Is irrational times rational always irrational? Is an irrational number times a rational number always irrational?
If the rational number is zero, then the result will be rational. So can we conclude that in general, we can't decide, and it depends on the rational number?
 A: Claim: If $x$ is irrational and $r \ne 0$ is rational, then $xr$ is irrational.
Proof: Suppose that $xr$ were rational.  Then, $x = \frac{xr}{r}$ would be rational (as the quotient of two rationals).  This clearly contradicts the assumption that $x$ is irrational.  Therefore, $xr$ is irrational.
The $r = 0$ case is special, and the above argument doesn't work.
A: Any nonzero rational number times an irrational number is irrational. Let $r$ be nonzero and rational and $x$ be irrational. If $rx=q$ and $q$ is rational, then $x=q/r$, which is rational. This is a contradiction.
A: Irrational times non-zero rational is irrational number.
If not, suppose a is a irrational number and b is non-zero rational number such that ab=c, where c is a rational number.As collection of all rational number forms field.so any non-zero rational is invertible.So that would imply a is rational number--which is not true.
A: If $a$ is irrational and $b\ne0$ is rational, then $a\,b$ is irrational. Proof: if $a\,b$ were equal to a rational $r$, then we would have $a=r/b$ rational.
