How to solve this $ (7/2)\bmod5$?

Question

I know its answer is $3\cdot5$ but I want to ask that is the following true-

$$(a/b)\bmod(p) = (a\bmod(p))\cdot((1/b)\bmod(p)))\bmod(p)$$

(where $a$ and $b$ are any integers and $p$ is a prime number.)

$(1/b)\bmod(p)$ can be solved by Euler's formula but i don't get correct answer if i solve it this way, how to solve this?

Done that way the answer is $,1\equiv 7\cdot 3,,$ not $,5\cdot 3,,$ since $,3\equiv 2^{-1}\pmod 5\ \ $ — Bill Dubuque, Jul 13 '14 at 18:43

score 2 · Accepted Answer · answered Jul 13 '14 at 17:11

2

As $\displaystyle7\equiv2\pmod5, \frac72=7\cdot2^{-1}\equiv2\cdot2^{-1}\pmod5\equiv1$

Alternatively, As $\displaystyle2\cdot3\equiv1\pmod5\implies 2^{-1}\equiv3$

$\displaystyle\implies\frac72=7\cdot2^{-1}\equiv7\cdot3\equiv1\pmod5$

answered Jul 13 '14 at 17:11

lab bhattacharjee

I know how to find $(1/2)mod(5)$ but please tell me is the equation i wrote above correct? @lab bhattacharjee – user164013 Jul 13 '14 at 17:13
@user164013 $$ab\pmod p\equiv a\pmod p\cdot b\pmod p,$$ right? – lab bhattacharjee Jul 13 '14 at 17:15
thanks! I got my answer :) – user164013 Jul 13 '14 at 17:19

score 0 · Answer 2 · 2014-07-13T20:11:30.463

0

I do not really understood what you have done but this is how i would do...

If for some $0\leq x\leq 4$ i have $\dfrac{7}{2}\mod5=x$ then see that $5$ divides $\frac{2x-7}{2}$ which implies $5$ divides $2x-7$

Check what $x$ would work in this case...

edited Jul 13 '14 at 20:11

answered Jul 13 '14 at 17:17

I cannot make any sense of the above. Please elaborate. – Bill Dubuque Jul 13 '14 at 18:39
@BillDubuque : Does this make any sense to you now? – Jul 13 '14 at 20:11
Ok, now I see. You had written $,5/2x - 7,$ when you meant $,5\mid 2x-7.\ $ – Bill Dubuque Jul 13 '14 at 20:18
@BillDubuque : Yes Yes... I should have been careful :) – Jul 13 '14 at 20:20

2 Answers2