Today I stumbled upon this question: Prove $\binom{2p+1}{p}\equiv2$ mod $p$ when $p$ is any prime. While reading it, I was quite surprised that the equivalence could be true, because $\binom{2p + 1}{p} = \frac{(2p + 1)!}{p! (p + 1)!}$ and the denominator includes a $p$, which should be division by zero according to my understanding.

Let me pick a simpler example: according to that question, I should have $\frac{2p}p \equiv 2 \pmod{p}$. What I say is that, according to me, $\frac{2p}p$ is division by zero, so it cannot have any result.

If I check the definition of division in modular arithmetic, I find definitions like this from stanford.edu:

$x$ divided by $y$ modulo $n$ is only defined when there is a unique $z \in \mathbb{Z}$ such that $x = yz$

In my case, the expression $x = yz$ becomes $2p = pz$, but because we are mod $p$, the expression is valid for every $z \in \mathbb{Z}$, hence there's no unique $z$, hence division is undefined.

So my question is: where is my mistake and what is the correct definition of division I should use in such cases?

  • $\begingroup$ $x$ modulo $y$ is uniquely defined, but there need not be a number $z\in \mathbb Z$ with $x=yz$. The mentioned definition means that $y$ divides $x$. $\endgroup$ – Peter Mar 19 '17 at 16:50

You're correct about the definition of division modulo $n$, but it doesn't matter - just because division and modular arithmetic are both present in the statement doesn't mean you're doing both at once. To take an analogy, dividing an integer by $2$ to get an integer only works if the original integer is even; nevertheless, it makes perfect sense to talk about $\frac{3 - 1}{2}$.

In the example you noted, it's certainly not possible to prove the claim by doing everything modulo $p$. But nothing says you have to. For example, $\frac{p}{p} \equiv 1 \mod p$, not because we can divide $p$ by $p$ mod $p$, but because we can divide $p$ by $p$ in integers and get $1$, which is $1$ mod $p$.

The crux of the matter is this: to say that $\frac{a}{b} \equiv c \mod p$ isn't saying that $\frac{a \mod p}{b \mod p} = c \mod p$. It's saying that the number obtained by dividing $a$ by $b$ (which is perfectly well-defined as long as $b$ isn't zero) has a remainder of $c$ when further divided by $p$. The technique of considering the components of an equation modulo $p$ is a useful tool, but it loses some information and therefore sometimes doesn't give an answer; in cases like that, we have to return to the original definition.


Of course we have to cancel $p$ in the term $2p/p$ in the ring of integers, before we take the value modulo $p$. If we write $x\equiv y \bmod p$ for integers $x$, $y$, then we first need $x\in \mathbb{Z}$. So $x=\frac{2p}{p}=2$. Now $\binom{2p+1}{p}$ is indeed an integer, because the factorials in the denominator cancel.

  • $\begingroup$ But I can write $\frac{2}3 \equiv 3 \pmod{7}$. Here $\frac{2}3$ is not an integer $\endgroup$ – user1807 Mar 19 '17 at 16:57
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    $\begingroup$ The question is about integers, you said, namely $2$ and $\binom{2p+1}{p}$. Of course we can generalise the modulo notation to other rings. $\endgroup$ – Dietrich Burde Mar 19 '17 at 16:58

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