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I will prove the false statement, that n\geq a\Rightarrow n!\geq a^n, n\in \mathbb{N}-\left \{ 0 \right \} with induction
For n=1 1\geq a\Rightarrow 1!\geq a^1\Rightarrow 1 \geq a which is true.
Suppose that n\geq a\Rightarrow n!\geq a^n, n\in \mathbb{N}-\left \{ 0 \right \}
Then,
n\geq a\Rightarrow (n+1)!\geq nn!\geq aa^n=a^{n+1} which yields that n+1\geq a\Rightarrow(n+1)!\geq a^{n+1}
Therefore, n\geq a\Rightarrow n!\geq a^n
But for n=3,a=2 using the inequality we just proved 3\geq 2\Rightarrow3!\geq 2^3\Leftrightarrow 6\geq 8 Impossible!. Where is my mistake?
[EDIT] Don't bother answering. I have highlighted the mistake I made that rendered the inequality invalid for a greater than 1
For n=1 1\geq a\Rightarrow 1!\geq a^1\Rightarrow 1 \geq a which is true.
Suppose that n\geq a\Rightarrow n!\geq a^n, n\in \mathbb{N}-\left \{ 0 \right \}
Then,
n\geq a\Rightarrow (n+1)!\geq nn!\geq aa^n=a^{n+1} which yields that n+1\geq a\Rightarrow(n+1)!\geq a^{n+1}
Therefore, n\geq a\Rightarrow n!\geq a^n
But for n=3,a=2 using the inequality we just proved 3\geq 2\Rightarrow3!\geq 2^3\Leftrightarrow 6\geq 8 Impossible!. Where is my mistake?
[EDIT] Don't bother answering. I have highlighted the mistake I made that rendered the inequality invalid for a greater than 1
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