Can any two irrational numbers NOT of the form (m+A) and (n-A) be added to produce a rational number?

Question

$m$ 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?

Answer 1

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.

Answer 2

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: $x=\frac{\pi}{8}$

Answer 3

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!

Answer 4

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.")