Real Analysis - Natural Number Induction

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SUMMARY

The discussion centers on proving that if n is a natural number greater than 1, then n-1 is also a natural number using mathematical induction. Participants emphasize the importance of establishing the base case, specifically that 2-1=1 is a natural number, before proceeding to the inductive step. The correct formulation of the induction hypothesis is crucial, with the participants correcting each other's approaches to summation and induction. The solution involves confirming the inductive step by demonstrating that if n is in the natural numbers, then n-1 must also be in the natural numbers.

PREREQUISITES
  • Understanding of mathematical induction
  • Familiarity with natural numbers and their properties
  • Basic knowledge of summation notation
  • Experience with proofs in real analysis
NEXT STEPS
  • Study the principles of mathematical induction in detail
  • Review the properties of natural numbers and their operations
  • Practice constructing proofs using induction with various examples
  • Explore "Advanced Calculus by Fitzpatrick" for deeper insights into real analysis
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Students studying real analysis, particularly those learning mathematical induction, as well as educators seeking to clarify concepts related to natural numbers and proof techniques.

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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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