Prove that $2^{n} > n$ for all positive integers $n$ .

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Solution

To prove:
$2^{n} > n$ , for all $n \in Z^{+}$

For $n = 1$ ,
$L H S = 2^{1} = 2$
$R H S = 1$

Thus,
$L H S > R H S$

So, the above expression is true for $n = 1$

Therefore, we assume that it is true for $n = k$ , where $k \in Z^{+}$ . So,
$2^{k} > k$

So,
$2 \cdot 2^{k} > 2 \cdot k$
$\Rightarrow 2^{k + 1} > k + k \geq k + 1$ , because $k \in Z^{+}$

Thus, the above equation is true for $n = k + 1$ .

Since, the inequality is true for $n = k$ and $n = k + 1$ . Therefore, it is true for all values of $n \in Z +$ .

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