Can any two irrational numbers NOT of the form (m+A) and (n-A) be added to produce a rational number? $m$ and $n$ being rational numbers, A being an irrational number.
I was wondering if two irrational numbers when added always yield an irrational number. All the counter-examples I could find were of the form $(m+A) + (n-A) = m+n$.
Are there any counter-examples NOT of this form?
 A: Since $\sin(x)^2+\cos(x)^2=1$ for all $x \in \mathbb R$ and the $\sin$ and $\cos$ functions are irrational most of the time. Therefore, two irrational numbers added together can yield a rational number.
example: $x=\frac{\pi}{8}$
A: Here's another way to look at it. Let $r$ and $s$ be two irrational numbers that add up to some rational number $q$.  Let $m = (r+s)/2$ be the average of those two irrational numbers.  Note that $m = q/2$ so $m$ is rational.  Now if you define $A = s - m$ then we have the following:


*

*$r = m - A$

*$s = m + A$

*$m$ is rational and $A$ is irrational


This shows that any pair of irrational numbers that have a rational sum can be written in the form given in the OP -- with the bonus that the rational numbers $n$ and $m$ can be chosen to be equal!
A: Let $i_1$, $i_2$ be irrational, and $i_1 + i_2 = a$; with $a$ rational (the assumption).
Then
$$
i_1 = \frac{a}{2} - \frac{b}{2}\\
i_2 = \frac{a}{2} + \frac{b}{2}
$$
where $$b \equiv i_2-i_1$$
Case 1: $b$ is irrational.
If $b$ is irrational, then $i_1$ and $i_2$ are of the form that we are trying to avoid.  This is because if $a$ is rational, so is $a/2$; if $b$ is irrational, so is $b/2$.
Case 2: $b$ is rational.
Substituting $a-i_2$ for $i_1$ into the definition of $b$, we get
$$i_2 = \frac{a+b}{2}$$
But, since $a$ and $b$ are rational in Case 2, then their sum must be rational.  This would mean $i_2$ is rational, which violates our original assumption.  (A similar line leads to $i_1$ having to be rational in Case 2, as well.)  This is a contradiction, so we learned $b$ must be irrational.
So, putting it together, if two irrationals sum to a rational, then they are of the form we are trying to avoid.  (So, the answer to the question as asked is "no.")
A: Let $a,b$ be irrational numbers such that
$$
r=a+b\text{ is rational.}
$$
Then
$b=r-a$, $a=0+a$, and $0$ is rational.
