Help with understanding definition of divisibility in this case.

Question

I have a proof that shows that if $5 \mid xy$ then $5 \mid x$ or $5 \mid y$.

It's pretty clear to me that I can just say that suppose $5 \mid x$, then $x=5a$, where $a$ is an integer.

then $xy = 5(ay)$, and thus $5 \mid xy$.

However, I'm having trouble understanding how this shows that $5 \mid xy$. I understand how it works for like $5 \mid x$, but am unsure how the $(ay)$ proves that $5 \mid xy$.

Answer 1

You appear to be mixing up things.

On the one side, you appear to want to prove that divisibility is transitive, in the particular case $5 \mid x$ and $x \mid xy$ implies $5 \mid xy$.

Now to say that $u$ divides $v$ it means there exists an integer $w$ such that $v = u w$. So when you get $xy = 5 \cdot (a y)$ you have that $u = 5$ divides $v = xy$, here $w = a y$.

On the other side, if $5$ divides $xy$, that is, there exists $z$ such that $x y = 5 z$, factor $x, y, z$ as products of primes. Since the prime $5$ appears in $5 z$, by the uniqueness of factorization it has to appear also in $x y$, that is, either as a factor of $x$, or as a factor of $y$.

Answer 2

Assume 5 does not divide x then we have to establish that 5|y If 5 does not divide x then the G.C.D of 5 and x is 1. i.e., (5,x)=1 Then 5*a+x*b=1 where a and b are unique integers Multiply both sides by y then y*5*a+y*x*b=y 5|x*y implies 5|y*x*b and 5|y*5*a

5|(y*5*a+y*x*b) Therefore 5|y

Answer 3

You're trying to show $ 5 \mid xy \implies 5 \mid x \text{ or } 5 \mid y$.

We know $5$ is prime. If $5 \nmid x,$ then we know $5$ and $x$ are relatively prime. Then $5 \mid xy$ implies that $5 \mid y$. Else if $5 \nmid y,$ then $5 \mid xy$ implies that $5 \mid x$ in like manner. So $ 5 \mid xy \implies 5 \mid x \text{ or } 5 \mid y$.