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In the expression $\frac{1}{b}\pmod m$, where $(b,m)=1$, is $\frac{1}{b}$:

a) a rational number (and so rational numbers are defined in modulo arithmetic using multiplicative inverses)?

b) just created notation for $($the inverse of $b)\pmod m$ that looks like division just to confuse us (and is used because of similarities between division and division $\pmod m$)? (same for $b^{−1}$)

Is it a) or b)?

Here it says it is a).

Bill Dubuque from M.SE seemingly claims it is b). So does a comment here, also this blog.


Edit: now that I thought about it, either

1) the notation $\frac{a}{b}$

2) or the definition $a\equiv b\pmod {m}\iff m\mid a-b$

is misleading. 1) seems a lot more likely, since I'm not sure how the $\bmod$ function could be defined otherwise.

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I'm not sure if there is a general consensus, but I would like to add another interpretation different from the ones already mentioned.

I tend to say it is

  • not a number, since we want it to be 'an inverse of $b$ modulo $m$', without specifying which inverse. I mean, why should an inverse of $b$ modulo $m$ be defined as $\frac1b$ and not $\frac{1+m}b$ or $\frac1{b+m}$?
  • not a congruence class modulo $m$. This is because in high school the relation '$\equiv\pmod m$' is usually introduced without even mentioning ring theory, we simply define it as some equivalence relation on integers: $a\equiv b\pmod m$ and not $(a+m\mathbb{Z})\equiv(b+m\mathbb{Z})\pmod m$.

If not a number nor a congruence class, then what should it be?

I'd say it isn't a thing, it's just a notation used in congruences representing an unspecified inverse of $b$ modulo $m$. I.e, $\frac ab\equiv c\pmod m$ is a notation meaning as much as $a\equiv bc\pmod m$ (provided $\gcd(b,m)=1$), but $\frac ab$ has no meaning to me when pulled out the congruence.
Note that this point of view still conserves the expected properties suggested by the rational numbers, such as $b\cdot\frac ab\equiv a\pmod m$ and $\frac ab+\frac cd\equiv\frac{ad+bc}{bd}\pmod m$ (provided the inverses exist).

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  • $\begingroup$ Can I claim that $\frac{ac}{bc}\pmod m$ is defined when $\gcd(b,m)=1$, but $\gcd(c,m)> 1$? $\endgroup$ – user26486 Jan 28 '15 at 18:36
  • $\begingroup$ I'd say yes. For example, the Wikipedia page about Fermat quotients uses things like $\frac{a^{p-1}-1}p\pmod p$. So here the fraction would be an ordinary fraction. In the case of $\frac{ac}{bc}$ one could say it means $\frac{\frac{ac}c}b$, where the upper fraction is an ordinary one. Though I should say I'd be very happy to learn that someone else has already made the effort to set those things clear. $\endgroup$ – punctured dusk Feb 18 '16 at 20:50
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One of the difficulties when kids learn the arithmetic of fractions is moving from thinking about 1/5 in terms of slices of pie to thinking of 1/5 as a number - the thing you multiply 5 by to get 1.

I have found that modular arithmetic actually sometimes helps. Kids enjoy understanding how 1/5 on an 11 hour clock is 9. Then of course 1/9 is 5. They like the weird fact that on the usual 12 hour clock 1/5 is 5 - and that many reciprocals don't exist.

So I think the answer to your question is "It's (b)" - but the notation is intended to enlighten, not confuse.

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If you have a modular expression

$$a \equiv b \pmod c$$

Then you can characterize it as:

$$f(a, c) = f(b, c)$$

Where $f$ is the function that describes the remainder operation (and its generalizations),

Or the expression can be characterized as

$$a \in \mathbb Z / c\mathbb Z ~\land ~b \in \mathbb Z / c\mathbb Z ~\land ~a = b$$

Where $\mathbb Z / c\mathbb Z$ is the structure (commutative ring) of $c$ elements that defines addition and multiplication as you would expect.

Both characterizations are equivalent, so there is no answer to the question of "which one is it".

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Generally this is notation for multiplicative inverse.

To 'mod' by $m$ simply means saying that $m = 0$. When we just consider integers, we get modular arithmetic. If we consider $\{\frac{a}{b}:a,b\in\mathbb{Z},(b,m)=1\}$ we get a subset of the rationals which we can mod by $m$ and arithmetic works in the same way. This means in some sense that a) and b) are true, just looking at the notation in different ways.

We can't simply mod all the rationals, because things like $\frac{1}{m}$ don't make sense.

In practice, considering it as multiplicative inverse is easiest, but both approaches are essentially the same.

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