Prove that the product of an irrational number and a rational number is irrational. 
If $x$ is an irrational number and $r$ is a rational number then $xr$ is an irrational number.

Proof. Suppose that $xr$ is a rational number. By defintion of a rational number $xr= m/n$ where $m,n$ are some integers...
That's all I have so far since this topic really confuses me. Can someone please help me out?
 A: This is false because if you take $x = \sqrt{2}$ and $r = 0$,
$$
x \cdot r = \sqrt{2} \cdot 0 = 0,
$$
which is rational, not irrational.
However, suppose that $r \ne 0$.
Then suppose towards contradiction that $x$ is irrational and $r$ is rational, but $rx$ is not irrational, i.e. $rx$ is rational.
Then write $rx = s$, where $s$ is rational.
Since $\boldsymbol{r \ne 0}$, this implies $x = \frac{s}{r}$, which is a contradiction because...
A: $\pi$ is irrational and $\dfrac{10}\pi$ is irrational, but their product is rational.  In other words, you are right that it's not easy to prove that the product of two irrational numbers is rational.
Proving $\pi$ is irrational is not so easy: https://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational
However, if $\dfrac{10}\pi$ is rational, then $\dfrac{10}\pi$ is $\dfrac n m$ for some integers $n$ and $m$, so $\pi = \frac{10m} n$, a rational number.  Thus if $\pi$ is irrational, then $\dfrac{10} \pi\vphantom{\dfrac{\displaystyle\sum}\sum}$ must also be irrational.
PS: It would appear that I answer the wrong question.  The question was about the product of an irrational number and a rational number.
Say $x$ is irrational and $r$ is rational.  Suppose $xr$ is rational.  Then for some integers $a,b,c,d$ we have
$$
x r = x \frac a b = \frac c d.
$$
Consequently
$$
x = \frac {cb}{da},
$$
and thus $x$ is rational.
A: r is rational so r = a/b for some integers a and b and assume r not equal zero. (If r is zero xr is zero and so rational)
Assume $xr$ is rational and so $xr = m/n$ for some integers m and n
lets divide $xr$ by $r$
ie $xr/r = (m/n)/(a/b)$
so   $x  = mb/na$
and $x$ would be rational which is a contradiction.
So the statement "rational r by irrational x is irrational" is true if r is not equal to zero
