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?
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asked Nov 6, 2014 at 21:01
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2$\begingroup$ If $q\ne0$ is rational and $x$ is irrational then $qx$ is irrational. Is this your question? $\endgroup$– DidNov 6, 2014 at 21:04
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$\begingroup$ Hey, zero is the only one that breaks the rule :) $\endgroup$– brickNov 6, 2014 at 21:04
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5$\begingroup$ why people are down voting? $\endgroup$– Firas Abd El GaniNov 6, 2014 at 21:17
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$\begingroup$ Well the idea falls apart very quickly, if $x$ is irrational, $x^{-1}$ is irrational too. $\endgroup$– orionNov 7, 2014 at 12:34
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5$\begingroup$ possible duplicate of Prove that the product of a rational and irrational number is irrational; see also Proof verification: Let a be an irrational number and r be a nonzero rational number… $\endgroup$– Scott - Слава УкраїніNov 7, 2014 at 15:34
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4 Answers
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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.
answered Nov 6, 2014 at 21:04
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$\begingroup$ great answer, Matt. How about $irratonat^{annother irrational}$? $\endgroup$– shma2001Dec 11, 2014 at 15:53
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1$\begingroup$ 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. $\endgroup$– jdcJan 27, 2015 at 0:42
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$\begingroup$ @mattsamuel Does this assume an irrational times one is irrational? Should this also be proved, or is it just assumed? $\endgroup$ Oct 16, 2018 at 21:39
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1$\begingroup$ @Rasmus It is not assumed, but it is practically tautological. $1x=x$ for all $x$, so if $x$ has property $P$ then so does $1x$. $\endgroup$ Oct 16, 2018 at 22:26
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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.
answered Nov 6, 2014 at 21:04
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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.
answered Nov 6, 2014 at 21:06
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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.
