Real Analysis - Natural Number Induction

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Homework Help Overview

The discussion revolves around proving that if n is a natural number greater than 1, then n-1 is also a natural number. The problem is situated within the context of mathematical induction in real analysis.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the use of mathematical induction, questioning the validity of their summation approaches and the base case. Some suggest establishing the trivial case before proceeding with induction.

Discussion Status

There is an ongoing exploration of the problem, with participants reflecting on their previous attempts and seeking clarification on the foundational aspects of the proof. Some guidance has been offered regarding the importance of the base case and the next natural number.

Contextual Notes

Participants express uncertainty about their methods and the implications of their assumptions, particularly regarding the properties of natural numbers and the structure of the proof.

mliuzzolino
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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 \mathbb{N} and n - 1 in \mathbb{N}} is inductive.)

Homework Equations


The Attempt at a Solution



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

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

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

2 + 3 + \cdots + (k - 1)

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

2 + 3 + \cdots + (k - 1) = (2 + 3 + \cdots + k) - 1 = \dfrac{(k-1)(k+2)}{2} - 1 <br /> = \dfrac{k^2 + k - 2 - 2}{2} = \dfrac{k^2 +k -4}{2}.<br />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!
 
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mliuzzolino said:

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 \mathbb{N} and n - 1 in \mathbb{N}} 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.
 
Mandelbroth said:
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...

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

S(2) = (2 - 1) = 1 = \dfrac{2(2-1)}{2} = 1

?
 
mliuzzolino said:
Would it be...

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

S(2) = (2 - 1) = 1 = \dfrac{2(2-1)}{2} = 1

?

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

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