Re: Proof that the sum of two consecutive integers is always an odd number.



On 10 Nov, 01:30, Kenneth Doyle <nob...@xxxxxxxxxxx> wrote:
I saw this last week as a question here. I've never been good at proofs,
but for some reason a "sketch" of this proof just popped into my head.

We want to prove that the sum of two consecutive integers is always an
odd number, that is:

x + y = n, where y = x + 1 and n is an odd number.

This can be re-written as:

x + x + 1 = n, because y = x + 1.

or:

2x + 1 = n.

Any integer multiplied by two is an even number and an even number plus
one is an odd number, therefore the sum of two consecutive integers is
always an odd number.

In this example, would I also have to prove that an integer multiplied by
two is always even? Something like:

2x = m, where m is even, because m / 2 always gives us
the original integer x and a number is even if it can be divided by two
leaving no remainder.

Would the proof that m is always even, have to be included in the proof
or can it be just a reference to a "sub-proof"; if you see what I mean.
Sorry about the clumsy language.

Try formulating the definition of an even number. Or ask yourself,
what is an even number if not a multiple of 2?

.



Relevant Pages