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I'm a bit new to modular arithmetic on congruence classes, and I'm aware of two axioms that let me do this:

  • $[a]\cdot[b]=[a \cdot b]$
  • $[a]+[b]=[a+b]$

I'm also aware that when working with $\mathbb Z_4$ for example we would have:

  • $[3]=[3]$
  • $[4]=[0]$
  • $[5]=[1]$

But what would this look like?

  • $[2]^{-1}\cdot[3]$
  • $[2]^{-1}+[3]$

I'm just not sure how to work with inverses. I know that $[a]\cdot[a]^{-1}=[1]$, but I'm not sure what happens if both $a$'s are different instead of the same.

Anyway, thanks.

2 Answers 2

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Not every element has a multiplicative inverse modulo $n$. Indeed, since $[2] \cdot [2] = 0$, if $[2]$ had an inverse than $[0] = [2] \cdot [2] \cdot [2]^{-1} = [2]$, a contradiction.

In general, in $\mathbb Z/ n\mathbb Z$ a residue class $[m]$ has a multiplicative inverse if and only if $m$ and $n$ are relatively prime: $[m]$ has a multiplicative inverse if and only if there is a $[k]$ such that $$km \equiv 1 \mod n,$$ that is, a $k$ such that there exists $t$ such that $km -nt = 1$. This is equivalent to $n$ and $m$ being relatively prime by Bézout's lemma.

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    How would we prove this?2017-01-09
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    @user2965071 What in particular would you like to prove?2017-01-09
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    That $[m]$ has a multiplicative inverse if and only if $m$ and $n$ are relatively prime.2017-01-10
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    @user2965071 A proof is given in the second paragraph of my answer, is there a point you would like me to clarify?2017-01-10
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The multiplicative inverse of $[a]\in \Bbb Z_n$ only exists if $\gcd(a,n)=1$. That's not the case for your $[2]\in \Bbb Z_4$.

As for how to calculate something like, say, $[2]^{-1}\cdot [3]$ in $\Bbb Z_5$, you first have to find $[2]^{-1}$, which happens to be $[3]$. This makes $[2]^{-1}\cdot[3]=[3]\cdot[3]=[4]$.

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    Maybe I should have reworded my question, but I'm really looking for how to evaluate something like $[2]^{-1}$ and why it happens to be $[3]$2017-01-08
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    $[2]^{-1}=[3]$ because $[2]\cdot[3]=[1]$. We also have $[4]^{-1}=[4]$, because $[4]\cdot[4]=[1]$.2017-01-08