Is 1 always a modular inverse of 1?

Question

I have been looking for an answer online but I am unable to find one. Are inverses defined as to not include the number 1?

Would this mean that 1 is an inverse of itself in

$1*1^{-1}=1 (\textrm{mod p})$

where p is any number?

lhf · Accepted Answer · 2018-09-25T14:16:10.660

3

Yes: $1 \cdot 1 = 1 \equiv 1 \bmod m$ for all $m$.

Recall the definition: $b$ is an inverse of $a$ mod $m$ iff $ab \equiv 1 \bmod m$.

edited Sep 25 '18 at 14:16

answered Sep 25 '18 at 14:12

lhf

Yes, but is it properly defined as an inverse, or does the term inverse not include 1 within the definition for what it means to be/have an inverse? Thank you for the edit! – S5amuel Sep 25 '18 at 14:15

1 Answers1