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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?

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    $\begingroup$ You're probably allowed to use the fact that the product of rational numbers is yet a rational number. Do you see how this helps? $\endgroup$ – Git Gud Nov 3 '15 at 20:29
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    $\begingroup$ First of all the claim as stated is false. Can you think of a counter example? With a minor adjustment the claim can be patched, and contradiction is the way to go. Start by using the definition of rational number? $\endgroup$ – Zach Stone Nov 3 '15 at 20:31
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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...

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$\pi$ 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.

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    $\begingroup$ It would appear that I answer the wrong question. I will edit further. $\endgroup$ – Michael Hardy Nov 3 '15 at 20:54
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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

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