Use mathematical induction to prove that the statement is true for every positive integer n. Show your work. 2 is a factor of n2 - n + 2

Let n = 1

then f (1) = 1^1 - 1 + 2 = 2 so it is true for n = 1

for the next number after n (n+1) we have f (n+1) =

(n+1) ^2 - (n+1) + 2

= n^2 + 2n + 1 - n - 1 + 2

= n^2 + n + 2

= n (n+1) + 2

Now n (n+1) must be divisible by 2 because either n is odd and n+1 is even OR n is even and n+1 is odd and odd & even always = an even number.

So the function is divisible by 2 for n+1 We have shown that its true for n = 1 Therefore it must be true for n = 1,2,3,4 ...

True for all positive integers