TheFerruccio
- 216
- 0
I made two attempts at proofs. I feel the second one is ok, but the first one feels lacking. I'm not sure if I could represent it in a better way.
Prove the following statements
a) If x is real, and x > 1, then x^n > 1
b) If x is real, and x > 1, then x^m < x^n with m < n
a) Using induction:
1: assume true for n=1
x > 1, x^1 > 1 since x^1 = x
2: Assume true for n=k, let x^k > 1
x^{k+1} = x\bullet x^k
We know that x > 1, so x^k > x \forall k > 1
so x x^k is a number greater than 1, multiplied by another number greater than 1, so x^{k+1} > 1
Therefore, by the principle of mathematical induction... original statement
b) if m < n then there exists some integer k such that m+k=n, x^n = x^{m+k}, \frac{x^n}{x^m} = \frac{x^{m+k}}{x^m} = \frac{x^m x^k}{x^m} = x^k, since x > 1, x^k > 1, so if \frac{x^n}{x^m} > 1, then x^n > x^m
Homework Statement
Prove the following statements
Homework Equations
a) If x is real, and x > 1, then x^n > 1
b) If x is real, and x > 1, then x^m < x^n with m < n
The Attempt at a Solution
a) Using induction:
1: assume true for n=1
x > 1, x^1 > 1 since x^1 = x
2: Assume true for n=k, let x^k > 1
x^{k+1} = x\bullet x^k
We know that x > 1, so x^k > x \forall k > 1
so x x^k is a number greater than 1, multiplied by another number greater than 1, so x^{k+1} > 1
Therefore, by the principle of mathematical induction... original statement
b) if m < n then there exists some integer k such that m+k=n, x^n = x^{m+k}, \frac{x^n}{x^m} = \frac{x^{m+k}}{x^m} = \frac{x^m x^k}{x^m} = x^k, since x > 1, x^k > 1, so if \frac{x^n}{x^m} > 1, then x^n > x^m