Prove that: odd + odd = even even + even = even odd + even = odd even + odd = odd

All even numbers are multiples of 2 and so can be written as 2n, where n is an integer. All odd numbers are one more or one less than an even number so can be written as 2n + 1 or 2n – 1, where n is an integer. It doesn’t matter what letter you use, as long as the letter is defined as an integer. i.e. 2 × integer = even 2 × integer ± 1 = odd

Look carefully at the following expressions. All the letters represent integers. Sort out which are odd, even, or could be either. 2x 2k - 1 3a a + b 2a + 2b 4r + 1 2(a+ p) + 1 2kr 2kr + a odd even either

Proof that odd + odd = even p and q are integers (2p + 1) and (2q + 1) are both odd numbers (2p + 1) + (2q + 1) =2p + 2q = 2p + 2q + 2 = 2(p + q + 1) which is 2 × integer and therefore even Can you write out a proof that: odd + even = odd even + odd = odd even + even = even Also try to prove that: odd × odd = odd odd × even = even even × odd = even even × even = even