# Thread: For what k is a relation equivalent

1. ## For what k is a relation equivalent

Hello,

I have three relations on the set of natural numbers defined like this.

1) $x$ is in relation with $y$, when $x+y$ can be divided by $k$.
2) $x$ is in relation with $y$, when $x-y$ can be divided by $k$.
3) $x$ is in relation with $y$, when $x-y$ is equal to $k$.

I need to find $k$ such that $k$ is a positive number and when put into those relations they are equivalent.

So for the first one I think that $k$ can only be either 1 or 2, because it has to be reflexive, and any number is divisible by 1, and any sum of two same numbers is divisible by two.

For the second one k can be any number except 0 because every number can divide 0, which we get when we test the reflexiveness of the relation, and for any number $n$ we can find $m$ such that m is a sum of any multiple of k and $k-$ $n$ MOD $k$, and then $n+m$ is divisible by k.

In the third one the only possible value of k is 0, because the relation has to be reflexive and when we substract two of the same numbers we get a 0. When $k$ is 0 the relation is also symetrical and transitive.

Is this correct?

2. ## Re: For what k is a relation equivalent

k can only be equal to 0. Suppose k different from 0. by 3) x is in relation only wih x+k. But by 2) he is in relation with x+k,x+2k,x+3k....

3. ## Re: For what k is a relation equivalent

We can choose different k for each of those relations. It doesn't have to be just one k. And k can't be equal to 0 in 2) because we can't divide things by 0.

4. ## Re: For what k is a relation equivalent

We can't divide things by 0 but we can say if a given number "a" is or not divided by 0. the definition of divisibility is "a is idivisble by b if there exists c such that a=bc". The definition of divisibility doesn't involve division. The only number divisible by 0 is 0. So for the 2) k can also be 0. I think you're right for 1) and 3).