Mathematical Induction with an Inequality

In summary: We have (k + 2)k = (k + 2)(k + 2)k - 1 = (k + 2)( (k + 1) + 1)k -1Now we use the fact that k > 1 to get (k + 2)k > (k + 1)k + 1
  • #1
yellowsnow
4
0

Homework Statement



Prove that (n + 1)n - 1 < nn for n ∈ Z+. [Hint: Induction is suggested. Write out the induction statement explicitly. Make one side of the inequality look like your induction hypothesis.]

Homework Equations



The Attempt at a Solution



PS2-1.png


^ That's what I have so far. I'm good with induction, for the most part, but not really for inequalities.

I tried to follow the hint, but I'm not sure if I did that right.

If anyone can help me out I'd really appreciate it.

Thanks!
 
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  • #2
Your image is way too big (1996 pixels X 1869 pixels). Please shrink your image to about 800 x 600. Better yet, write your inequalities right in the text entry window.
 
  • #3
Mark44 said:
Your image is way too big (1996 pixels X 1869 pixels). Please shrink your image to about 800 x 600. Better yet, write your inequalities right in the text entry window.
Yeah, I resized it soon after posting (shows 613 x 573 now)

I'll try typing it up though :)
 
  • #4
You don't include it, but you also need to show that the statement is true in a base case, such as when n = 1. This is easy to show, since 20 <= 11.

You have for the case n = k + 1, (k + 2)k <= (k + 1)k + 1. You seem to be assuming that this is true. Instead you need to show that it is true, using the statement in the induction hypothesis (i.e., (k + 1)k - 1 <= kk).

I don't have any other advice or tips right now, but I'll give it some thought.
 
  • #5
I haven't taken this all the way through, so can't guarantee this is the way to go.

You need to show that (k + 2)k <= (k + 1)k + 1

Working with the left side, we have
(k + 2)k = (k + 2)(k + 2)k - 1 = (k + 2) ( (k + 1) + 1)k -1

Now expand the last factor using the Binomial Theorem, which says that
[tex](a + b)^n = {{n}\choose{0}}a^nb^0 + {{n}\choose{1}}a^{n - 1}b^1 + {{n}\choose{2}}a^{n - 2}b^2 + ... + {{n}\choose{n}}a^{0}b^n[/tex]
 

1. What is mathematical induction with an inequality?

Mathematical induction with an inequality is a proof technique used to demonstrate that a statement holds for all natural numbers greater than or equal to a starting value. It involves breaking down a problem into smaller subproblems and proving that the statement holds for each subproblem, which then implies that it holds for the larger problem as well.

2. How is mathematical induction with an inequality different from regular mathematical induction?

Mathematical induction with an inequality is a variation of regular mathematical induction. The main difference is that instead of proving that a statement holds for a specific starting value and all natural numbers greater than that value, we prove that it holds for a starting value and all natural numbers greater than or equal to that value. This allows us to prove statements involving inequalities, such as "for all n ≥ 5, n^2 ≥ 5n".

3. What is the basis step in mathematical induction with an inequality?

The basis step in mathematical induction with an inequality is the first part of the proof where we show that the statement holds for the starting value. This is usually done by substituting the starting value into the statement and showing that it is true. For example, to prove the statement "for all n ≥ 5, n^2 ≥ 5n", the basis step would involve showing that 5^2 ≥ 5(5), which simplifies to 25 ≥ 25, which is true.

4. What is the induction hypothesis in mathematical induction with an inequality?

The induction hypothesis in mathematical induction with an inequality is the assumption that the statement holds for a particular value, say k. This allows us to prove that the statement also holds for the next value, k+1. The induction hypothesis is used in the induction step of the proof.

5. How do you prove the induction step in mathematical induction with an inequality?

The induction step in mathematical induction with an inequality involves showing that if the statement holds for a particular value, k, then it also holds for the next value, k+1. This is usually done by using the induction hypothesis and the properties of inequalities to manipulate the statement. For example, to prove the statement "for all n ≥ 5, n^2 ≥ 5n", we would assume that the statement holds for some k ≥ 5, and then use this assumption to show that the statement also holds for k+1, which would be (k+1)^2 ≥ 5(k+1).

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