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In the book I'm reading it says that there exist elements with multiplicative inverses in a integer ring if $gcd(a,m)=1$ where $a$ is the element and $m$ is the modulo.

The inverse $a^{-1}$ is defined such that $a \times a^{-1} \equiv 1$ mod $m$

Also it says that if an inverse exist for $a$ we can divide this element since $b/a \equiv b \times a^{-1}$ mod $m$

I'm confused as to what $b$ is in this case? My guess right now is that it is another member in one of the equivalence classes for the modulos $m$.

J. W. Tanner
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KetDog
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    If you're working with congruences then $a$ and $b$ are any integers, and $b/a := a^{-1}b$ denotes the (unique) solution of $,ax\equiv b\pmod{!m}.,$ OTOH if you're working with cosets $,[a] = a+m\Bbb Z,$ in the quotient ring $,\Bbb Z/m,$ then you should already know how to translate between congruences and quotient rings (else you put the cart before horse). – Bill Dubuque Mar 30 '20 at 03:56

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Yes. For example, let $m=7$ and $a=[2]$. Then $a^{-1}=[4],$ because $2\times4\equiv1\pmod7$.

For another example, if $m=7$ and $a=[3]$ then $a^{-1}=[5],$ because $3\times5\equiv1\pmod7$.

$b$ is any other element. Say you want to solve for $x$ in $ax\equiv b\pmod m$.

The answer is $x\equiv b\times a^{-1}\pmod m$.

For example, the solution to $2x\equiv5\pmod7$ is $x\equiv5\times2^{-1}\equiv5\times4=20\equiv6\pmod7$.

J. W. Tanner
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  • I think I understand. So, when we deal with modular arithmetic the inverse of a certain element is completely different from the inverse that we learn in high school math where the inverse of a = 1/a ? – KetDog Mar 30 '20 at 04:13
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    In the context of real numbers, the inverse of $2$, which could be written $1/2$, is the real number $0.5$; in the context of integers modulo $7$, the inverse of $[2]$, which could be written $1/[2]$, is $[4]$ – J. W. Tanner Mar 30 '20 at 04:24
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    @Papaya It's not "completely different". Rather, in both cases $, a^{-1}$ denotes the solution of $,ax = 1,,$ which is necessarily unique. – Bill Dubuque Mar 30 '20 at 04:30