Prove that if $ r,p \in \Bbb{N} $, then $ \gcd(r,rp) = r $.

Question

Problem. Prove that if $ r,p \in \Bbb{N} $, then $ \gcd(r,rp) = r $.

I tried solving this. If $ \gcd(r,p) = 1 $, then $ \gcd(r,rp) = 1 \times r $. Is that right?

Answer 1

Use the fact that if $a,b$ are positive integers so that $a|b$ and $b|a$ then $a=b$ :

Since $r| r$ and $r|rp$ it follows that $r |gcd(r,rp)$.

Also by definition $\gcd(r,rp) |r$.

Answer 2

The GCD is the greatest common divisor. Now $r$ is the greatest common divisor of itself, and $r$ divides $rm$. So $\gcd(r, rm)$ is divisible by $r$, so it's less than or equal to $r$, but $\gcd(r, rm)$ divides $r$, hence it's less or equal to $r$. Hence $\gcd(r, rp) = 1$. Your method might not work if $\gcd(r, p)$ is not $1$!

Answer 3

since $r,p$ belongs to the Natural numbers. Then we know that the biggest number that divides $r$ is $r$ itself. And also the biggest number that divides $p$ is $p$ itself.

That is if and only if $r,p > 0$

I will talk about the case when $r$ or $p$ is zero

Now since $r$ clearly divides $rp$ and $r$ is the greatest divisor of $r$ then $\gcd(r,rp) = r$

It will work even if the $\gcd(r,p) \neq 1$

For example $r=6,p=8$ . Now $\gcd(6,8) = 2$ and $\gcd(6,6\times8 = 48) = 6$

I challenge you to find a counter example, if $r,p \in \mathbb{N}$ and $r,q \neq 0$

The only problem happens when either $r,p = 0$.

For example, $r=2,q=0$ Now $\gcd(2,0) = 2$ and $\gcd(2,2\times0) = 2$ .

However, $rq = 0$ and so it doesn't hold for that case.

If we are only considering the set of Positive numbers ($\mathbb{N} - 0$)

Then the statement will always hold