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$m$ 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?

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  • $\begingroup$ $m$ and $n$ can be rational. $\endgroup$ – Jack Yoon Sep 19 '14 at 12:10
  • $\begingroup$ Thank you. I have edited the question to reflect same. $\endgroup$ – Akash C Sep 19 '14 at 12:12
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    $\begingroup$ Actually all the counter-examples you've seen are of the simpler form $(A,n-A)$ where $A$ is irrational, and $n$ is rational (that is, one can take $m=0$). They must be, because if two irrational numbers $A,B$ add up to a rational number $n$, then $B=n-A$. $\endgroup$ – Marc van Leeuwen Sep 19 '14 at 14:21
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    $\begingroup$ Any pair of irrationals whose sum is rational can be turned into something of the form you are trying to avoid. (Of course, there are usually more interesting interpretations of the two numbers than through this form) $\endgroup$ – Christian Chapman Sep 19 '14 at 18:26
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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.

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  • $\begingroup$ But $b=r-a$ falls into the class of $n-A$ (i.e., a rational minus an irrational) - the question is about irrational numbers that are NOT of this form. $\endgroup$ – Akash C Sep 19 '14 at 12:24
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    $\begingroup$ @AkashC, the author is hinting that any counter example can be reduced to this form. $\endgroup$ – hrkrshnn Sep 19 '14 at 12:25
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    $\begingroup$ In short, the answer to your question is no. All counterexamples are of that form. $\endgroup$ – Julián Aguirre Sep 19 '14 at 12:27
  • $\begingroup$ @Moron Which must mean that the answer to the original question is no? $\endgroup$ – Akash C Sep 19 '14 at 12:27
  • $\begingroup$ @Julián Aguirre - thank you :) $\endgroup$ – Akash C Sep 19 '14 at 12:27
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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: $x=\frac{\pi}{8}$

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    $\begingroup$ But the OP's query is different. $\endgroup$ – hrkrshnn Sep 19 '14 at 12:14
  • $\begingroup$ That's overkill. Just take for instance $\sqrt 2$ and $1-\sqrt 2$. $\endgroup$ – TonyK Sep 19 '14 at 12:14
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    $\begingroup$ @TonyK but that's the specific counterexample the o/p already gave $\endgroup$ – Tom Tanner Sep 19 '14 at 13:25
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    $\begingroup$ sqrt(2) and 1-sqrt(2) is boring, there is clearly sqrt(2) on both sides. It looks as if it is explicitly designed to produce 1 on summation. The sin(1)*sin(1) and cos(1)*cos(1) pair is more interesting, as if you don't know the theorem, you can be surprised getting 1. $\endgroup$ – Vi0 Sep 19 '14 at 18:37
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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!

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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.")

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  • $\begingroup$ Great proof. Thanks! $\endgroup$ – JWT Aug 27 '19 at 15:10

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