If $r \ne 0$ is rational and $i$ is irrational, then $ri$ is irrational 
Prove the following:
The product of a nonzero rational number and an irrational number is also irrational.

I assumed the following:
Let $r = c/d$ be rational, where $c$ and $d$ are integers and $r$ is nonzero, so $c$ and $d$ are nonzero as well.  Let $i$ be irrational.
Then I tried proving by contradiction that:
Suppose that $ri = a/b$.  Then $(ci)/d = a/b$. I assume that $ci$ is also irrational because multiplying an irrational number by an integer results in an  irrational number (but I don't know why this is, to be honest).
Is that correct?
 A: You have to go a bit further and show that, with your assumptions, $i$ would be a rational number $\frac{ad}{cb}$, thus showing the contradiction. ($ci$ is indeed irrational, but this fact is not needed for the proof.)
A: We only need to know that the rationals form a field: Suppose that $ri$ is rational (i.e. $ri \in \Bbb Q$)
then $i= \frac{ri}{r} \in \Bbb Q$ because we can divide a rational by a non-zero rational and the result is rational. But $i$ is given not to be rational. This contradiction shows $ri$ is not rational, so irrational.
A: If $ri = \frac ab$ where $a,b$ are integers and $r = \frac cd\ne 0$ then $c \ne 0$ and
$ri = \frac cdi = \frac ab$ and 
$i = \frac {ad}{cd}$ which is rational.
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Basic props.
Prop 1: If $r$ is rational and $n$ is an integer then $nr$ and $r+n$ and $r-n$ and, if $n \ne 0$ then $\frac rn$ are rational.  
Pf:  If $r$ is rational then there are integers, $a,b$ so that $r= \frac a b$.  So $nr = \frac {an}b$ and $r + n = \frac {a + bn}n$ and $r-n = \frac {a-bn}b$ and $\frac rn = \frac a{bn}$ and as $an, a+bn, a-bn,$ and $bn$ are all integers those are all rational.
Prop 2:  If $r$ is rational and $q$ is rational then $qr$ and $q+r$ and $q-r$ and, if $q\ne 0$ the $\frac rq$ are rational..
Pf: $r$ is rational then $r =\frac ab$ form some integers $a,b$ and $q=\frac cd$ for some integers $c,d$.  $qr = \frac {ac}{bd}, q+r = \frac{ad + bc}{bd}, q-r = \frac{ad-bc}{bd}$ and $\frac rq = \frac {ad}{bc}$ and as $ac,bd,ad+bc, ad-bc, ad,$ and $bc$ are all integers those are all rational.
Corollary:  If $r$ is rational and $q$ is rational and if $xr = q$ or $\frac rx = q$ or $\frac rx = q$ or $r+x=q$ or $r-x = q$.  If any of those hold then $x$ is rational.
Pf: If any of those hold then $x = \frac qr$ or $x=\frac rq$ or $x=q-r$ or $x = r-q$.  And by 1, and 2) above those are all rationals.
Prop:  If $i$ is irrational and $p$ is rational then $pi, \frac ip$ (assuming $p\ne 0$), $\frac pi, p+i, p-i, i-p$ etc are all irrational.  Pf: If any of the results were rational the the corolary above would prov to us that $i$ is rational.  Which it isn't.
THE ONE EXCEPTION.
If $i$ is irrational and $j$ is irrational, then we have utterly NO way of knowing whether $ij, \frac ij, \frac ji, i+j, i-j, j-i$ are rational or irrational or what.
