1
$\begingroup$

The relation F on Z is defined by F = {(n,m)| (2n+3m) is divisible by 5}
I need help proving this relation is symmetric.
I know I should choose two members of Z and assume F(x,y).
and I know I need to prove F(y,x).
But I keep getting stuck on the algebra. I'm not sure how to manipulate the equation to get 2y + 3x = 5*(some int)


Solution:
Choose x,y ∈ Z and assume F(x,y)
So 5|2x+3y
Which means 2x+3y = 5 * k for some integer K
-1(2x + 3y) = -1(5k) (mult both sides by -1)
-2x - 3y = -5k (distribute the negative)
-2x + 5x - 3y + 5y = -5k + 5x + 5y (add 5x and 5y to both sides)
2y + 3x = 5(-k + 5x + 5y)
Since -k, 5, x, y ∈ ℤ then -k + 5x + 5y ∈ ℤ (since integers are closed under addition)
Therefore 5|2y + 3x

$\endgroup$
2
$\begingroup$

One observation is that

$$5x+5y=(2x+3y)+(2y+3x).$$

Since $5$ divides the left side always, it follows that $5\mid 2x+3y$ iff $5\mid 2y+3x$.

$\endgroup$
1
$\begingroup$

You should assume $F(x,y)$ and prove $F(y,x)$, not $F(y,z)$

There are only five cases for $x \bmod 5$. For each one, you can compute $y \bmod 5$. Now note that you have $F(y,x)$ as well.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.