Mathematical Induction step $2^n < n!$

Question

I am trying to learn discrete mathematics on my own and have been reading Rosen's discrete math textbook, there is one step in his proof that I do not understand.

Where $<2*k!$ becomes $(K + 1)K!$

he states that $2 < k + 1$ but how is he determining that?

I have attached an image of the proof for viewers to see exactly, it is the second last line of the inequalities.

Rosen Description Snapshot

$$2<k+1\iff k>1$$ right? – lab bhattacharjee Dec 05 '16 at 16:18 — lab bhattacharjee, Dec 05 '16 at 16:18

score 1 · Answer 1 · answered Dec 05 '16 at 16:21

1

The basis step says, it is true for $k=4$. And then the inductive step assumes that it is valid for a $k$ with $k\geq 4$.
It's quite easy to deduce $k+1 > 2$ then.

answered Dec 05 '16 at 16:21

tylo

261

score 0 · Accepted Answer · answered Dec 05 '16 at 16:20

0

Check that he has clearly mentioned $k \ge 4$ in the last line of the inductive step. And also it must be kept in mind that the inequality holds only for $k\ge 4$.

So $k+1 \ge 5 > 2$

Hope this helps you.

answered Dec 05 '16 at 16:20

SchrodingersCat

24,924

Why is 2 < (k + 1) used, is the statement 2 < k also valid by the inductive step? And how is it possible to replace the 2 with (k + 1)? Is it because if you do, the inequality still holds? Since 2 != equal (k + 1) but since 2 < (k + 1) it doesn't break the inequality – user2076774 Dec 05 '16 at 16:31
@user2076774 Yes, in response to your first and third questions. – SchrodingersCat Dec 05 '16 at 16:32
So you could also say 2 < k but the choice to use 2 < k + 1 was made since it completed the proof? – user2076774 Dec 05 '16 at 16:40
@user2076774 Exactly. Thats the main reason. – SchrodingersCat Dec 05 '16 at 16:41

score 0 · Answer 3 · answered Dec 05 '16 at 16:21

You have to have a base case. Note that your inequality is false until $n=4$, so you should say for all $n\ge 4, 2^n \lt n!$ You prove the $n=4$ case by computation. Then assume you have proved it for $k$ where $k \ge 4$. Since you have $k \ge 4$, you have $2 \lt k+1$

Mathematical Induction step $2^n < n!$

3 Answers3