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....
Hello,
I have three relations on the set of natural numbers defined like this.
1) is in relation with , when can be divided by .
2) is in relation with , when can be divided by .
3) is in relation with , when is equal to .
I need to find such that is a positive number and when put into those relations they are equivalent.
So for the first one I think that 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 we can find such that m is a sum of any multiple of k and MOD , and then 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 is 0 the relation is also symetrical and transitive.
Is this correct?
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).
Most working mathematicians do not allow that.
See this page.
OR this page.