Real Analysis - Natural Number Induction

[mliuzzolino]

Homework Statement

Prove that if n is a natural number greater than 1, then n-1 is also a natural number. (Hint: Prove that the set {n | n = 1 or n in [itex] \mathbb{N} [/itex] and n - 1 in [itex] \mathbb{N} [/itex]} is inductive.)

Homework Equations

The Attempt at a Solution

[itex] S(n) = \sum_{j = 2}^{n} j = 2 + 3 + \cdots + n = \dfrac{(n-1)(n+2)}{2} [/itex]

Checking the induction hypothesis, S(2) = 2 = (1)(4)/(2) = 2. True.

Suppose the statement [itex] P(k): 2 + 3 + \cdots + k = \dfrac{(k-1)(k+2)}{2} [/itex] is true. Then let P(k-1) be the statement:

[itex] 2 + 3 + \cdots + (k - 1) [/itex]

We want to show that this equals [itex] \dfrac{(k-2)(k+1)}{2} [/itex].

[itex] 2 + 3 + \cdots + (k - 1) = (2 + 3 + \cdots + k) - 1 = \dfrac{(k-1)(k+2)}{2} - 1
= \dfrac{k^2 + k - 2 - 2}{2} = \dfrac{k^2 +k -4}{2}.
[/itex]I'm not sure where I'm going wrong with this, but I'm feeling quite lost at this point. I'm self studying using "Advanced Calculus by Fitzpatrick," and my university does not have upper division math tutoring over the summer. I greatly appreciate any points in the right direction!

 

[Mandelbroth]

Homework Statement

Prove that if n is a natural number greater than 1, then n-1 is also a natural number. (Hint: Prove that the set {n | n = 1 or n in [itex] \mathbb{N} [/itex] and n - 1 in [itex] \mathbb{N} [/itex]} is inductive.)

Forget what you were doing.

After 1, what's the next greatest natural number? Is that number minus 1 a natural number? Establish that it is true for the trivial case before you attempt induction.

 

[mliuzzolino]

Forget what you were doing.

After 1, what's the next greatest natural number? Is that number minus 1 a natural number? Establish that it is true for the trivial case before you attempt induction.

Would it be...

[itex] S(n) = \sum_{j = 2}^{n} j - 1 = 1 + 2 + \cdots + (n - 1) = \dfrac{n(n - 1)}{2} [/itex]

[itex] S(2) = (2 - 1) = 1 = \dfrac{2(2-1)}{2} = 1 [/itex]

?

 

[Mandelbroth]

Would it be...

[itex] S(n) = \sum_{j = 2}^{n} j - 1 = 1 + 2 + \cdots + (n - 1) = \dfrac{n(n - 1)}{2} [/itex]

[itex] S(2) = (2 - 1) = 1 = \dfrac{2(2-1)}{2} = 1 [/itex]

?

No. No sums. Forget what you were thinking. I don't know where you're going with it, but it's not correct.

What is the NEXT natural number after 1?

 

[mliuzzolino]

Ok. I see where I'm going wrong with this idiotic summation stuff. I don't know what I was thinking.

 

[Mandelbroth]

Ok. I see where I'm going wrong with this idiotic summation stuff. I don't know what I was thinking.

It's okay. I do stupid math things all the time. In fact, I literally just tried to say that a Cauchy sequence does not necessarily converge to a constant when considering a discrete metric, which is obviously wrong. I make mistakes all the time.

2 is the next natural number, correct? And, as we know, 2-1=1, which is a natural number. Next, we prove the n+1 case (the inductive step).

Here's the set up:

$$(n+1)-1=n\stackrel{?}{\in}\mathbb{N}.$$

How might we prove this?