# Even vs. Odd

#### Hyperreality

Here is a question I suddnely thought of.

The natural number series: 1,2,3...

If we double the series, we get: 2,4,6...

In other words get all the even numbers.

My question is there any ways to determine all odd number series?

#### Doctor Luz

If you add one more step and substract 1 after multpliying by 2 to the series you obtain the odd number series.

Staff Emeritus
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Dearly Missed
Or add 1 and treat 1 as a special case. So every even number is of the form 2n and every odd number is of the form 2n+1. Where n can be read as "something".

Now look what you can do. (2n)^2 = 4n^2 = 2(2n^2) so the square of an even number is of the form 2 times something, and so it is even too.

And (2n+1)^2 = 4n^2 + 4n +1
= 2(2n^2+2n) + 1.
So the square of an odd number is one more than an even number, so it is odd.

You have just proved a theorem: The square of an even number is even and the square of an odd number is odd.

#### quartodeciman

Take this another step.

Pretend that you are IBM Corporation{1} in the mid-1950s. Start your sequence with 0 instead of 1.

0 1 2 3 4 5 .....

Now, square the members of this sequence.

0 1 4 9 16 25 .....

This is the derived sequence of perfect squares.

Now, subtract adjacent members of this sequence.

Voilá!

1 3 5 7 9 .....

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{1}IBM computer systems numbered peripherals devices and their plug ports beginning with 0 instead of with 1 (which is what other computer manufacturers started with at that time).

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#### quartodeciman

One more bit.

0 1 2 3 4 5 .....

0 2 6 12 20 30 .....

Now, subtract adjacent terms of this.

2 4 6 8 10 .....

whoopie!

#### Integral

Staff Emeritus
Gold Member
Originally posted by quartodeciman
One more bit.

0 1 2 3 4 5 .....

0 2 6 12 20 30 .....

Now, subtract adjacent terms of this.

2 4 6 8 10 .....

whoopie!
n(n+1) - n(n-1)=n2+n - n2+n =2n

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