So I am learning about induction and I am trying to prove that
$(n^3 - n^2 + 1)$ is an odd number for all n ∈ N (natural number)
I want to see if I am on the right track to understanding this and if my reasoning makes sense :)
STEP 1: Base Case
So we first need to solve for the base case which we can say is n = 1.
$(1^3 - 1^2 + 1 = 1 - 1 + 1 = 1)$
Since 1 is odd, the base case holds! Onward to induction step.
STEP 2: Inductive Step:
We have some arbitrary positive integer k.
We can assume $ (k^3 - k^2 + 1) $ is odd. Using the natural number of $ k+1$ we can write
$(k+1)^3 - (k+1)^2 + 1 $
Expanding it out.....
$(k+1)^3 - (k+1)^2 + 1$
= $k^3 + 3k^2 + 3k + 1 - (k^2 + 2k + 1) + 1$
= $(k^3 - k^2 + 1) + 3k^2 + 3k$
Now, we know that $k^3 - k^2 + 1$ is odd and $3k^2 + 3k$ is always even (since it's divisible by 3).
Given that we assume $(k^3 - k^2 + 1)$ is odd
...and the sum of an odd number and an even number is always odd
...$ (k+1)^3 - (k+1)^2 + 1)$ is also odd
We proved the induction!
In line with @Marco, it remains to be proven that $3k^2+k$ is even.
There are two cases:
In all cases, $3k^2+k$ is even
This type of reasoning is called case-disjunction reasoning.
On the other hand, it is desirable to conclude(step 3):
"We have proved by induction that : $$\forall n\in \mathbb N^*, n^3-n^2+1\text{ is odd.}"$$