# Mathematic Inductive Proof question

## Homework Statement

Give a proof by Mathematical Induction of the following:

For all integers n>=3, (n^2 - 3n + 2) is positive.

## The Attempt at a Solution

Hey guys, this is a problem from my discrete mathematics study guide. Here's what I got so far:

Proof: n-initial=3
Basis step: If n=3, then LHS=3 and RHS = (3^2 - 3(3) + 2) = 2 -> positive
Induction: Assume that (n^2 - 3n + 2) is positive for arbitrary n>=3

Now I'm not sure about how to actually go about the proof. I understand that we then show the induction hypothesis working for n+1 but I'm not sure how to put this together.

Something like: (n+1)^2 - 3(n+1) + 2 = n^2 - n....

EDIT: I see, so then using (n^2 - n) and plugging in 3 I get (3^2 - 3 = 6) which is equal to (4^2 - 3(4) + 2 = 6) -> positive. Would this be a complete proof?

Last edited:

SammyS
Staff Emeritus
Homework Helper
Gold Member
Assume that n2 - 3n + 2 > 0 . Now, show that (n+1)2 - 3(n+1) + 2 > 0 .

It may also be the case that you need to use the fact that n ≥ 3.

Well (n+1)^2 - 3(n+1) + 2 simplifies to n^2 - n.

How do I show that n^2 - n > 0 ?

n^2 - n > 0 <=> n^2 > n, and because your n is greater than or equal to 3, there is no case that shows n^2 < n, or even equal.

HallsofIvy
$n^2- n= n(n- 1)$.
n> 0 and, since $n\ge 3$, n-1> 0.