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Prove that $n < 2^n$ for all $n ∈ N$. By induction. I know how simple is this, but could anyone help and give detailed explanation?

Edit: Its $2^n$ NOT 2n

Mohamed
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$0 \lt 2^0=1$ and $1 \lt 2^1=2$.

If it is true for for $n=k$

then $k \lt 2^k$ so $k+1 \lt 2^k +1 \le 2^k + 2^k = 2^{k+1}$

so it is true for $n=k+1$,

and so by induction it is true for all $n \in \mathbb{N}$.

Henry
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  • Thanks. So you assumed that $1<2^k$, right? – Mohamed Feb 06 '15 at 10:13
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    no, he uses that $1 \leq 2^k$ which follows from the first sentence (assuming $k \in \mathbb{N}$). That's why he uses the $\leq$ symbol in the inequality of the second step in the induction. – Ronald Feb 06 '15 at 13:35
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Consider $n=1$. It's true that $1 < 2$.
Now you have to prove that if the statement is true for $n$, then it's true for $n+1$: $n < 2^n \Rightarrow (n + 1) < 2^{n + 1} \Leftrightarrow n + 1 < 2 * 2^n \Leftrightarrow n < 2 * 2^n - 1$.
So, you have to prove that $n < 2^n \Rightarrow n < 2*2^n - 1$ which is the same as prove $2 * 2^n - 1 > 2^n$ or $(2-1)2^n > 1 \Leftrightarrow 2^n > 1$, which is true for all $n \in N$.

Andrei Rykhalski
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