# Analysis Proof

• Sep 18th 2006, 02:32 PM
JaysFan31
Analysis Proof
How would you go about proving this:

Let c be greater than 1. Show that c^n is greater than or equal to c for all n greater than or equal to 1 and c^n is greater than c for all n greater than 1. Further than c^m is greater than c^n for all m greater than n.
• Sep 18th 2006, 03:12 PM
Plato
If c>1 then c>0 therefore multiply by c and get c^2>c>1.
Now you can do it by induction.
If c^k>c^(k-1) then c^(k+1)>c^k; we multiply by c.
• Sep 18th 2006, 03:57 PM
topsquark
Quote:

Originally Posted by JaysFan31
How would you go about proving this:

Let c be greater than 1. Show that c^n is greater than or equal to c for all n greater than or equal to 1 and c^n is greater than c for all n greater than 1. Further than c^m is greater than c^n for all m greater than n.

Quote:

Originally Posted by Plato
If c>1 then c>0 therefore multiply by c and get c^2>c>1.
Now you can do it by induction.
If c^k>c^(k-1) then c^(k+1)>c^k; we multiply by c.

Plato, you are probably correct. However the original problem statement does not specify that m, n are positive integers. (Though I suspect it's an oversight in the statement of the problem.)

-Dan
• Sep 18th 2006, 04:07 PM
Plato
Oh come Quark!
In the first post it was stated “for all n greater than or equal to 1”.
In fact, I only addressed the first of his/her questions.
However, I think it is logical to assume that in the second question that both n & m are positive integers. But again I did not give an answer to the second question.
• Sep 18th 2006, 04:11 PM
ThePerfectHacker
Quote:

Originally Posted by JaysFan31
Further than c^m is greater than c^n for all m greater than n.

The exponent function is defined as,
a^n=exp( n ln a) for a>0.
This since,
exp (x) is an increasing function (look at is derivative sign) it follows that if,
n>m
then,
n ln a > m ln a , since a>1 thus, ln a>0
Thus,
exp(n ln a)> exp( m ln a) using the increasing function theorem
Thus,
a^n>a^m
Q.E.D.
• Sep 18th 2006, 04:24 PM
topsquark
Quote:

Originally Posted by Plato
Oh come Quark!
In the first post it was stated “for all n greater than or equal to 1”.
In fact, I only addressed the first of his/her questions.

(Shrugs) There is nothing in the statement of even the first part of the problem that states that n is an integer either. Yes, I was generalizing, but my comment still applies.

Quote:

Originally Posted by Plato
However, I think it is logical to assume that in the second question that both n & m are positive integers. But again I did not give an answer to the second question.

I believe I was clear in stating that I also had assumed the original question was intended to have m, n integral. However, as the problem statement (both statements in fact) are true for real m, n I didn't feel comfortable in making the assumption that the problem had to intend this.

Just making sure all bases are covered. (Uh, no pun intended there!)

-Dan