# Is irrational times rational always irrational? [duplicate]

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?

• If $q\ne0$ is rational and $x$ is irrational then $qx$ is irrational. Is this your question? – Did Nov 6 '14 at 21:04
• Hey, zero is the only one that breaks the rule :) – brick Nov 6 '14 at 21:04
• why people are down voting? – Firas Ali Abdel Ghani Nov 6 '14 at 21:17
• Well the idea falls apart very quickly, if $x$ is irrational, $x^{-1}$ is irrational too. – orion Nov 7 '14 at 12:34
• – Scott Nov 7 '14 at 15:34

Any nonzero rational number times an irrational number is irrational. Let $r$ be nonzero and rational and x irrational. If $rx=q$ and $q$ is rational, then $x=q/r$, which is rational. This is a contradiction.

• great answer, Matt. How about $irratonat^{annother irrational}$? – shma2001 Dec 11 '14 at 15:53
• No. The classic argument is that if ${\sqrt 2}^{\sqrt 2}$ isn't a counterexample, then $2 = ({\sqrt 2}^{\sqrt 2})^{\sqrt 2} = {\sqrt 2}^2$ must be. It's much harder to show, but apparently ${\sqrt 2}^{\sqrt 2}$ is irrational. – jdc Jan 27 '15 at 0:42
• @mattsamuel Does this assume an irrational times one is irrational? Should this also be proved, or is it just assumed? – Rasmus Larsen Oct 16 '18 at 21:39
• @Rasmus It is not assumed, but it is practically tautological. $1x=x$ for all $x$, so if $x$ has property $P$ then so does $1x$. – Matt Samuel Oct 16 '18 at 22:26

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.

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.

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.