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As stated, I tried to prove the following:

The theorem seems to be very incompletely phrased, since one can obtain non integer sums of the form

$$\frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{7} + \frac{1}{9}$$

or

$$\frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{{18}} + \frac{1}{{20}}$$

so further detail is needed. Maybe this should be closed as no longer relevant until I come up with a better phrasing and I consider more initial conditions. The big question would be

Given the set $S$ of $n$ integers

$$S=\{x_1,x_2,\dots,x_n\}$$ what are sufficient conditions on $x_1,x_2,\dots,x_n$ so that $$\eta = \sum_{k \leq n }x_k^{-1}$$ is not an integer?

Though I don't know if this is an important/relevant question to be asking.

THEOREM If $x_1,\dots,x_n $ are pairwise coprime, $x_i\neq 1$, let

$$\mu =\sum_{k=1}^n \frac 1 x_k $$

Then $\mu$ can't be an integer.

PROOF By induction on $n$. Asume the theorem is true for $2, \dots, n-1$. I'll analize the case $k=n$.

$(1)$ It is true for $n=2$. If $$(x_1,x_2)=1 \Rightarrow (x_1 x_2,x_1+x_2)=1$$

The proof is simple. We have that $(x_1,x_2)=1$. Let $d \mid x_1+x_2 , d \mid x_1x_2$. Then

$$d\mid x_1(x_1+x_2)-x_1x_2 \Rightarrow d\mid x_1^2$$

$$d\mid x_2(x_1+x_2)-x_1x_2 \Rightarrow d\mid x_2^2$$

So $$d \mid (x_1^2,x_2^2)=(x_1,x_2)=1 \Rightarrow d=1$$

This means $$\frac{1}{x_1}+\frac{1}{x_2}=\frac{x_1+x_2}{x_1 x_2}=\phi$$

is not an integer.

$(2)$ Let

$$\mu = \frac{1}{x_1}+ \frac{1}{x_2}+\cdots+ \frac{1}{x_{n-1}}+ \frac{1}{x_n}$$

Then

$$x_n \mu-1 = x_n\left(\frac{1}{x_1}+ \frac{1}{x_2}+\cdots+ \frac{1}{x_{n-1}}\right) =x_n \omega$$

By hypothesis, $(x_1,\dots,x_{n-1})=1$ so $\omega$ is not an integer. Thus, if $x_n \mu-1$ were an integer, it must be the case:

$$ x_n\left(\frac{1}{x_1}+ \frac{1}{x_2}+\cdots+ \frac{1}{x_{n-1}}\right) =k \text{ ; } k \text{ an integer }$$

$$ x_n \frac{\tau}{x_1 x_2 \cdots x_{n-1}} =k \text{ ; } k \text{ an integer }$$

$\tau$ is the numerator obtained upon taking a common denominator.

But since $\omega$ is not an integer, then it must be the case

$$x_1 x_2 \cdots x_{n-1} \mid x_n$$

which is imposible. Then $x_n \mu -1$ is not an integer. But since $x_n$ and $1$ are, this means $\mu$ isn't an integer, this is,

$$\mu =\sum_{k=1}^n \frac 1 x_k $$

is not an integer. $\blacktriangle$

NOTE The hypothesis that $x_k \neq 1$ is necessary to avoid sums like

$$\frac{1}{1}+\overbrace{\frac{1}{n}+\cdots +\frac{1}{n}}^{n }=1+n\frac{1}{n}=2$$

however, if $(x_1,\dots,x_n)=1$, the sum

$$\nu =\sum_{k=1}^n \frac 1 x_k +1 $$

will clearly not be an integer.

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1  
The notation $(x_1,\dots, x_k)=1$ does not mean that the numbers are pairwise relatively prime. Presumably you want pairwise relatively prime, because of $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$ and many others. –  André Nicolas Jun 13 '12 at 20:20
    
How did you get $x_1\cdots x_{n-1}\mid x_n$? –  anon Jun 13 '12 at 20:25
    
@AndréNicolas Sure. I'll make that clear. It is good you understood it, though. –  Pedro Tamaroff Jun 13 '12 at 20:25
    
In the step 1. of induction you go by reduction to contradiction. Let $p$ prime which $p|x_1x_2$ and $p|(x_1+x_2)$. Then $p$ divides to $x_1$ or $x_2$. Put $p|x_1$. As $p|(x_1+x_2)$ too, $p|x_2$. But then $p|(x_1,x_2)=1$ (Contradiction). –  H. Kabayakawa Jun 13 '12 at 20:26
    
@anon $\tau$ can't be divided by $x_1 \cdots x_{n-1}$ by hypothesis, so for $k$ to be an integer, one needs $x_1 \cdots x_{n-1} \mid x_n$. –  Pedro Tamaroff Jun 13 '12 at 20:26

4 Answers 4

up vote 3 down vote accepted

Suppose that $(a_i,b_i)=1$ for $1\le i\le n$, and that $(b_i,b_j)=1$ for $1\le i< j\le n$. Then the denominator, in lowest terms, of $$ \sum_{i=1}^n\frac{a_i}{b_i}\tag{1} $$ is $$ \prod_{i=1}^nb_i\tag{2} $$ This follows by induction from the case $n=2$, and that is true because $$ (a_1b_2+a_2b_1,b_1b_2)=1\tag{3} $$ Suppose not, and there is some prime $p$ that divides both $b_1b_2$ and $a_1b_2+a_2b_1$. If $p\,|\,b_1$, then $p\,|\,a_1b_2$, but since $(a_1,b_1)=1$ and $(b_1,b_2)=1$, $p$ can divide neither $a_1$ nor $b_2$. If $p\,|\,b_2$, then $p\,|\,a_2b_1$, but since $(a_2,b_2)=1$ and $(b_1,b_2)=1$, $p$ can divide neither $a_2$ nor $b_1$. Therefore, $(3)$ must be true.

Therefore, since harmonic sums with pairwise coprime denominators fall under this case, no harmonic sum with pairwise coprime denominators (other than the singleton sum $1$) can be an integer.

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Edit: The comment about $(x_1,\dots,x_k)$ refers to an earlier iteration of the question.

The notation $(x_1,\dots, x_k)=1$ does not mean that the numbers are pairwise relatively prime. Presumably you want pairwise relatively prime, because of $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$ and many others.

If the $x_i$ are pairwise relatively prime, and greater than $1$, there is an easy non-induction proof. Suppose to the contrary that the sum of the reciprocals is an integer. Let $M$ be the product of all the $x_i$ except one, say $x_n$. Multiply our sum of reciprocals by $M$. We get an integer. But that is impossible, since $x_n$ does not divide $M$, and all the other terms $\frac{M}{x_i}$ are integers.

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Wouldn't the proof work for, for example, $$\frac{1}{2}+\frac 1 3 +\frac 1 4 +\frac 1 5$$? –  Pedro Tamaroff Jun 13 '12 at 20:30
    
I mostly want to use this to prove the $n$th harmonic number is not an integer. Clearly some of the numbers in $1+\frac 1 2 +\frac 1 3 +\cdots +\frac 1n $ are not coprime, but in a whole, they are. –  Pedro Tamaroff Jun 13 '12 at 20:32
    
The idea of the proof would work for many non-pairwise relatively prime situations. We just have to be more careful about what to multiply by. –  André Nicolas Jun 13 '12 at 20:32
1  
For the harmonic number, the standard argument is to consider the largest power of $2$ less than or equal to $n$. –  André Nicolas Jun 13 '12 at 20:33
    
I'm looking to polish this proof. Maybe word it better, be more precise, that's what I need. =) –  Pedro Tamaroff Jun 13 '12 at 20:34

As I noted in the comments, even if $x_n\frac{\tau}{x_1\cdots x_{n-1}}$ is an integer and $\frac{\tau}{x_1\cdots x_{n-1}}$ is not, this is not sufficient to establish that $x_1\cdots x_{n-1}\mid x_n$. However we do not need this divisibility to obtain a contradiction.

Here's how I would salvage your argument. Again we have the induction hypothesis. Then:

  • If $x_n\mu $ is not an integer, then neither is $\mu$ and we are done.
  • If $x_n\mu$ is an integer, then writing $\tau=e_{n-2}(x_1,\cdots,x_{n-1})$ we have $$x_n\frac{\tau}{x_1\cdots x_{n-1}}\in \Bbb Z,~~\frac{\tau}{x_1\cdots x_{n-1}}\not\in\Bbb Z ~\implies~ (x_n,x_1\cdots x_{n-1})>1$$ (justify this!) which contradicts pairwise coprimality (justify!), hence $x_n\mu$ is not an integer.
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$\rm b_i\:$ pair coprime, $\rm\displaystyle \: \sum_i \dfrac{a_i}{b_i} = n\in\mathbb Z\:\Rightarrow\: mod\ b_j\!:\ a_j = b_j\left[n- \sum_{i\,\ne\, j} \dfrac{a_i}{b_i}\right]\!\equiv 0\:\Rightarrow\,\dfrac{a_j}{b_j}\in \mathbb Z,\, \forall\, j$

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@Peter Be sure to understand why those fractions exist mod $\rm\:b_{\,j},\:$ viz. $\rm\:1/b_{\,i}\:$ exists by $\rm\:(b_{\,i},b_{\,j})=1.\quad$ –  Bill Dubuque Jun 17 '12 at 1:01
1  
@Peter Numbers coprime to the modulus are invertible, for if $\rm\: (b,m) = 1\:$ then by Bezout $\rm\: j\,b+k\,m = 1,\:$ i.e. $\rm\:mod\ m\!:\ j\,b\equiv 1,\:$ so $\rm\: j \equiv 1/b,\:$ so $\rm\: aj\equiv a/b.\:$ Thus fractions with denominator coprime to the modulus always exist. –  Bill Dubuque Jun 17 '12 at 1:22

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