# name of rule that says: x(x+1) is even number

What is the name of the rule that states that $x(x+1)$ is always even number? Mathworld says: "The product of an even number and an odd number is always even" But it does not state any name: http://mathworld.wolfram.com/OddNumber.html

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Why should it have a name? – Bruno Joyal Nov 8 '13 at 2:04
@BrunoJoyal Everything seems to have a name, so why does odd*even=even not have one? – user106280 Nov 8 '13 at 2:07
@user106280 If it not true that each and every property in mathematics has a name. Obviously, it would be utterly meaningless to have a name for every little property. Especially trivial and very simple and straightforward properties (like the one you mention) do not have a name. – Ittay Weiss Nov 8 '13 at 2:11
Usually names are given to non-trivial ideas which are used a great deal. It's much easier to say "Bolzano-Weirstrass" theorem than to write it all out, and it also leaves people the option of looking it up if they don't know it. Anyone who knows the barest amount of mathematics can see right away that n(n+1) is even (assuming n is an integer). If something has a name, you can be pretty sure your teachers will mention it. – Betty Mock Nov 8 '13 at 2:24
This world is each day worse because we give names to everything. – ILoveMath Dec 2 '13 at 13:20

If you want to say this on an olympiad, I'd say it'd be pretty safe to just say "by parity", since this is a pretty trivial result. Others might also state it as follows:

Since $x$ and $x+1$ have opposing parity, the product $x(x+1)$ must be even.

or

Either $x$ or $x+1$ must be even, since they have opposing parity. Thus, their product is even.

I wouldn't really assert a name to it.

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It's called the very famous "Even-odd Multiplication Property" :) – imranfat Nov 8 '13 at 2:56
I wouldn't recommend quoting that on an Olympiad. – Ahaan S. Rungta Nov 8 '13 at 3:03
I certainly wouldn't dare to! – imranfat Nov 8 '13 at 3:04

$0(0+1) = 0$, given that $0\cdot2=0$ then $0$ is an even number.

Let's assume $n(n+1)$ is even as hypothesis of induction. Then there is $k\in\Bbb N$ such as \begin{align} n(n+1)&=2k &&\text{Let's add $2(n+1)$ at both sides:}\\ n(n+1)+2(n+1)&=2k+2(n+1) &&\text{Let's apply distributive rule:} \\ (n+2)(n+1)&=2(k+n+1) &&\text{and conmutiative rule in products:} \\ (n+1)((n+1)+1)&=2(k+n+1) \end{align} Where $k+n+1\in\Bbb N$, therefor $(n+1)((n+1)+1)$ is even, and the result has been proved by mathematica induction.

But if you want a name for a rule that even times odd is even, it is just definition of even and transitivity of the relationship “divides”. If $n$ is even, then $2$ divides $n$. Now, if $nm$ is the product of $n$ and $m$ (even or odd) it means $n$ divides $nm$, as divides is transitive, then $2$ divides $nm$.

For the specific part of $x$ and $x+1$ (with $x\in\Bbb N$) then we need a theorem of one of them being even. That can be granted by modularity and can be extended that for every $n\in\Bbb N$, the set $\{x,x+1,\ldots,x+(n-1)\}$ will include a multiple of $n$. Modularity for $n=2$ has a special name: parity.

So parity grants that either $x$ or $x+1$ is even, and transitivity of the divides relationship grants that the product of an even number with any other natural number is even.

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Or you could just prove it (if you are afraid of using it because it doesn't have a name):

• if $x=2k$ then

$x(x+1)=(2k)(2k+1)=2*[k(k+1)]=2m$, where $m=k(k+1)$

• if $x=2k+1$ then

$x(x+1)=(2k+1)(2k+2)=(2k+1)*[2(k+1)]=2(2k+1)(k+1)=2m$, where $m=(2k+1)(k+1)$

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So you say that k(k+1) = (2k+1)(k+1)? and k=-1? – user106280 Nov 8 '13 at 3:50

You can think of this as follows.

The product $x(x+1)$ is basically the addition of $x$ with itself $(x+1)$ times.This is same as the addition of $(x+1)$ with itself $x$ times.[that is how we define the MULTIPLICATION of two natural numbers] If $x$ is odd add $x$ to itself $(x+1)$ times.
Now $x$ being even it can be written in the form of $2j$ where $j$ is any natural number.

Thus $(x+x+x+....+x)$ [$(x+1)$ times]= $(2j+2j+2j+...+2j)$ [$(x+1)$ times]=$2(j+j+j+...+j)$[$(x+1)$times] which is obviously even.
you can then extend the proof when $x$ is odd that is $(x+1)$ is even.

Having proved these two you can then easily extend the proof for negative integers.

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