# Find two irrational numbers $x,y$ such that $x+y$ and $xy$ are both rational.

I know how to satisfy one of the statements but never both together.

$$(a+b)*(a-b)=a^2-b^2$$ so taking $$a=\sqrt k_1$$ and $$b=\sqrt k_2$$, $$k_1,k_2\in\mathbb{Q}$$ such that $$a,b \notin\mathbb{Q}$$ would suffice for the second one. Also, for every $$x\in\mathbb{R} - \mathbb{Q}$$, taking $$y=x^{-1}$$ would also work for the product.

For the sum one could do something boring like given $$x\in\mathbb{R}-\mathbb{Q}$$, take $$-x\in\mathbb{R}-\mathbb{Q}$$ and $$x-x=0\in\mathbb{R}$$.

Ideally only "easy-to-define functions" (like square root, sum, subtraction...) should be used since stuff like $$log,e,cos$$... are not formally defined until later stages of my real analysis book (this is a question from the real numbers chapter).

What about $$a=\sqrt k$$, $$b=-\sqrt k$$?

• where $k$ is not a perfect square Jan 13, 2019 at 17:08

$$-\sqrt a+b, \sqrt a+b$$ are examples

• where $a$ is not a perfect square Jan 13, 2019 at 18:34

It is not difficult to show what is the general solution here.

If $$a+b=2r$$ and $$ab=q$$ we find (eg by substitution, or vieta) that $$a$$ and $$b$$ are the roots of the quadratic equation $$x^2-2rx+q=0$$ which has the roots $$x=r\pm \sqrt{r^2-q}$$giving the two values of $$a$$ and $$b$$. If $$r^2-q$$ is not a rational square then the values of $$a$$ and $$b$$ will be irrational.

$$x=\text{Golden Ratio}={1+\sqrt 5\over 2}\\y={1-\sqrt 5\over 2}$$More generally, take the roots of $$x^2+ax+b=0$$ when $$a,b$$ are rational and the equation has two irrational roots. For example, my $$x,y$$ are roots of $$x^2-x-1=0$$

Take any polynomial $$f(x)=x^2+ax+1$$ where $$a\in\Bbb Z$$ and $$a^2>4$$. Let $$\alpha\in\Bbb R$$ such that $$f(\alpha)=0$$. By the rational root test $$\alpha=1$$, but this is impossible since $$f(1)=a+2\neq 0$$. As the roots comes in pairs there is another root $$\bar\alpha$$, and $$\alpha\bar\alpha=1$$, $$-\alpha-\bar\alpha=a$$.