Proof Related to the Binomial Theorem

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Homework Help Overview

The discussion revolves around proving a statement related to the Binomial Theorem, specifically showing that for a rational number \( a > 1 \) and another rational number \( A \), there exists a natural number \( b \) such that \( a^b > A \). The participants are exploring the implications of the Binomial Theorem and its application in this context.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to use the Binomial Theorem to establish a relationship between \( a \) and \( A \). There are questions about how to choose \( N \) or \( b \) to satisfy the inequality \( a^N > A \). Some participants are considering the implications of the inequality \( (1+a)^n \geq 1+na \) and questioning the relationship between the variables involved.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations and approaches. Some have suggested specific values for \( b \) and are reflecting on the implications of their choices. There is no explicit consensus yet, but various lines of reasoning are being examined.

Contextual Notes

Participants are navigating the distinction between the variables \( a \) and \( A \), and there is an emphasis on ensuring the correct application of the Binomial Theorem. The discussion acknowledges the challenge of finding a suitable \( b \) given the conditions of the problem.

icestone111
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Homework Statement


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.

Homework 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

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?
 
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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?
 
Office_Shredder said:
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?


Thanks for the reply!

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?
 
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
 
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:
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.
 

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