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I don't understand how a modular inverse is unique. $3^{-1}≡7$$(mod\hspace{.1cm}20)$.

But $3*27≡1(mod\hspace{.1cm}20)$

So, $3^{-1}≡27$$(mod\hspace{.1cm}20)$.

What do I not understand here?

Can I think of $3^{-1}$ in its "normal sense," as a fraction? For example:

$\frac{1}{3}≡7$$(mod\hspace{.1cm}20)$

Mulitply both sides by $8$ and get: $8*\frac{1}{3}≡8*7$$(mod\hspace{.1cm}20)$

and finally claim: $\frac{8}{3}≡56$$(mod\hspace{.1cm}20)$.

Is this valid?

Please help me understand the questions above. If you perceive any additional holes in my understanding that may not be directly present in my post, please do elaborate.

Thanks!

B_math
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    Note $27 \equiv 7 \pmod{20}$. – John Omielan Aug 11 '21 at 03:18
  • @B_math: you might like to review https://math.stackexchange.com/questions/3001818/what-is-the-meaning-of-modular-fractions – Moo Aug 11 '21 at 03:46
  • @JohnOmielan I agree that that is true, but why is this relevant? – B_math Aug 11 '21 at 03:47
  • @Moo yes thank you. I have actually already found and reviewed this post before asking my question, but I am struggling to apply it to my specific questions above. – B_math Aug 11 '21 at 03:49
  • @B_math The statement a modular inverse is "unique" means only up to the same congruence element, e.g., $3^{-1} \equiv 7 \equiv 27 \pmod{20}$. – John Omielan Aug 11 '21 at 03:55
  • @JohnOmielan, thank you for the reply. So, the inverse of three is only that which is 7 mod 20- that what it means by unique? All those things are in there are unique equivalence class? What about what I have done with the fractions-is that valid? – B_math Aug 11 '21 at 04:43
  • The inverse is unique $!\bmod 20,,$ i.e. if $x$ and $x'$ are inverses then $,x\equiv x'\pmod{20}.\ $ See the first linked dupe for more on that. Yes, inverses are a special case of modular fractions - see the other dupes for introductions to modular fractions. – Bill Dubuque Aug 11 '21 at 11:45

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Given a modulus $m$ and a number $x$ where $x \not \equiv 0 \pmod m$, a modular inverse is a number $y$ such that $x y \equiv 1 \pmod m$.

Suppose we have two modular inverses $y$ and $y'$, let's try to show they are equivalent mod $m$. So we know that:

  • $x y \equiv 1 \pmod m$.
  • $x y' \equiv 1 \pmod m$.

putting these together

$x y \equiv x y' \pmod m$

this is equivalent to

$m | x y - x y'$

we can rewrite using distributivity

$m | x (y - y')$

and we know that $m \not | x$ so we get $m | y - y'$ which is equivalent to $y \equiv y' \pmod m$.