Question

use induction to prove that for all n > 0, n!>=n^(2-1), having trouble with the inductive step

Answer 1

What is the equation? \[
n! \geq n^{2-1}
\]?

Answer 2

yes, not familiar with equation generator, sorry

Answer 3

I mean isn't that the same as: \[
n! \geq n^1 \\
n! \geq n
\]

Answer 4

dang, sorry, wrote it wront, it's supposed to be n!>=2^(n-1)

Answer 5

basis step is 0! = 1, then we assume true the hypothesis

Answer 6

First plug in \(n+1\) into \(n\)

Answer 7

(n+1)! >= 2^((n+1) - 1)
>= 2^n

Answer 8

Now you can pull out \(n+1\) and \(2\)?

Answer 9

or the left side may be n(n-1)(n-2)...(1)(n+1)

Answer 10

not sure what you mean pull out

Answer 11

\((n+1)! = (n+1)n!\)

Answer 12

yes, now i see, how does that relate to the right side?

Answer 13

Now you can divide both sides by \(n+1\)

Answer 14

\[ \large
n! \geq \frac{2}{n+1}2^{n-1}
\]

Answer 15

Now you want to use the \[
n<0
\]inequality

Answer 16

can you demonstrate that inequality?

Answer 17

oops, I mean \(n > 0\)

Answer 18

so you mean the original hypothesis now relates to this equation?

Answer 19

yeah, it always did.

Answer 20

remember that \[
n > 0 \implies n+1 > 1 \implies n+1 \geq 2 \implies \frac{2}{n+1} \geq 1
\]

Answer 21

\[
n+1>2 \implies n+1 \geq 2
\]Comes from the fact that \(n\) can't be a non-integer.

Answer 22

does that prove our hypothesis?

Answer 23

now you combine the two equations we used.

Answer 24

wio, can you help with this, is the inequality transitive, and can therefore be combined?

Answer 25

\[
\frac{2}{n+1} \geq 1 \implies \frac{2}{n+1}2^{n-1} \geq 2^{n-1}
\]

Answer 26

thanks, gonna have to practice more of this to get it to sink in...

Answer 27

Basically the steps are:
1) Plug in \(n+1\)
2) Try to get it in terms of \(n\) again.

Answer 28

thanks again, signing off until tomorrow

Answer 29

one question about this n>0⟹n+1>1⟹n+1≥2⟹2n+1≥1

Answer 30

shouldn't the last part of it be 1>=2/(n+1)?