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When $(a,b)=1$, $\cfrac{1}{a+b}\binom{a+b}{a}$ refers to the number of paths from one corner to its opposite corner of an $a\times b$ lattice that lies completely above (or below) the diagonal. Therefore, it must be an integer.

But does anyone know if there is an arithmetical proof of this?

There is an arithmetical proof for $\binom{a}{b}$ is integer. See this post.

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NECing
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    $a\dbinom{a+b}{a} = \left(a+b\right)\dbinom{a+b-1}{a-1}$ and $b\dbinom{a+b}{a} = \left(a+b\right)\dbinom{a+b-1}{a}$ are integers, and $a$ and $b$ are coprime... Take care of corner cases ($a=0$ and $b=0$). – darij grinberg Sep 07 '13 at 01:42

2 Answers2

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You know that $\binom{a+b}a$ is an integer, and from its formula involving factorials (or otherwise), we have that $\binom{a+b}{a}=\frac{a+b}a\binom{a+b-1}{a-1}$, which means that $$\frac1{a+b}\binom{a+b}{a}=\frac1a\binom{a+b-1}{a-1}. $$ Now, since $\mathrm{gcd}(a+b,a)=1$, and $\frac{a+b}a\binom{a+b-1}{a-1}$ is an integer, it follows that $a$ divides $\binom{a+b-1}{a-1}$, and we are done. (In general, if $\mathrm{gcd}(\alpha,\beta)=1$ and $\alpha$ divides $\beta\gamma$, then $\alpha$ divides $\gamma$.)

Andrés E. Caicedo
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  • From the equation in the first sentence, you have that $\frac{a}{a+b} X$ is an integer (the other binomial coefficient), and $(a,a+b)=1$, so that $X/(a+b)$ is an integer, where $X$ is the binomial coefficient in the problem. The factor of $a$ is not helping it be integral. – zyx Sep 07 '13 at 03:12
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Combinatorial proof may be clearer in this case.

Realize $a+b$ as the integers mod $a+b$ and rotate the $a$-subset by adding $1$ mod $a+b$ to every element. This process is periodic, of period $a+b$ and (because $a$ and $a+b$ have no common factor) has no smaller no period. The number of rotation-equivalence classes of $a$-subsets from $a+b$ is then $\frac{{a+b} \choose a}{a+b}$.

For proving number theoretic properties of binomial coefficients, arithmetic methods are stronger but combinatorial proofs (which are not always available) are easier to understand and generalize.

zyx
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  • "and generalize": that's not true. The OP's statement holds for $a$ and $b$ of any sign (as long as $a+b\neq 0$), but the combinatorial proofs get seriously complicated in that context. – darij grinberg Sep 07 '13 at 04:21
  • If you want negative parameters an applicable definition of the binomial coefficients should be specified. Some definitions invalidate the arithmetic arguments, in some definitions it is trivial to note that the combinatorial argument works. And one does get all sorts of generalizations from the combinatorial/group action argument that are hard to see arithmetically. – zyx Sep 07 '13 at 18:05