The hint I was given was to simply prove that $y=xz$ is irrational given that $x$ is nonzero, $x$ is rational and $z$ is irrational. Here's how I did it:
Claim: $y=xz$ is irrational.
Proof: Assume $x\neq0$, $x$ is rational and $z$ is irrational.
By contradiction assume that $y=xz$ is rational. This means $y$ can be expressed as $m/n$, $m$ and $n$ being integers; $y$ can be expressed similarly as $p/q$, $p$ and $q$ being integers. By substitution, we have that $$ p/q=mz/n$$
and $$z=pn/qm, qm \neq 0.$$
Since $pn$ and $qm$ are integers $z$ has to be rational.
In addition to this it seems like there's a part 2 as follows:
Proof: Given an interval $(x,y)$ we will choose a positive irrational number, $z$, say. By density of the rationals there is a rational $p$ in the interval $(x/z, y/z)$ s.t. $$ x/z <p< y/z.$$ From this we see that $pz$ is irrational since it is the product of a rational and irrational number.
Is the $pz$ the $xz$ that we proved is irrational in the first proof? So ideally when presenting a full proof like this, should we do part 2 then part 1?
Best Answer
Great proof! You can apply your result about what you called $xz$ in part 1 to $pz$ in part 2 because it obeys your assumptions: $p \ne 0$, $p$ is rational and $z$ is irrational. So from part 1 we know that $pz$ is irrational.
We also know that $pz$ lies in the interval $(x,y)$, so irrationals are dense.
What is ideal to present when presenting a full proof depends on what your reader already knows. It sounds like in this case you want to present both part 1 and part 2, and that will be a complete proof. But in some other case your reader (or lecturer or marker) might not know or want to assume, for example, that rationals are dense in the real numbers, so you might have to include a part 1.5 that proves that.
EDIT: Oops, I think I misunderstood you. Yeah, I would present part 2 then part 1 when presenting this proof; I think the sequence of ideas flows better. But it's up to you!
Also, just to point out a few typos:
Hope that helps!