# Thread: mathematical induction

1. ## mathematical induction

1. Use math induction to show that (n2 - n + 2) is divisible by 2 for all natural numbers n.

2. the equation is n^2-n+2
let n=1 then
1^2-1+2=2 which is divisible by 2 hence its true for n=1.
now let it be true for n=k therfor
k^2-k+2=2r
now let us consider n=k+1
(k+1)^2-(k+1)+2
=k^2+1+2k-k-1+2
=[k^2-k+2]+2k
=2r+2k
=2(r+k) which is divisible by 2
hence whenever it is true for k it is also true for k+1
hence by PMI n^2-n+2 is divisible by 2

3. 1. Case for 1: (1^2 - 1 + 2) = 2 is divisible by 2
2. Case for n + 1
((n+1)^2 - (n+1) + 2) =
(n^2+2n+1 - n - 1 +2) =
(n^2+n+2) =
(n^2-n+2)+2n
Assume (n^2-n+2) is divisible by two, which gives 2*(term + n), which evidently is divisible by 2.

I'm not used to proving these things, so don't use this as your proof, however, this is the general idea.

4. ## Simpler answer

n2 - n = n(n - 1)

The product of an even number with an odd number is always even as the above equation is and dividing 2 by 2 is self evident. This completes the proof.