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TheFerruccio
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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 [tex]x[/tex] is real, and [tex]x > 1[/tex], then [tex]x^n > 1[/tex]
b) If [tex]x[/tex] is real, and [tex]x > 1[/tex], then [tex]x^m < x^n[/tex] with [tex]m < n[/tex]
a) Using induction:
1: assume true for n=1
[tex]x > 1, x^1 > 1[/tex] since [tex]x^1 = x[/tex]
2: Assume true for n=k, let [tex]x^k > 1[/tex]
[tex]x^{k+1} = x\bullet x^k[/tex]
We know that [tex]x > 1[/tex], so [tex]x^k > x \forall k > 1[/tex]
so [tex]x x^k[/tex] is a number greater than 1, multiplied by another number greater than 1, so [tex]x^{k+1} > 1[/tex]
Therefore, by the principle of mathematical induction... original statement
b) if [tex]m < n[/tex] then there exists some integer k such that [tex]m+k=n[/tex], [tex]x^n = x^{m+k}[/tex], [tex]\frac{x^n}{x^m} = \frac{x^{m+k}}{x^m} = \frac{x^m x^k}{x^m} = x^k[/tex], since [tex]x > 1[/tex], x^k > 1, so if [tex]\frac{x^n}{x^m} > 1[/tex], then [tex]x^n > x^m[/tex]
Homework Statement
Prove the following statements
Homework Equations
a) If [tex]x[/tex] is real, and [tex]x > 1[/tex], then [tex]x^n > 1[/tex]
b) If [tex]x[/tex] is real, and [tex]x > 1[/tex], then [tex]x^m < x^n[/tex] with [tex]m < n[/tex]
The Attempt at a Solution
a) Using induction:
1: assume true for n=1
[tex]x > 1, x^1 > 1[/tex] since [tex]x^1 = x[/tex]
2: Assume true for n=k, let [tex]x^k > 1[/tex]
[tex]x^{k+1} = x\bullet x^k[/tex]
We know that [tex]x > 1[/tex], so [tex]x^k > x \forall k > 1[/tex]
so [tex]x x^k[/tex] is a number greater than 1, multiplied by another number greater than 1, so [tex]x^{k+1} > 1[/tex]
Therefore, by the principle of mathematical induction... original statement
b) if [tex]m < n[/tex] then there exists some integer k such that [tex]m+k=n[/tex], [tex]x^n = x^{m+k}[/tex], [tex]\frac{x^n}{x^m} = \frac{x^{m+k}}{x^m} = \frac{x^m x^k}{x^m} = x^k[/tex], since [tex]x > 1[/tex], x^k > 1, so if [tex]\frac{x^n}{x^m} > 1[/tex], then [tex]x^n > x^m[/tex]