Irrationality or Rationality of p+q given that pq=1

Take $$p=\sqrt { 2 } ,q=\frac { 1 }{ \sqrt { 2 } } \\ \\ p=2-\sqrt { 3 } ,q=2+\sqrt { 3 } \\ $$

The question then becomes, if $p$ is irrational, then is $p+\frac{1}{p}$ irrational?

In other words, you want to know if $$ \frac{p^2+1}{p}=\frac{a}{b} $$ for some integers $a$ and $b$.

Let's just try a value for $\frac{a}{b}$. Let's try $\frac{a}{b}=3$. Therefore, we are trying to solve:

$$ \frac{p^2+1}{p}=3. $$ This can turned into $$ p^2-3p+1=0 $$ and solved with the quadratic formula: In this case, $$ p=\frac{3\pm\sqrt{9-4}}{2}=\frac{3}{2}\pm\frac{\sqrt{5}}{2}, $$ which is irrational. Therefore, the answer to your question is that $p+q$ can be rational.

In general, we can find a $p$ that satisfies the conditions if $\frac{p^2+1}{p}\geq 2$. Most of the results from this are irrational. We can certainly replace $\frac{a}{b}$ with an irrational number greater than $2$ and get possibilities for $p$ and $\frac{1}{p}$ where their sum is irrational.

Hint: consider for example:

• $\;p=\sqrt{2}, q= 1 / \sqrt{2}$

• $\;p=3+2\sqrt{2}, q= 3 - 2 \sqrt{2}$

• irrational sum: let $p$ be a transcendental number (such as $\pi$ for example) and $q=1/p\,$, then $p+q$ will necessarily be transcendental, thus irrational; the same goes for $p$ being the sum between a transcendental number and an algebraic irrational (such as $\pi+\sqrt{2}$ for example);

• rational sum: let $p,q$ be the roots of the equation $x^2 - n \,x + 1 = 0\,$ for integer $n \gt 2$. Both $p,q$ are irrational by the rational root theorem, and $pq=1\,$, $p+q=n\,$ by Vieta's relations.

Let consider $S=p+q$, $P=pq=1$ and $p>q$

The roots of the polynom $P[X]=X²-SX+P=X²-SX+1$ are $p$ and $q$.

$discriminant = \delta = S²-4 > 0$ given that $P[X]$ has two roots and they are irrational numbers. Then, $ p = \frac{S + \sqrt{S²-4}}{2}$ and $ q = \frac{S - \sqrt{S²-4}}{2}$

As soon as $|S| > 2$ and $S²-4$ is not a perfect square, you can choose $S$ as you want, rational or irrational. $p$ and $q$ will be irrational and the sum will be what you have chosen.