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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).

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

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    $\begingroup$ where $k$ is not a perfect square $\endgroup$ – J. W. Tanner Jan 13 at 17:08
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$-\sqrt a+b, \sqrt a+b$ are examples

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    $\begingroup$ where $a$ is not a perfect square $\endgroup$ – J. W. Tanner Jan 13 at 18:34
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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.

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$$x=\text{Golden Ratio}={1+\sqrt 5\over 2}\\y={1-\sqrt 5\over 2}$$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 $$x^2-x-1=0$$

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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 $\alpha\bar\alpha=1$, $-\alpha-\bar\alpha=a$.

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