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Sum and product of irrationals is not necessarily irrational.

by RoRi

Prove that the sum of two irrational numbers need not be irrational and the product of two irrational numbers need not be irrational.

Proof. Let y be irrational. Then, by an argument we made in I.3.12, Exercise #7, we know that -y and y^{-1} are both irrational. However,

    \[ y + (-y) = 0 \in \mathbb{Q} \qquad \text{and} \qquad y \cdot \left( \frac{1}{y} \right) = 1 \in \mathbb{Q}. \]

Therefore, the sum and product of two irrationals need not be irrational. \qquad \blacksquare

(Note: we could also just say \sqrt{2} \cdot \sqrt{2} = 2 is not irrational, but we have not yet established that \sqrt{2} is irrational. In fact, I’m not sure we’ve even established that an irrational number actually exists yet.)

5 comments

  1. Daniel Fugisawa says:

    Sure you answer exactly what was asked. You gave a counterexample that disproves the statement. I’m very new to Mathematics, and I’m glad you can discuss the possibilities of solving these types of exercises with me.

    • Rori says:

      Yes, we can certainly do that, and then really we can get any rational number we want, as you say. I just went with the simplest examples in each case. It’s definitely good that you’re thinking about how much more we can do though.

      Of course, I’m glad to discuss any of these things. I started the blog hoping people will ask questions about the problems and solutions so that everyone can benefit by seeing the comments.

      Also, if you want to include latex in comments just type “[ latexpage ]” (without the spaces in the brackets) at the beginning of your comment. I need to include instructions for that somewhere.

  2. Daniel Fugisawa says:

    But since we proved \$ (-y) $ is still irrational we can rename it say $\ a $, and the same goes for \$ (-r) $, we can call now \$ b $. Is it possible, then?

  3. Daniel Fugisawa says:

    Exercise 7 allows us to generalize the argument for some arbitrary rational x, not only to the identity elements.

    • Rori says:

      Hi Daniel,

      I’m not 100% certain I follow what you want to generalize. You want to say if r \in \mathbb{Q} then there is some pair of irrational numbers, say y and w such that y + w = r? (And similarly, for the multiplication.) That definitely is true, in that case w = -y + r. To solve the problem, I just found the simplest example since we just need to show that there exists some pair of irrational numbers such that the sum is rational, and some pair such that the product is rational.

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