Question

In a finite group z/5z (mod 5),
why 2^-1=3 ?

Answer 1

We say that :
\[a^{-1}=b\]
iff
\[ab=1\]

Now we have :
\[2\times 3=6=5+1=0+1=1\pmod{5} \]
so :
\[2^{-1}=3\]

Answer 2

oh, so basically 2 times what number will give me 1?

Answer 3

Yeah now i get it. Thanks a bunch

Answer 4

I like euclidean's algorithm
but what @Noura11 has done is really classy also
\[2b=1 \mod 5 \\ 2b-1=5k \\ 2b-5k=1 \\ 5=2(2)+1 \\ 2=1(2) + 0 \\ 5-2(2)=1 \\ 5+2(-2)=1 \\ b=-2 \\ \text{ if you get the inverse is negative } \\ \text{ I always keep adding +5 since we have \mod 5} \\ b=-2+5=3 \]

Answer 5

you know until we get the first positive number

Answer 6

Looks like euclidean's algorithm might come in handy for big numbers.

Answer 7

@Ankh Yes. Exactly !

Answer 8

Thank you both!