Prove that $2^n > n$ for all positive integers $n$.
Let $P(n): 2^n > n$.
Step $1$: For $n = 1$,$2^1 = 2 > 1$. Hence,$P(1)$ is true.
Step $2$: Assume that $P(k)$ is true for some positive integer $k$,i.e.,$2^k > k$ ..............$(1)$.
Step $3$: We shall now prove that $P(k+1)$ is true whenever $P(k)$ is true.
Multiplying both sides of $(1)$ by $2$,we get:
$2 \cdot 2^k > 2k$
$2^{k+1} > 2k$
Since $k \geq 1$,we have $k + k \geq k + 1$.
Thus,$2k = k + k > k + 1$.
Therefore,$2^{k+1} > k + 1$.
Conclusion: $P(k+1)$ is true whenever $P(k)$ is true. Hence,by the principle of mathematical induction,$P(n)$ is true for every positive integer $n$.
Step $1$: For $n = 1$,$2^1 = 2 > 1$. Hence,$P(1)$ is true.
Step $2$: Assume that $P(k)$ is true for some positive integer $k$,i.e.,$2^k > k$ ..............$(1)$.
Step $3$: We shall now prove that $P(k+1)$ is true whenever $P(k)$ is true.
Multiplying both sides of $(1)$ by $2$,we get:
$2 \cdot 2^k > 2k$
$2^{k+1} > 2k$
Since $k \geq 1$,we have $k + k \geq k + 1$.
Thus,$2k = k + k > k + 1$.
Therefore,$2^{k+1} > k + 1$.
Conclusion: $P(k+1)$ is true whenever $P(k)$ is true. Hence,by the principle of mathematical induction,$P(n)$ is true for every positive integer $n$.
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