Proofs with integers.......q1?

Question

Prove that for all integers n, n^2 - n + 3 is odd.

Answer 1

n^2 - n + 3 = (n-1) n + 3. Out of the consecutive integers n-1 and n, one is even, so that (n-1)n is even for every integer n. Since 3 is odd and the sum of an even number with an odd one is always odd, it follows n^2 - n + 3 is odd.

Answer 2

All integers may be either odd or even. Therefore, if we can prove that for any 2 integers p and q, where p is odd and q is even, n^2 - n + 3 is odd, it stands proved.

Lets take an even integer p
p^2 is even.
p+3 is odd(even + odd = odd no.)

(p^2)-(p+3) is like Even - odd = odd
This is true even if p is a negative integer.

Now, lets take an odd integer q.
q^2 is odd. (odd * odd = odd no.)
Q+3 = even no. (odd+even = even no.)
(q^2) -(q+3) is like odd-even = odd no.
This is true even if q is a negative integer.

Therefore, for all integers, the above holds true.

Answer 3

Ok, we just need to proove that n^2 - n is even (since adding 3 to an even number leaves an odd number)

Assume n is even. Then n=2m for some m

n^2-n = 4m^2-2m = 2 x (2m^2 - m)
and this is an even number

Now assume n is odd. Then n = 2m +1 for some m
n^2-n =
(2m+1)^2 - 2m - 1 =
4m^2 + 4m +1 - 2m - 1 =
4m^2 + 2m =
2 (2m^2 +m)
which is again an even number

Answer 4

If n is even then n^2 is also even (square of an even no is even). n^2 - n is an even number as even - even = even. n^2 - n + 3 is odd coz even + odd = odd

If n is odd then n^2 is also odd.(square of an odd no is odd) n^2 - n is an even number as odd - odd = even. n^2 - n + 3 is odd coz even + odd = odd

Hence n^2 - n + 3 is always odd

Hope that helped

Answer 5

n^2-n+3 = n(n-1) + 3
if n is odd 1st term is even as n-1 is even so sum is odd
if n is even then again 1st tern is even so sum is odd