# The sum (or difference) of two irrational numbers

So far I that for any irrational number without a real part (that $-n=\overline{n}$) plus/minus any irrational number with the same restrictions equals another irrational number. However, I want to know if there is a concrete rule that states what the properties of a sum/difference made up of irrational numbers. My intuition says that two irrational numbers will only add up to a rational number if their irrational parts are opposites (for instance $\sqrt{2}$ and $-\sqrt{2}$), but I'm not really sure.

Thank you!

MAJOR EDIT: I'm sorry, I forgot to mention something very important. The irrational numbers in this problem are all in the form $a+\sqrt{b}$ where a and b are both positive integers and $\sqrt{b}$ is irrational.

• How do you define the irrational part of a real number? – JimmyK4542 Jul 27 '14 at 22:30
• Isn't any nonzero complex number with no real part irrational? The rationals are a subset of the reals. – Samuel Yusim Jul 27 '14 at 22:30
• Hmmm... I'm not very good at this type of math, but here's what I was trying to say: If we have an irrational number (like $2+\sqrt{2}$, then in my (possibly convoluted and incorrect reasoning), the 2 would be the rational part and the $\sqrt{2}$ would be the irrational part. – hyperdo Jul 27 '14 at 22:32
• But why wouldn't you have $3$ be the rational part, and $\sqrt 2 - 1$ the irrational part? Or $4$ and $\sqrt 2 - 2$, or so on? – user61527 Jul 27 '14 at 22:33
• With your new restriction, your intuition is more or less correct. A more concrete way to say this is the observation that the set of square-free radicals $\{1,\ \sqrt{2},\ \sqrt{3},\ \cdots \}$ is linearly independent over $\mathbb{Q}$. – EuYu Jul 27 '14 at 22:56