This is a homework problem for my Real Analysis course and I am having trouble getting started in the right direction. I understand the definition of the set of rational numbers and how $\mathbb{R}-\mathbb{Q}$ is the set of irrational numbers, but I am having trouble making the leap to where q+x is an element of the set of irrational numbers. Is there something that I'm missing?

I started with $a,b∈Q$ where either $a=0$ or $b=0$ which then gives us either $a+b=a$ or $a+b=b$. We know from this that $a+b∈Q$. It's the next part I'm struggling with.


You know $q$ is rational, so you can write $q = \dfrac{a}{b}$ for some integers $a,b$ where $b \neq 0$.

You also know that $x$ is irrational, so you cannot write $x$ as the ratio of two integers.

You need to show that $q+x$ is irrational. Suppose $q+x$ is rational, for the sake of contradiction.

Then, you can write $q+x = \dfrac{c}{d}$ for some integers $c,d$ where $d \neq 0$.

What does this tell you about $x$?

  • $\begingroup$ That x would be equal to zero? $\endgroup$ – aneorddot Nov 7 '14 at 4:31
  • $\begingroup$ No, $x$ does not necessarily have to be $0$. Once you've written $q = \dfrac{a}{b}$ and $q+x = \dfrac{c}{d}$, can you solve for $x$ in terms of the integers $a,b,c,d$? $\endgroup$ – JimmyK4542 Nov 7 '14 at 5:17
  • $\begingroup$ Yes, you're right. Now I'm getting it. I'll keep working on it now that it makes a bit more sense. Thank you for your help. $\endgroup$ – aneorddot Nov 9 '14 at 3:37

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