The sum of multiple irrational numbers can be rational, even when they're not conjugates. Is this normal?

Question

From my observation, a sum of a series of irrational numbers can be rational, even when none of them "cancels out" the irrational part of another. Consider the following sum:

$$\sum_{n=1}^9 \log_{10}(1+\frac{1}{n}) = 1$$

It is easy to proof that all terms in the sum are irrational. However, according to law of logarithms:

$$\log(a)+\log(b)=\log(ab)$$

Therefore, the expression becomes:

$$\log_{10}(\frac{2}{1}*\frac{3}{2}*\frac{4}{3}...\frac{9}{8}*\frac{10}{9})$$

which simply to be $\log_{10}(10) = 1$.

This appears quite strange, as two non-conjugate (not in the form of $a+b$ and $c-b$ for $a,c\in {\bf Q}$ and $b\notin {\bf Q}$) irrational numbers never sum up to rational numbers (correct me if I'm wrong), so the sum of the series should be irrational.

Is there a specific reason why this happens, or is there any proof that a sum of non-conjugate irrational numbers can be rational? Can this only happen with logarithms?

Answer 1

For another example, consider $a = \sqrt{2} - \sqrt{3}$, $b = \sqrt{3} - \sqrt{5}$, $c = \sqrt{5} - \sqrt{2}$, whose sum is the rational number $0$, although $a$, $b$, $c$, $a+b$, $a+c$ and $b+c$ are all irrational.

You could replace $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{5}$ by any three irrationals $u$, $v$, $w$ such that $1, u, v, w$ are linearly independent over the rationals.

Answer 2

$1=0.999\dots \\= 0.18118881111\dots + 0.81881118888\dots \\=0.13113331111\dots +0.05005550000\dots + 0.818881118888\dots$

and now we have three irrational numbers without the sum or difference of any two being rational. Note that all I needed was a non-repeating pattern since I can always pair off digits that sum to $9$ then decompose whichever one is higher than $1$ into smaller pieces.

Answer 3

Take $u_1$ and $u_2$ to be two irrational numbers whose sum is not rational.

Look at \begin{align} a &= 3 - u_1 \\ b &= 6 - u_2 \\ c &= 5 + u_1 \\ d &= 11 + u_2 \end{align} Your theorem tells us that $a+b$ isn't rational. And then we know by computing the sum, which is $14 - u_2$, that $(a+b) + c$ isn't rational. And yet $a + b + c + d$ is rational, and it's hardly surprising when you compute the sum, which is $25$.

The problem with the little theorem you cited is that it only applies to two items at a time, and those items need to have a specific property. WHen you sum up more and more terms of a sum with more than 2 numbers, that property no longer holds, and whammo -- we can get a rational sum.

If you'd like more examples, just replace my $3, 6, 5,$ and $11$ with any other rational numbers. That gives you an infinite set of examples, and suddenly your phenomenon doesn't seem so weird.

I challenge you think of a set of three irrationals that add up to a rational, perhaps motivated by my silly example.

Answer 4

tl;dr.

For a general example, consider

$$ t_0 = e^1 - e^2 \\ t_1 = e^2 - e^3 \\ t_2 = e^3 - e^4 \\ t_3 = e^4 - e^1. $$

Since $e$ is transcendental, the only way to get a rational sum is to sum all terms; no partial sum will work. This works for any $n$.

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This is a general mathematical phenomenon that has nothing to do with irrational numbers.

If you have access to pairs $(a_k, b_k)$ that - in some sense - cancel themselves out (like $q$, $-q$ for $q$ irrational), you can always build items that are all non-reducible (irrational in your case), but so that their sum is reducible (rational).

For an example with $n+1$ terms, take $t_k = a_k + b_{n+1}$ ($0\leq k <n$) and $t_n = a_n + b_0$. Once you sum all $t_k$, everything cancels.

You just have to make sure that the $a_n$ don't interfere with each other. You can, for example, consider powers $a_n = \phi^{n+1}$ for a transcendental number $\phi$. Euler's number $e$ would work.

This even works in non-commutative systems.

It's basically like matter/antimatter particles that don't interfere with each other - you seperate them and they only annihilate everything if you mix them all together.

Answer 5

Given that there infinitely more irrational numbers than there are rational numbers, there are more possible sums of irrationals that come to any number than there are possible sums of rationals for that same result.

Answer 6

Choose five numbers $p,q,r,s,t$ uniformly at random from $[0,1]$. Then the five numbers $$ \frac{p}{p+q+r+s+t}, \frac{q}{p+q+r+s+t}, \frac{r}{p+q+r+s+t}, \frac{s}{p+q+r+s+t}, \frac{t}{p+q+r+s+t} $$ have sum $1$, but (with probability one) there is no algebraic relation among any four of them.

Answer 7

Suppose you have $n$ irrational numbers that sum to S. Either S is rational or irrational. If it's rational, then there's an example of irrational numbers that add up to a rational number. If it's irrational, we can take any rational number Q, subtract S from it, and we will then end up with a number that when appended to our original list, results in a list that adds up to the rational number Q.

That is, if $x_1+x_2+...x_n =S$, then $ x_1+x_2+...x_n+(Q-S)=Q$. If S is irrational, then Q-S is also irrational.

Another way to get an example is what's called a "telescoping series", which your example can be written as.

$\log(1+\frac 1n) = \log(\frac{n+1}{n})=\log(n+1)-\log(n)$

When we write it like this, we get

$[\log(2)-\log(1)]+[\log(3)-\log(2)]+[\log(4)-\log(3)]+...$

The $\log(2)$ from the first term cancels with the $-\log(2)$ from the second term, etc. In the end, the only terms that don't cancel are the $-\log(1)$ from the first term and the $\log(10)$ from the last term. So all the irrational parts cancel, leaving rational ones.