Proving inequality by mathematical induction

In summary: The induction hypothesis is a premise that you are trying to prove, and the goal is to prove that the hypothesis is true for a certain number of cases. In this case, you are trying to prove that 2(k+1) < [(k+1)+1]!, or that P(k+1) follows from P(k).To do this, you need to know P(k). P(k) is the probability that k is less than or equal to 1. So, you need to find a way to prove that 2(k+1) < [(k+1)+1]!, or that P(k+1) follows from P(k).To do this, you can use the laws of mathematics
  • #1
dustbin
240
5

Homework Statement



I am asked to prove:
2n < (n+1)! , where n≥2

The Attempt at a Solution



Base step: set n=2, then test 22 < (2+1)!

22 = 4
(2+1)!= 3! = 3(2)(1) = 6
so 4 < 6 , which is true.

Induction hypothesis is 2k < (k+1)!
Using this, prove 2(k+1) < [(k+1)+1]! = (k+2)!

Attempt to solve:

starting with what I know: 2k < (k+1)!
Multiplying both sides by 2: 2(2k) = 2(k+1) < 2(k+1)!

I know that 2(k+1)! < (k+2)!
since (k+2)! = (k+2)(k+1)! and because k≥2, (k+2) will be greater than 2. Thus, multiplying (k+1)! by 2 on the LHS is less than multiplying (k+1)! by (k+2) on the RHS.

Thus, since 2(k+1) < 2(k+1)! is true, then 2k+1 < [(k+1)+1]!.

P(k+1) follows from P(k), completing the induction step. By mathematical induction, P(n) is true for n≥2.Thanks for any help!
EDIT: fixed a couple of type-o's.
 
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  • #2
dustbin said:

Homework Statement



I am asked to prove:
2n < (n+1)! , where n≥2

The Attempt at a Solution



Base step: set n=2, then test 22 < (2+1)!

22 = 4
(2+1)!= 3! = 3(2)(1) = 6
so 4 < 6 , which is true.

Induction hypothesis is 2k < (k+1)!
Using this, prove 2(k+1) < [(k+1)+1]! = (k+2)!

Attempt to solve:

starting with what I know: 2k < (k+1)!
Multiplying both sides by 2: 2(2k) = 2(k+1) < 2(k+1)!

I know that 2(k+1)! < (k+2)!
since (k+2)! = (k+2)(k+1)! and because k≥2, (k+2) will be greater than 2. Thus, multiplying (k+1)! by 2 on the LHS is less than multiplying (k+1)! by (k+2) on the RHS.

Thus, since 2(k+1) < 2(k+1)! is true, then 2k+1 < [(k+1)+1]!.

P(k+1) follows from P(k), completing the induction step. By mathematical induction, P(n) is true for n≥2.Thanks for any help!
EDIT: fixed a couple of type-o's.

Welcome to PF, and very nice first post! Looks good!
 
  • #3
Thank you very much for your response! It was posted as an extra credit problem by my professor. I wanted to make sure my reasoning was correct before posting it on the board, as I've been having a little difficulty grasping what we covered about mathematical induction.
 
  • #4
It looks to me like you have the idea of induction proofs.
 

1. How does mathematical induction prove inequalities?

Mathematical induction is a proof technique that is used to prove mathematical statements, including inequalities. It follows a three-step process:
1. Base case: Prove that the statement is true for the first value.
2. Inductive hypothesis: Assume that the statement is true for some arbitrary value.
3. Inductive step: Use the assumption to prove that the statement is also true for the next value.
By repeating this process, the statement can be proven for all values, including the inequality.

2. What is the difference between induction and strong induction?

Induction and strong induction are both proof techniques that are used to prove mathematical statements. The main difference between them is the inductive hypothesis. In induction, the hypothesis is that the statement is true for a single arbitrary value, while in strong induction, the hypothesis is that the statement is true for all values up to the arbitrary value. This allows for a stronger proof, as it covers more cases and can be used to prove more complex statements.

3. Can inequalities be proved using only strong induction?

Yes, inequalities can be proved using only strong induction. In fact, strong induction is often used to prove more complex inequalities where induction may not be sufficient. However, in some cases, induction may be more straightforward and efficient for proving inequalities.

4. Are there any specific inequalities that can only be proved using mathematical induction?

No, there are no specific inequalities that can only be proved using mathematical induction. Both induction and strong induction can be used to prove any inequality, but the approach may differ depending on the complexity of the statement.

5. What are some common mistakes to avoid when using mathematical induction to prove inequalities?

Some common mistakes to avoid when using mathematical induction to prove inequalities include:
- Not clearly stating the base case
- Not following the inductive hypothesis
- Using incorrect algebraic manipulations
- Assuming the statement is true for all values instead of just the next value
- Not considering all cases in the inductive step
To avoid these mistakes, it is important to carefully plan and organize the proof, paying attention to each step and checking for errors.

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