$$\frac{a}{a}\pmod a,\ \ \ a\in\mathbb Z\setminus\{-1,0,1\}$$

There are two cases:

$1)$ $\frac{a}{a}$ is the notation for the real number $1$.
Then the expression is equivalent to $1$ modulo $a$.

E.g., see this, where the expression on the LHS would not even exist if what would be meant by $99$ in the denominator is modular inverse instead of regular integer division.

$2)$ $\frac{a}{a}$ is the notation for $ax$, where $x$ is in the class of solutions to $ax\equiv 1\pmod{a}$.
Since $x$ does not exist, $\frac{a}{a}$ does not exist too in this case.

So $\frac{a}{a}\mod a$ can be thought of as either equivalent to $1$ modulo $a$ or not existing at all.

Now, $aa^{-1}$ is possibly more commonly used than $\frac{a}{a}$ in modular arithmetic and $\frac{a}{a}$ is more commonly used than $aa^{-1}$ in division in integers, but surely not always.

Such notation was probably created due to its similarities to division in integers, but I think less ambiguous notation should've been created instead.

Also see this.
There's a comment there that points out my problem. Consider:

$$\frac{ac}{bc}\pmod m,$$

where $$\gcd(b,m)=1,\ \ \ \gcd(c,m)>1,\ \ \ m\in\mathbb Z\setminus\{-1,0,1\},\ \ \ c\in\mathbb Z\setminus\{0\},\ \ \ b\in\mathbb Z\{0\},\ \ \ a\in\mathbb Z$$

can either be equivalent to $\frac{a}{b}\pmod{m}$ if the fractional notation denotes integer division or it could not exist at all if the fractional notation denotes modular inverses.

  • 1
    $\begingroup$ Sorry I didn't see your earlier question since I was not active on MSE then. I will address both shortly. $\endgroup$ – Bill Dubuque Feb 24 '15 at 19:04

First, a simple example: note that $\, 1+2+\cdots +n\, =\,\dfrac{n(n+1)}2\ $ remains true modulo $\,2,\,$ but it would be more cumbersome to state without the very convenient fractional notation.

Let's consider an analogous but simpler example of the question in your first link, namely $\qquad {\rm mod}\ 9\!:\ \ \dfrac{10^n-1}9\, =\, \overbrace{\color{#c00}{11\cdots 11}}^{\large n\ 1's} \,\equiv\, n\ $ by $10\equiv 1,\,$ i.e. by casting out nines

Here, the context is congruences on integers, so the fraction denotes an $\rm\color{#c00}{integer,}$ viz. the integer obtained by cancelling $\,9\,$ from the fraction. The mod applies to that integer. More generally

$\qquad {\rm mod}\ x\!-\!a\!:\ \ \dfrac{f(x)-f(a)}{x-a} \equiv\, f'(a)$

Here the context is polynomials over some ring, so the fraction denotes the unique polynomial obtained by the division (which is exact by the Factor Theorem). This yields a purely algebraic definition of polynomial derivatives.

In all cases, the fraction can be considered to be a convenient notation used to denote a certain element $\,r\,$ of the ambient ring. The notation proves convenient because it conveys information about the ring-theoretic properties of that element, e.g. in the first example above it denotes the (unique!) solution of $\ 9r = 10^n-1\,$ in $\,\Bbb Z.$

Such (universal) cancellation (before evaluation) can prove quite powerful, e.g. we can prove Sylvester's determinant identity $\rm\,\det(1+AB) = \det(1+BA)\,$ by computing the determinant of $\rm\ (1+A\ B)\ A\ =\ A\ (1+B\ A)\ \ $ then cancelling $\rm\ det(A),\,$ all done in $\,\Bbb Z[a_{ij}],\,$ where the matrix entries are indeterminates. Similarly one can compute the adjugate of the adjugate. Beware: it can be perplexing at first glance, esp. understanding why the proof works even when $\,\rm\det(A) = 0.$

  • $\begingroup$ Your point is that using integer division notation in modular arithmetic is useful, which is of course true. The problem is that the notation for modular inverses is the same, so you can mistake it for integer division, so $\frac{a}{a}\pmod{a}$ may be equivalent to $1$ or it may not exist at all. $\endgroup$ – user26486 Feb 25 '15 at 16:10
  • $\begingroup$ @user314 Can you please provide an example (with context!) where you think the denotation is ambiguous. Modular fractions are only well-defined when the denominator is coprime to the modulus, so any fraction not satisfying that does not denote a modular fraction. $\endgroup$ – Bill Dubuque Feb 25 '15 at 16:13
  • $\begingroup$ Surely it is not likely that a problem "find $\frac{a}{a}\mod a$" would want you to conclude that $\frac{a}{a}\mod{a}$ actually does not exist instead of wanting you to calculate $1\mod {a}$. The slight ambiguity is minimal. I'll just accept your answer and leave it at that, since I suppose no context would actually mislead anyone. $\endgroup$ – user26486 Feb 25 '15 at 17:00
  • $\begingroup$ Slightly off-topic: can I say $0.5\equiv 0.5\pmod{3}$, because it's true by the definition of mod operation: $3\mid 0.5-0.5=0$. But $0.5$ does not belong to a congruence class mod $3$. Another question: is there any rule about whether it's allowed to use equality symbols inbetween equivalences $a\equiv b=c\equiv d\pmod{m}$? $\endgroup$ – user26486 Sep 25 '15 at 2:36

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