Answer to Question #341159 in Discrete Mathematics for calvin

Question

prove that n! > 2n for n a positive integer greater than or qual to 4 what is the base step

Answer 1

We prove the required result by mathematical induction

The result is true for n = 4

$24>16\implies 4!>2^4$

Let the result be true for n = k. That is

$k!>2^k$ Where $k\ge4$

Now we need to prove that the result is also true for n = k + 1. That is

$(k+1)!>2^{k+1}$

By our assumption

$k!>2^k$

$\implies k!(k+1)>2^k(k+1)$

$\implies (k+1)!>2^k(2)$ $\because k+1>2$ replacing $k+1$ with 2 will not effect the inequality

$\implies (k+1)!>2^{k+1}$

Hence the result is true for $n=k+1$ . Hence by the principle of mathematical induction the result is true for all $n\ge4\isin \Z^+$