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1) For every rational number $x$ and every irrational number $y$, $x+y$ is irrational.

2) For every rational number $x$ and every irrational number $y$, $xy$ is irrational.

3) For every rational number $x$ and every irrational number $y$, $y^x$ is irrational.

Apparently, only 1) is true. Can someone give me reasons why is 1) correct and 2) and 3) are wrong? 2) is the one that confuses me the most because numbers like $2\pi,3\pi,4\pi,...$ are in the form $xy$, yet they are irrational. Does this mean that some $xy$ numbers are irrational while some aren't? As for 3), it's easy for me since I know at least one example when it's false: $(\sqrt{2})^2=2$ (but some general explanation would be nice).

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    $\begingroup$ 2) Hint: $\,x=0\,$ is a rational number. $\endgroup$ – dxiv Jun 14 '18 at 22:03
  • $\begingroup$ Ohhhh.... yeah, I forgot about $0$. $\endgroup$ – Hanlon Jun 14 '18 at 22:04
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For $(2)$ you can take $x=0$ and $y$ to be your favorite irrational number. For $(1)$ here is a proof. Towards a contradiction suppose that $x+y$ is rational. Then since $x$ is rational, $x+y-x=y$ is rational, a contradiction.

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Look at how the proof goes: let $x$ be rational and $y$ be irrational. Suppose $x+y$ were rationals. Rationals are closed under the field operations, so $(x+y) - x = y$ would then be rational, contradiction. So $x+y$ must be irrational.

The same idea almost works for 2: suppose that $xy$ rational then $y =\frac{(xy)}{x}$ is rational.. (ooops, if $x \neq 0$ or we cannot divide), which immediately gives a counterexample $x=0$, $y = \pi$. But the proof does give

If $x \neq 0$ is rational, and $y$ is irrational, then $xy$ is irrational

(explaining your $2\pi$ example etc.)

The last example you gave is fine, many variations can be imagined, like $(\sqrt[3]{2})^3 = 2$.

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Regarding 1) you can prove it with by contradiction.

Suppose there is a rational $x$ and irrational $y$ such that $x + y$ is rational.

If our statement is true we can write: $\frac ab + y = \frac cd$ (with integers $a,b,c,d$)

multiply through by $bd$

$ad + bdy = cb$

and $bdy$ must be an integer.

But if $y$ is irrational that is not possible.

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