Every real number is a sum of irrational numbers

Let $A = \mathbb{R}\setminus\mathbb{Q}$. Then it can be shown that $A + A = \mathbb{R}$, for example by using the fact that $A$ is $G_{\delta}$. Let $q\in\mathbb{Q}$. This means that $q = r_1+r_2$ where $r_1,r_2$ are irrational numbers. But this is not too surprising, as every rational can be written $q = \left(\frac{q}{2} + r_1\right) + \left(\frac{q}{2} - r_1\right)$. The question is, is this the only way? More precisely, if $q = r_1 + r_2$ is rational and $r_1,r_2$ are irrational, does this mean that there exists $r\in\mathbb{R}\setminus\mathbb{Q}$ and $q_1,q_2\in\mathbb{Q}$ such that $q_1+q_2 = q$, and $r_1 = q_1 + r, r_2 = q_2 - r$?

• Yes, take $q_1=0,q_2=q$ and $r=r_1$. – Dietrich Burde Dec 18 '17 at 15:25

Yes, the set of solutions to your equations are:

\begin{align} q_2 = q - q_1 \\ r = r_1 - q_1 \end{align}

where $q_1$ is a free variable allowed to be any rational number.

You are engaging in a slippery terrain.

When you write $r_1=q_1+r$, do you imply that $r_1$ is the sum of a rational $q_1$ and a "rational free" irrational $r$ ? We shall see that this concept is not so simple.

In fact since $(\mathbb Q,+)$ being a normal subgroup of $(\mathbb R,+)$ we can define the quotient space $\mathbb R/\mathbb Q$ of the reals modulo the rationals.

And thanks to the axiom of choice we can construct a Vitali set $V$ such that $V\subset[0,1]$ and $V$ contains exactly one representative for each class of $\mathbb R/\mathbb Q$.

This may seems a convenient construction but this set belongs to a freak show, it is not measurable and we have no way of knowing what's exactly inside.

Let the representative of a real $x$ be noted $[x]$.

Obviously for any rational $q$ then $0$ seems a suitable choice for $[q]$.

But what would be $[\sqrt{2}]$ ? Would this be $\sqrt{2}-1\in[0,1]$ or maybe $\sqrt{2}-\frac 34$ which is also an element of $[0,1]$.

We could say then that since $\mathbb Q$ is countable, let just have an ordering $\phi$ based on the bijection with $\mathbb N$ and select $[x]=x+\phi(\min\{n\mid x+\phi(n)\in[0,1]\})$.

Unfortunately $[\sqrt{2}]$ and $[\sqrt{2}+17/113]$ would not have the same representative via this process although they belong to the same class and this is quite annoying. This is why the axiom of choice is required to pick a suitable representative for every real.

Now let's come back to your question:

First let's notice that for $(v_1,v_2)\in V^2$ then $v_1+v_2$ is a rational means that $[v_1]=[-v_2]$ ($v_1+v_2=q\iff v_1=q-v_2=-v_2\pmod{\mathbb Q})$

Thus if you have two irrationnals $r_1+r_2=q$ then $\begin{cases} r_1=[r_1]+q_1\\r_2=[r_2]+q_2\end{cases}$ and $[r_1]+[r_2]=q-q_1-q_2$

So $[r_1]=[-r_2]$, let's call it $v$.

Consequently we have $\begin{cases}r_1=v+q_1\\r_2=(q_3-v)+q_2\end{cases}\iff\begin{cases}r_1=q_1+v\\r_2=(q_3+q_2)-v\end{cases}$ with $q_1+q_2+q_3=q$

So indeed the construction you have envisaged with $r_1,r_2$ being the sum of rationals and $\pm$ the same irrational number makes sense.

However, as we have seen there is no way in practice to exhibit $v\in V$. All we can access are elements in this class which are already $q_i+v$.

You should really see $r_1+r_2=q$ as $[r_1]=[-r_2]$ which is just $r_2=q-r_1$ and nothing more really that can be said about the "inners" of $r_1$ or $r_2$. They just sum up to a rational...