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

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To prove that n^3 > n^2 for n ≥ 2 using mathematical induction, one can start by establishing the base case for n = 2, where 2^3 > 2^2 holds true. The inductive step involves assuming n^3 > n^2 for some integer n and proving it for n + 1. The expression n^3 - n^2 can be factored as n^2(n - 1), which is positive for n > 1, confirming that n^3 > n^2. This approach effectively demonstrates the inequality without needing to delve into more complex arguments. The discussion highlights the simplicity and sufficiency of the proof method.
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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

\forall\alpha>1:\alpha n^2 > n^2

we must conclude that

\forall n>1: n^3 > n^2
 
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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