Proving n^3>n^2: Using Mathematical Induction and Set of Natural Numbers

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Discussion Overview

The discussion revolves around proving the inequality n^3 > n^2 for n greater than or equal to 2, specifically using mathematical induction and properties of natural numbers. Participants explore different approaches to establish this inequality and express their thoughts on the sufficiency of various proofs.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant inquires about proving n^3 > n^2 and expresses uncertainty about demonstrating that n^3 - n^2 is a natural number without generalizing that it is always greater than or equal to 1.
  • Another participant suggests that since for all α > 1, αn^2 > n^2, it follows that for all n > 1, n^3 > n^2.
  • A different participant points out that n^3 - n^2 can be factored as n^2(n - 1), asserting that this expression is positive for n > 1, thus supporting the inequality.
  • A later reply acknowledges the previous point about the brevity of the proof and expresses a sense of reassurance about the sufficiency of the argument.

Areas of Agreement / Disagreement

Participants present multiple approaches to the proof, indicating that there is no consensus on a single method being superior or more complete. The discussion remains open with various viewpoints being explored.

Contextual Notes

Some participants express concern that certain proofs may seem too brief or lacking in detail, suggesting that the completeness of the arguments could be questioned. There is also an implicit assumption that n is a natural number greater than or equal to 2, which is central to the discussion.

Ed Quanta
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How do you go about proving a statement like n^3>n^2 for n is equal to or greater than 2? I can prove this using mathematical induction, but I am unsure how to show n^3-n^2 is an element of the set of natural numbers without just saying in general that this must always yield a number equal to or greater than 1.
 
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Can't you just say that because

[tex]\forall\alpha>1:\alpha n^2 > n^2[/tex]

we must conclude that

[tex]\forall n>1: n^3 > n^2[/tex]
 
Why not just note that n3- n2= n2(n-1)? As long as n> 1, this will be a positive integer.
 
Yeah, Halls of Ivy, that's what I was thinking. It just looked like too short a proof so I thought I was missing something. Thanks dudes.
 

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