Prove the following equivalence (rationals)

Question

I'm working on trying to prove the following equivalence for all x in real numbers:

$x$ is rational $\leftrightarrow x - 5$ is rational $\leftrightarrow x/3$ is rational

I know i need to prove each one individually such as: $$x\,\,is\,\,rational \rightarrow x - 5 \,\,is\,\,rational\\ x-5\,\,is\,\,rational \rightarrow x/3\\ x/3 \,\,is\,\,rational \rightarrow x \,\,is\,\,rational\\$$

1. I started by assuming if $x = \frac{p}{q} $ then $x-5= \frac{p}{q}-5$ which is still a rational.
2. Then if $x - 5 = \frac{p}{q}$ then $\frac{x}{3} = \frac{\frac{p}{q}}{3}$ which is still a rational (Is there a proper way to prove this algebraically, I feel like it might be insufficient proof to just keep saying it is a rational).
3. After this I am very unsure what to do to get from step 2 to 3. I've looked at a bunch of examples but I suck at proving rationals. Would it be something like if $3*\frac{x}{3} = \frac{\frac{p}{q}}{3}*3$ then $x = \frac{p}{q} $

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Edit:

1. Assume x is rational   $p \rightarrow q$
   x = $\frac{p}{q}$
   Prove x-5 is rational
   substitute x :  $x-5 = \frac{p-5q}{q}$   : x-5 is a rational
2. Assume x-5 is rational   $q \rightarrow r$
   x-5 = $\frac{p}{q}$
   solve for x:   $x=\frac{p-5q}{q}$
   Prove $\frac{x}{3}$
   substitude for x:   $\frac{x}{3} = \frac{\frac{p-5q}{q}}{3}$   : $\frac{x}{3}$   is rational because the numerator and denominator are rationals
   
3. Assume $\frac{x}{3}$ is rational   $r \rightarrow p$
   $\frac{x}{3} = \frac{p}{q}$
   solve for x = $\frac{\frac{p-5q}{q}}{3}$
   Prove x is a rational
   we already proved x is a rational in the second step, because the quotient of two rationals is always a quotient.

Answer 1

$(i)\implies (ii)$

Suppose $x$ is rational. Then $x=\frac{p}{q}$ for some integers $p,q$ with $q\neq 0$.

We have then $x-5=\frac{p}{q}-5=\frac{p}{q}-\frac{5q}{q}=\frac{p-5q}{q}$ and noting that $p-5q$ is an integer (since integers are closed under addition and multiplication) and $q\neq 0$ as before, we have successfully written $x-5$ as a ratio of two integers with the integer in the denominator nonzero and is therefore rational.

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$(ii)\implies (iii)$

Suppose $x-5$ is rational. Then $x-5=\frac{p}{q}$ for some integers $p,q$ with $q\neq 0$.

Then $\frac{x}{3}=\frac{1}{3}\left((x-5)+5\right)=\frac{1}{3}\left(\frac{p}{q}+5\right)=\frac{1}{3}(\frac{p+5q}{q})=\frac{p+5q}{3q}$ and noting that $p+5q$ is an integer since integers are closed under addition and multiplication and that $3q$ is a nonzero integer since both $3$ and $q$ are both nonzero integers and the product of two nonzero integers is again a nonzero integer, we see then that we can write $\frac{x}{3}$ as a ratio of two integers with the integer in the denominator nonzero and is therefore rational.

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$(iii)\implies (i)$

Similar to the other cases.