Using PMI prove that $2^{n} > n$ is true for all natural numbers.

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Solution

To prove any statement using principle of mathematical induction the first step is to check if it is true for n=1, then check the same statement for n = k and finally for n = k+1. If the LHS and RHS is same for these then that statement can be considered to be true for every n.
$P (n) = 2^{n} > n$
Let it is true for n = 1
$2^{1}$ > 1
Therefore it is true for n = 1Let it is true for n = k
$2^{k}$ > k
Multiply both sides by 2,
$2^{k}$ .2 > 2k
$\Rightarrow 2^{k + 1} > k + k > k + 1$ (2k=k+k, Also k+k is always greater than k + 1 which ultimately induces that $2^{k + 1}$ is greater than k + 1)
Therefore it is true for n = k + 1Hence By Principle of Mathematical induction, P(n) is true for all n belongs to N.

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Using PMI prove that $2^{n} > n$ is true for all natural numbers.