Thursday, 21 July 2016

Fractions in Modular Arithmetic





Suppose we have $$\frac{2}{3} \equiv x \bmod 5$$



I understand that the first step that needs to be made is: $$2 \equiv 3x \bmod 5$$



But from there I'm having a hard time understanding the logic of how to solve for $x$. Obviously, with simple numbers like this example, the answer is $4$, but how can I abstract the process to solve for $x$ when the numbers become very large?



Answer



Modulo arithmetic generally deals with integers, not fractions. Instead of division, you multiply by the inverse. For instance, you would not have $\frac2 3\equiv x\mod 5$, you would have $2*3^{-1}\equiv x\mod 5$. In this case, $3^{-1}\equiv 2\mod 5$, so you would have $2*2\equiv 4\mod5$. The inverse of a number $a$ in modular arithmetic is the number $a^{-1}$ such that $a*a^{-1}\equiv 1\mod n$.


No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...