My homework has this problem:

Prove that there exist irrational numbers $x$ and $y$ such that $x + y$ is rational.

There is an easy solution that I found on mathbitsnotebook.com:

\begin{align} \left(2+6\sqrt{7}\right) + \left(-6\sqrt{7}\right) = 2.\end{align}

Can we prove the same without using subtraction?


1 Answer 1


To give one answer, let $\varphi=\frac{\sqrt{5}+1}2$. Notice that both $\frac{1}{\varphi^2}$ and $\frac{1}{\varphi}$ are irrational, but their sum is $1$. That is $$\frac{1}{\varphi^2}+\frac{1}{\varphi}=1.$$ This stems immediately from the fact that $\varphi$ satisfies (and very often stems from) the equation: $$1+\varphi=\varphi^2$$ and if we divide through by $\varphi^2$, we get the first equation.

Strictly speaking, however, for any irrational $x$ and $y$ such that $x+y$ is rational, there exists a rational $a_1,a_2$ and irrational $b$ such that $x=a_1+b$ and $y=a_2-b$ meaning that we could write, for rational $p$: $$x+y=p$$ $$(a_1+b)+(a_2-b)=p$$ where the irrational terms obviously cancel. So, in some sense, we can't avoid subtraction here - no matter what example we choose, we can always express it in a way that makes the cancellation obvious.

(Though this is probably not helpful to the original poster, for others, it's worth noting that the last paragraph in a sense expresses that $\mathbb R$ is a vector space over $\mathbb Q$ - we can consider any irrational $x$ as being on a plane, including sums of the form $px+q$ for rational $p,q$. If we take the quotient space $\mathbb R/\mathbb Q$, we are left with a notion of the "irrational" component of $x$ - and we must undo it to get back to the rationals)

  • $\begingroup$ Thanks for your fast answer. I will study it. $\endgroup$
    – Beginner
    Feb 13, 2015 at 1:13
  • $\begingroup$ Great solution! Quick and simple. +1 $\endgroup$
    – bjd2385
    Feb 13, 2015 at 1:18
  • $\begingroup$ But $\dfrac1\phi=\phi^{\color{red}-1}~$ ;-$)$ $\endgroup$
    – Lucian
    Feb 13, 2015 at 1:36
  • $\begingroup$ @Lucian Don't worry, we can write it as $\varphi^{i\cdot i}$ to avoid that issue! :) $\endgroup$ Feb 13, 2015 at 1:38
  • $\begingroup$ @Lucian What would be the problem with ϕ^(−1)? Thanks! $\endgroup$
    – Beginner
    Feb 13, 2015 at 2:25

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