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Does the concept of modular multiplicative inverse require a proof or is it taken as a definition?

Suppose $5/4 \equiv 3$ (mod $7$).

Can that even be written in the standard $a = bq + r$ notation and proven from there?

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  • $\begingroup$ Not sure what you mean. To demonstrate that $5\times 4^{-1}\equiv 3 \pmod 7$ you just need to show that $5\equiv 3\times 4 \pmod 7$ which is clear. $\endgroup$ – lulu Feb 17 at 17:05
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What the statement means is that $5 \times 4^{-1} \equiv 3 \pmod{7}$.

The modular inverse of $4$ is defined as the number $x$ such that $4x\equiv 1 \pmod{7}$ (if it exists), and it's easy to check that $x=2$ satisfies this.

So $5 \times 4^{-1} \equiv 5 \times 2 \equiv 3 \pmod{7}$.

Alternatively, $5/4$ can be defined as the number that when multiplied by $4$, gives $5$, and $3 \times 4\equiv 12 \equiv 5 \pmod{7}$ as required.

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$5/4 \equiv 3\pmod 7$ means

$5 \times 4^{-1} \equiv 3 \pmod 7$ or $5 \times 2 \equiv 3 \pmod 7,$

so $5 \equiv 12 \pmod 7$ (i.e., $5=-1\times7+12) $ or $10 \equiv 3 \pmod 7$ (i.e., $10=1\times7+3),$

but $4^{-1} \pmod 7$ is different from $1/4$ as a real number.

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  • $\begingroup$ Re: claim in last sentence: do you claim the same for $1/4$ the rational number, and do you also claim that $\,4\pmod 7\,$ differs from $4$ the integer? What precisely do you mean by "different from"? $\endgroup$ – Bill Dubuque Feb 17 at 17:39
  • $\begingroup$ @BillDubuque: yes, also different from 1/4 as rational number. $4 \pmod 7$ is an equivalence class, which contains the integer $4$ as well as $11, -3$, etc. By "different from" I meant you don't say something like $1/4=0.25 \equiv 2 \pmod 7$ or $5/4=1.25=7q+3$ with $q \in \mathbb Z$ $\endgroup$ – J. W. Tanner Feb 17 at 18:00
  • $\begingroup$ fyi: in algebra we do often overload / abuse notation in this way, e.g. we use $4$ to denote the equivalence class $[4]_7 = 4+7\Bbb Z\,$, and we use $1/4$ to denote $[4]_7^{-1} = 2+7\Bbb Z,\,$ which makes sense because the natural map $\,\Bbb Z \to \Bbb Z_7\,$ extends uniquely to the subring of rationals writable with denominator coprime to $7$ (universal property of localization), via $\, a/b\mapsto [a]_7[b]_7^{-1}.\,$ This implies that (ring) arithmetic of such fractions maps homomorphically into $\,\Bbb Z_7,\,$ e.g. $\,1/2- 1/3 = 1/6\,$ maps to $\, 4 - 5 \equiv 6.\ \ $ $\endgroup$ – Bill Dubuque Feb 17 at 18:31
  • $\begingroup$ Wouldn't you say $1/4$ means something different in the context of $\mathbb Z_7$ vs. $\mathbb Q$ or $\mathbb R$, and could you write $5/4=7q+3$ as OP asked? $\endgroup$ – J. W. Tanner Feb 17 at 18:44
  • $\begingroup$ The point is that in certain contexts such overloaded notation is fruitful because it helps us to smoothly import intuition from closely related structures. The context usually determines precisely how "different" we wish to view such related objects. In all the rings that you mention $\,1/4\,$ denotes the inverse of $4$, i.e. the unique solution of $\, 4x = 1,\,$ so precisely how does it "differ" in those rings? $\endgroup$ – Bill Dubuque Feb 17 at 19:03

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