1

$P(n)$ is the statement $n! < n^n$, where $n$ is an integer greater than $1$.

I found a solution online here (https://people.cs.umass.edu/~barring/cs2... But I don't understand how they got from one step to the next.

One user gave me the explanation:

So we start with: $(k+1)! = (k+1) \cdot k!$

Now, since $P(k)$ is true, $k! < k^k$, therefore: $(k+1) \cdot k! < (k+1) \cdot k^k $

Obviously, $k < (k+1)$, therefore: $(k+1) \cdot k^k < (k+1) \cdot (k+1)^k$

And finally: $(k+1) \cdot (k+1)^k = (k+1)^1 \cdot (k+1)^k = (k+1)^{k+1} $.

What I don't understand is how they got from the 2nd step to the third step.

Please explain it.

Ken
  • 3,811
PCRevolt
  • 77
  • 2
    They prove the chain of inequalities $$ (k+1)! < (k+1)k^k < (k+1)\cdot (k+1)^{k} $$ where the first inequality follows from the induction hypothesis, and the second one follows from $k < (k+1)$. The last term is equal to $(k+1)^{k+1}$. Which one exactly are you struggling with? What I mean is: Which exactly is "2nd step" and "third step"? Since you haven't numbered them, I'm left to guessing, and it could mean practically any part of the proof, depending on what you mean by "step". – Arthur May 14 '15 at 00:05

1 Answers1

0

The steps, and their justifications, are as follows.

$$\begin{align} 1:\; & (k+1)! = (k+1)\cdot k! & \text{by definition} \\ 2:\; & (k+1)! < (k+1)\cdot k^k & \text{assuming: }P(n) \\ 3:\; & (k+1)! < (k+1)\cdot(k+1)^k & \text{since: } k^k < (k+1)^k\text{ for all }k>0 \\ 4:\; & (k+1)! < (k+1)^{k+1} & \text{because: } a\cdot a^k = a^{k+1} \\ \hline \therefore\quad & \forall k>0 \;( P(k)\to P(k+1)) &\text{giving us the iterative step} \\[2ex]0:\; & 2!<2^2 & \text{the base case: } P(2) \\ \hline \therefore \quad& \forall k\geq 2 \; ( P(k) )& \text{by induction } \\ & & \Box \end{align}$$

Graham Kemp
  • 133,231