing something so simple. Looking too hard perhaps.