+0

# help with mod inverses

0
43
1

I've been stuck on this one for a while, I just don't know what to do with all of this info :(

Given $$m\geq 2$$, denote by $$b^{-1}$$ the inverse of $$b\pmod{m}$$. That is, $$b^{-1}$$ is the residue for which $$bb^{-1}\equiv 1\pmod{m}$$. Sadie wonders if $$(a+b)^{-1}$$ is always congruent to $$a^{-1}+b^{-1}$$ (modulo m). She tries the example a=2, b=3, and m=7. Let L be the residue of $$(2+3)^{-1}\pmod{7}$$, and let R be the residue of $$2^{-1}+3^{-1}\pmod{7}$$, where L and R are integers from 0 to 6 (inclusive). Find L-R.

Dec 8, 2019

#1
0

Let us see if we can make some sense out of this !!

I don't think that ( a +b)^-1 is always congruent to a^-1 + b^-1 mod m.

(2 + 3)^-1 mod 7 =

3 mod 7 = 3, which is your L

While 2^-1  +  3^-1 mod 7 =

4      +     5=9 mod 7 =2, which is your R

So, L - R = 3 - 2 =1

Dec 8, 2019