Solved

# Proof by induction -- Inequalities

Posted on 2009-04-03
798 Views
Ok I can't figure out how to inductively prove the following inequalities, I have the base case and inductive hypothesis, I just don't see how to prove this stuff. I've already turned in the assignment, I'm just curious cause they never tell us. At least there's points in it for you guys.

1.) Prove that n! > n^3 when n is large enough. (n has to be greater than or equal to 6)

2.) Prove that n! < n^n for all positive integers by induction.

I just can't seem to prove these inequalities for P(K+1). Any help would be appreciated.

-Jeff   P.S. (If you solve one I'll give you half the points, unless only one is ever solved than I'll give you them all! HAHAHA)
0
Question by:jeffiepoo

LVL 84

Expert Comment

Assuming that k! < k^k, what can you then say about (k+1)! and (k+1)^(k+1) ?

what do you know about 1! and 1^1
0

LVL 84

Expert Comment

would it be any easier to prove that
(n-1)! > n^2 ?
0

LVL 6

Author Comment

I don't get what you're getting at
0

LVL 25

Expert Comment

You first assume that k!<k^k.
Now consider (k+1)! = (k+1)k! < (k+1)k^k,
it's easy to show that (k+1)^(k+1) > (k+1)k^k, in which case you're done.

As for (n-1)! > n^2, note that this is equivalent to n!>n^3. However, you may find it easier to prove that instead..
0

LVL 22

Accepted Solution

1.  n! > n^3 for all n >= 6.

First prove P(6), which is 6! > 6^3.  This is just easy arithmetic.

Then you have to prove P(k+1) given P(k) and k >= 6.
Given k! > k^3, and k >= 6, prove that (k+1)! > (k+1)^3.

Have you gotten this far?  The first thing you should do is expand the factorial and the polynomial.

2.  n! < n^n for all n >= 1.
This is actually not true.  1! < 1^1 reduces to 1 < 1, which is false.
Maybe you meant n! <= n^n for all n >= 0, or maybe n! < n^n for all n >= 2.
0

LVL 31

Expert Comment

n^n > n!

Assume true for n

(n+1)^(n+1) =  (n+1)(n+1)^n > (n+1)n^n

>  (n+1)n!   ( by inductive assumption)

=  (n+1)!

Hence if true for n true for n+1

Clearly true for n=2.  Hence by the principle of induction true for all n.
0

LVL 31

Expert Comment

n^3  <  n!     (for n > 5)

Assume true for n

(n+1)^3 = n^3 + 3n^2 + 3n + 1

<  n!  + 3n^2 + 3n + 1     (by inductive assumption)

<  n!  + n^3 + n^3 +n^3    (if n > 3)

<   n!  + n! + n! + n!        (by inductive assumption)

<  4n!

<  (n+1)n!  =  (n+1)!        (if n >4)

Hence if true for n true for n+1.

Since true for n=6 by the principle of induction true for all n.
0

LVL 31

Assisted Solution

GwynforWeb earned 250 total points
...a bit tidier is

(n+1)^3 = n^3 + 3n^2 + 3n + 1

<  n^3 + n^3 + n^3 + n^3 = 4n^3     (if n > 3)

<  4n!    (by inductive assumption)

<  (n+1)n!  =  (n+1)!        (if n >4)

Hence if true for n true for n+1.
0

## Featured Post

### Suggested Solutions

This article will show, step by step, how to integrate R code into a R Sweave document
If you’re thinking to yourself “That description sounds a lot like two people doing the work that one could accomplish,” you’re not alone.
Viewers will learn how to properly install Eclipse with the necessary JDK, and will take a look at an introductory Java program. Download Eclipse installation zip file: Extract files from zip file: Download and install JDK 8: Open Eclipse and …
In this seventh video of the Xpdf series, we discuss and demonstrate the PDFfonts utility, which lists all the fonts used in a PDF file. It does this via a command line interface, making it suitable for use in programs, scripts, batch files — any pl…