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Given an integer $n$ and an irrational $r$, $n>r$, $n-r$ is irrational but $r + (n - r)$, the sum of two positive irrationals, is an integer. Is that the only way that two irrationals can sum to an integer?

What if the question is rephrased using rationals instead of integers? Is the only way two irrationals can sum to a rational is by using the form $r + (a/b - r)$?

Can $r_{1} + (a/b - r_{2})$ ever be a rational? An integer?

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If $x$ and $y$ are irrationals such that $x+y$ is an integer (say, $n$), then $y= n-x$. So yes, that's the only way in which you can write $n$ as the sum of irrationals: pick one irrational and the second is just $n$ minus the first. – Srivatsan Nov 18 '11 at 20:04
We don't even know if $e+\pi$ is rational or not... – J. M. Nov 18 '11 at 20:05
What on earth do you mean by "the only way two irrationals can sum to a rational"? Yes, $a+b=c$ if and only if $b$ can be written as $c-a$. That has nothing to do with being rational or irrational. – Henning Makholm Nov 18 '11 at 20:34

Examples like $(1 + \sqrt{2})$ + $(- \sqrt{2})$ indeed sum to a rational number, but relationship between the two quantities being summed need not be so obvious at first glance.

Here is a result with which you are probably familiar. For any real number $x$, $$ \cos^2(x) + \sin^2(x) = 1. $$ Note that, for almost all $x$, $\cos^2(x)$ and $\sin^2(x)$ are transcendental, and so we have infinitely-many examples of transcendental numbers summing to an integer.

Certainly, some algebra gives $$ \cos^2(x) + (1 - \cos^2(x)) = 1, $$ but this can be done with any triple of numbers. The purpose of my example is merely to show that, on the face of it, the relationship between the irrational numbers being summed need not be as readily apprehended in quite the way that one immediately sees that $\sqrt{2}$ and $-\sqrt{2}$ cancel each other.

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You are correct. – Charles Nov 18 '11 at 21:26
But then the two numbers you are summing are $\sin^2(x)$ and $1-\sin^2(x)$ ..... – N. S. Nov 19 '11 at 2:58
@N. S. Only because you're "familiar" with the Pythagorean identity. What are the odds that adding up the squares of two series that evaluate to transcendental numbers give a rational, nay, integer result? – J. M. Nov 19 '11 at 3:24
@N.S. Strictly speaking, what you say is true. I think my example addresses the spirit of the question, however, which I took to be "Can two irrationals sum to a rational, even if they don't appear to be related in quite the obvious way that $\sqrt{2}$ and $-\sqrt{2}$ are?". – Austin Mohr Nov 19 '11 at 3:32
@QED I have explained my interpretation more clearly in the answer. – Austin Mohr Nov 19 '11 at 5:29

Yes, it is the only way two positive irrationals can sum to an integer.

Say you have two positive irrationals, $r$ and $s$, and their sum $r+s=n$ is an integer. Then since $r+s=n$, we get $s=n-r$, so $r+s$ is $r+(n-r)$.

And if $n$ is assumed rational, but not an integer, the same argument applies.

Can $r_1+(a/b-r_2)$ be rational? I assume you mean $r_1$ and $r_2$ are irrational and positive, and $a/b$ is rational. Then you could have, for example, $r_1=1+r_2$, so that $r_1$ and $r_2$ are not equal, and $r_1+(a/b-r_2)= a/b-1$, which is rational. But this is still a case of $r+(n-r)$, where $n$ is rational and $r$ is not, namely $n$ is the rational number $a/b-1$.

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I think you meant to say that "since $r+s=n$, we get $s=n-r$, so $r+s$ is $r+(n-r)$." – daOnlyBG Oct 27 '14 at 17:13

See if this makes sense. We'll call your sum x. So our equation is

$r_1+(\frac ab-r_2)=x$

Let's subtract $\frac ab$ from both sides.

$r_1-r_2=x-\frac ab$

If $x$ and $\frac ab$ are rational (I assume you want a and b to be integers), the right side of this equation is rational. So $r_1-r_2$ must be rational as well. Is this what you're looking for?

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The question is about linear combinations of real numbers, where some of the numbers are irrational, and can be interpreted as asking whether there are non-trivial linear relations (where trivial ones are things like $\pi + (5-\pi) = 5$, that are general algebraic properties of all numbers).

In general the expectation is that that given $k$ different irrational numbers, there will not be any non-trivial relations between them unless the numbers are constructed from fewer than $k$ "truly different" sources of irrationality, with some redundancy added. Relations can be interpreted as linear or polynomial (algebraic) relations with rational coefficients. There could, hypothetically, be suprises, such as nonzero integers $a,b,c$ (if any do exist) for which $a + b\pi + ce=0$, and there are a few known types of surprising relations, such as $r_1 = (\sum 1/n^2)$ being related to $r_2 = \pi^2$ by $6r_1 = r_2$. Nevertheless, in concrete situations there is usually at least a specific conjecture as to whether or not a relation with rational coefficients can exist between the numbers. If there is no apparent reason why a relationship should exist, it usually does not.

The formal description of these ideas is in terms of linear independence (over the rational numbers) and transcendence degree (also relative to the rational numbers) of a finite set of real numbers. The case where all the real numbers satisfy algebraic equations is covered by algebraic number theory and is well understood, as is the case where one number is transcendental. The case where several essentially different transcendental numbers are involved is the subject of transcendence theory, where there are many difficult conjectures and fewer theorems defining what is expected in most cases. A typical irrational-looking expression like $\pi + e^e$ has "probability 1" of being irrational but no known proof that it is irrational. A set of several irrationals such as $\pi$, $e$, $\zeta(5)$ will usually be as transcendental as possible (maximum transcendence degree) but the proof is out of reach.

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