Prove that there exists irrational numbers $x$ and $y$ such that $x + y$ is rational, without using subtraction 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?

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)
