# Math Help - Question on Relations

1. ## Question on Relations

Question 1:
D on Z(all integers): mDn iff m|n

My thoughts:
Why is 0D0 not reflexive as stated by the solution? 0|0 --> 0=n.0
So that means it does not matter what value n is, the answer will be still be 0. So shouldn't it be reflexive?

Question 2:
P on N(natural numbers): mPn iff There exists p, p|m,p|n and p is prime.

My thoughts:
I am not sure how to apply the transitive rules on this where if xPy and yPz then xPz.

2. Hello,

Another way to say that 0|0 is that 0/0=an integer, and 0/0 is ... i'll say undetermined...

I am not sure how to apply the transitive rules on this where if xPy and yPz then xPz.
You can find out a counter example.

The relation here is :

If xPy <=> p|gcd(x,y), with p prime
and yPz <=> p'|gcd(y,z), with p' prime

Does p'', prime, exist such as p''|gcd(x,z) ?

Counter example :

3P6 because 3, prime, divides 3 and 6.
6P14 because 2, prime, divides 6 and 14.

Does a prime number divide 3 and 14 ? The answer is no, because they are coprime (1 is not a prime number !)

Then why is it not reflexive for question 2? It works out to be like this right?
xPx <=> p|gcd(x,x). Then shouldn't it be reflexive?

4. I would have said that it's reflexive too , because there always exists a prime integer which divides a natural integer...

5. Originally Posted by Moo
Hello,

Another way to say that 0|0
0 does not divide 0. But 0|0 is true. Because $a|b$ if and only if there exists $c$ so that $a=bc$. This is definitely true for $0|0$.

6. But dividing by 0 is not true ? How do we know if it can be a counterexample or not ? (just asking, don't screw up )

7. Originally Posted by Moo
I would have said that it's reflexive too , because there always exists a prime integer which divides a natural integer...
What prime divides 1?

8. Oh yeah
None.

Sorry ~

9. Originally Posted by Moo
But dividing by 0 is not true ? How do we know if it can be a counterexample or not ? (just asking, don't screw up )
What are you talking about? Remember $0|0$ if and only if there exists a $c\in \mathbb{Z}$ such that $0 = 0\cdot c$. Clearly, such a $c$ exists. Thus, $0|0$.

10. Just going the wrong way
Forgive me