If (3+x)(3-x) is rational, then x is rational.
how would I claim that (3+x)(3-x) = p/q
Would I put (3+x)(3-x)/1=p/q -> q(3+x)(3-x)/p?
I don't think this is true, take $\displaystyle (3-x)(3+x)=1$ for instance. We have $\displaystyle x^2=8 \implies x =2\sqrt{2}$
The converse, however, is true. If $\displaystyle x=\frac{p}{q}$ then $\displaystyle (3+x)(3-x)=9-x^2=9-\frac{p^2}{q^2}=\frac{9q^2-p^2}{q^2}$
I think you meant to write (3+x)/(3-x). Then this is true.
Assume that (3 + x)/(3 - x) is rational. Then (3 + x) / (3 - x) = p/q for some integers p and q where q =/= 0. Then cross multiply.
Thus, q(3 + x) = p(3 - x).
Then 3q + qx = 3p - px
Then px + qx = 3p - 3q
Then x(p + q) = 3(p - q)
Then x = 3(p - q)/(p + q)
Since p and q are integers, p - q and p + q are integers. Then 3(p - q) and p + q are integers. Then 3(p - q)/(p + q) is rational. Thus, x is rational.