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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.
odd
2(a+ p) + 1
a+b
2kr + a
2x
2a + 2b
even
3a
4r + 1
2kr
2k - 1
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 + 1 + 1
= 2p + 2q + 2
= 2(p + q + 1)
which is 2 × integer and therefore even
Can you write out a proof that:
Also try to prove that:
odd + even = odd
odd × odd = odd
even + odd = odd
odd × even = even
even + even = even
even × odd = even
even × even = even