# Thread: Relation symmetric or transitive?

1. ## Relation symmetric or transitive?

Let Q denote the relation on the set Z of integers, where integers x
and y satisfy xQy if and only if x2 + 5y2 is divisible by 6.
Determine whether or not the relation Q is symmetric and whether or not
the relation Q is transitive.

2. ## Re: Relation symmetric or transitive?

Originally Posted by Cira
Let Q denote the relation on the set Z of integers, where integers x
and y satisfy xQy if and only if x2 + 5y2 is divisible by 6.
Determine whether or not the relation Q is symmetric and whether or not the relation Q is transitive.

Do you plan on showing some work of your own ?

3. ## Re: Relation symmetric or transitive?

Originally Posted by Cira
Let Q denote the relation on the set Z of integers, where integers x
and y satisfy xQy if and only if x2 + 5y2 is divisible by 6.
Determine whether or not the relation Q is symmetric and whether or not
the relation Q is transitive.
Hints: $(x^2+5y^2)+(y^2+5x^2)=6(x^2+y^2)$ and $(x^2+5y^2)+(y^2+5z^2)=(x^2+5z^2)+6y^2$.

4. ## Re: Relation symmetric or transitive?

Thank you Plato and Nehushtan!
I think I now understand?

If we suppose xQy then x2 + 5y2 is divisible by 6.
Then (x2 +5y2) + (y2 + 5x2) = 6(x2 + y2)
and then (y2 + 5x2) = 6(x2 + y2) - (x2 +5y2)

6(x2 + y2) - (x2 +5y2) is divisible by 6 since the difference of 2 integers divisible by 6 is itself an integer divisible by 6.
Thus yQx, showing that Q is symmetric.

If we suppose xQy and yQz then (x2 +5y2) and (y2 + 5z2) are both integers divisible by 6.
Then (x2 + 5z2) = (x2 +5y2) + (y2 + 5z2) - 6y2
It follows that (x2 + 5z2) is divisible by 6.
Thus xQz, showing that Q is transitive.