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In modular division, what is the meaning that should be ascribed to the notation exemplified below (also given on p. 5 of this)?

$$\begin{align} \implies & 5\cdot8 \equiv 4\pmod {12} \tag{i} \\[2ex] \implies & 5 \equiv \frac{4}{8}\pmod {12} \tag{ii} \\[2ex] \implies & 8 \equiv \frac{4}{5}\pmod {12} \tag{iii} \end{align}$$

I think in terms of values reached by different residue classes, but I am unable to get any clue.
As a very simple example,
values taken by $4 \pmod{12}$ residue class are: $4, 16, 28, 40$
values taken by $5 \pmod{12}$ residue class are: $5, 17, 29, 41$
values taken by $8 \pmod{12}$ residue class are: $8, 20, 32, 44$

This lends no meaning to eqns. $\text{(ii), (iii)}$ above.

user21820
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jiten
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2 Answers2

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  • $5 \cdot 5 =25 \equiv 1 \pmod{12}$ so the multiplicative inverse of $5 \bmod 12\,$ is $5^{-1}=5$.

    Therefore $5 \cdot 8 \equiv 4 \implies 5^{-1}\cdot5\cdot8 \equiv 5^{-1} \cdot 4 \implies 8 \equiv 5^{-1}\cdot 4 \pmod{12}\,$. The latter may sometimes be written as $8 \equiv \frac{4}{5} \pmod{12}\,$ but that's arguably an abuse of notation, unless such notation was very explicitly and narrowly defined before being used.

  • $8$ has no multiplicative inverse $\bmod 12$, so $5 \equiv \frac{4}{8} \pmod{12}$ makes no sense whatsoever.

dxiv
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  • On page #7, the author of the source states two types of divisions - (1) Which have no answer (for us I feel the correct word is: logic, meaning), (2) Which have an answer. The author further states :"Even seemingly odd divisions like $\frac{0}{3}$ or even worse $\frac{0}{0}$ are legal $\mod 6$. Both have the answer $2$. Explain this?" – jiten Jan 16 '18 at 04:13
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    @jiten Indeed, quoting from the book: "Unlike the division of real numbers, mod division does not always yield an answer". IMHO that's a rather odd and un-pedagogical way to "define" an operator, but which "does not always yield an answer". – dxiv Jan 16 '18 at 04:18
  • So, the author states it to be 'odd' but still legal. – jiten Jan 16 '18 at 04:19
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    @jiten For certain fuzzy values of "legal", I guess ;-) The focus of the text looks to be more computational than mathematical, so I guess some liberties are to be expected, but can still be confusing if you are reading it with a math-oriented eye. – dxiv Jan 16 '18 at 04:21
  • Okay, I will be having that into view. – jiten Jan 16 '18 at 04:23
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Note that $3*8=24\equiv 0\pmod{12}$. Adding this equation to your equation $(i)$ yields $5*8\equiv 8*8=4\pmod{12}$. So we will get $\frac48\equiv5$ as well as $\frac48\equiv8$ (both mod 12) which is nonsensical (that is not well-defined).

So one should avoid assigning any meaning to division by something which is a zero divisor.

When operating modulo a given $m$, if we stick to $b$ with gcd$(b,m)=1$ all zero divisors will be avoided and fractions with such $b$ in the denominators will be meaningful (will have unique value).

P Vanchinathan
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  • I hope you are also answering my comments to @dxiv answer, as you referred to division by $0$. – jiten Jan 16 '18 at 04:17
  • Not clear in which situation should stick to $b$ with $\gcd(b,m) =1$ for modulo arithmetic, it happens never; all values are used up. – jiten Jan 16 '18 at 04:23
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    If you want to put $b$ in the denominator of a fraction, then the fraction will be meaningful (unique-valued) only when gcd$(b,m)=1$. – P Vanchinathan Jan 16 '18 at 04:32