Inequality Mathematical Induction

In summary, for n=4 the inequality is true because 4!>2^4. For n=k the equation is true because k!>2^k. For n=k+1 the inequality is true because (k+1)!>2^{k+1}.
  • #1
themadhatter1
140
0

Homework Statement


Prove the Inequality for the indicated integer values of n.

[tex]n!>2^n,n\geq4[/tex]

Homework Equations



The Attempt at a Solution



For n=4 the formula is true because

[tex]4!>2^4[/tex]

Assume the n=k

[tex]k!>2^k[/tex]

Now I need to prove the equation for k+1

I can multiply both sides by 2 and have

[tex]2(k!)=2(k!)>(2)2^k=2^{k+1}[/tex]

Is this what you would do next? I'm not quite sure what to do past this point.

[tex]2(k!)>2^{k+1}>k+1[/tex]
 
Physics news on Phys.org
  • #2
Why would you multiply by 2?
It's not 2k! you want to make a statement about, but (k + 1)! = (k + 1) k!
 
  • #3
CompuChip said:
Why would you multiply by 2?
It's not 2k! you want to make a statement about, but (k + 1)! = (k + 1) k!

I don't know. When dealing with regular equations you would just set k=(k+1) but that's not the case here I think.

Wouldn't you be trying to prove that [tex](k+1)!>2^{k+1}[/tex]? Why would I be making a statement about '(k + 1)! = (k + 1) k!'? Where is the 2 raised to the power of (k+1) and why is there 2 factorials when there isn't in the original problem?
 
  • #4
Sorry, you are approaching the problem from a different angle than I expected.

Once you have 2 (k!) > 2k + 1, you only need to prove that (k + 1)! > 2 k! to be done with it.

The hint I gave, that (k + 1)! is equal to (k + 1) times k!, is still useful.
 
  • #5
CompuChip said:
Sorry, you are approaching the problem from a different angle than I expected.

Once you have 2 (k!) > 2k + 1, you only need to prove that (k + 1)! > 2 k! to be done with it.

The hint I gave, that (k + 1)! is equal to (k + 1) times k!, is still useful.

Ahh, so you could rewrite

(k + 1)! > 2 k!

as

k!(k+1) > 2 k!

Here you can see the when k > 1, k!(k+1) > 2 k! and since the original formula is only supposed to work for numbers greater than or equal to 4 its fine.

Thanks.
 
  • #6
We are required to show that [tex]n!>2^{n}[/tex], for the k+1 term, multiply by k+1 to obtain:
[tex]
(k+1)k!=(k+1)!>2^{k}(k+1)>2^{k+1}
[/tex]
Since k+1>2 for k>2 and we are done
 

1. How does mathematical induction prove inequalities?

Mathematical induction is a proof technique that involves proving a statement for all natural numbers. To prove an inequality using induction, we must first show that the statement is true for the first natural number, usually 1. Then, we assume that the statement is true for some arbitrary natural number k. Finally, we use this assumption to prove that the statement is also true for the next natural number, k+1. This process is repeated indefinitely, proving that the statement is true for all natural numbers.

2. What is the principle of mathematical induction?

The principle of mathematical induction states that if we can prove that a statement is true for the first natural number, and we can also show that if the statement is true for some arbitrary natural number k, then it must also be true for the next natural number, k+1, then the statement is true for all natural numbers.

3. Can mathematical induction be used to prove all inequalities?

No, mathematical induction can only be used to prove inequalities involving natural numbers. It cannot be used to prove inequalities involving real numbers or other mathematical objects.

4. What are the steps for using mathematical induction to prove an inequality?

There are three main steps for using mathematical induction to prove an inequality:

  1. Prove the statement for the first natural number, usually 1.
  2. Assume the statement is true for some arbitrary natural number k.
  3. Use this assumption to prove that the statement is also true for the next natural number, k+1.

5. Can mathematical induction be used to prove equalities as well as inequalities?

Yes, mathematical induction can be used to prove both equalities and inequalities. The principle of mathematical induction can be applied to any statement that involves natural numbers.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
6
Views
1K
  • Precalculus Mathematics Homework Help
Replies
10
Views
780
  • Precalculus Mathematics Homework Help
Replies
1
Views
1K
  • Precalculus Mathematics Homework Help
Replies
13
Views
1K
  • Precalculus Mathematics Homework Help
Replies
3
Views
623
  • Precalculus Mathematics Homework Help
Replies
13
Views
1K
  • Precalculus Mathematics Homework Help
Replies
5
Views
2K
  • Precalculus Mathematics Homework Help
Replies
4
Views
2K
  • Precalculus Mathematics Homework Help
Replies
3
Views
2K
  • Precalculus Mathematics Homework Help
Replies
8
Views
1K
Back
Top