Question

In Z_7 find all square roots of 2; of 3; of 4; of 5.

Answer 1

I'm trying to understand the Z tables for modulus arithmetic.

Answer 2

I'll work out the first one, the rest will be similar. So you need to solve the equation:\[x^2\equiv 2\pmod{7}\]

Answer 3

X^2-2=7q

Answer 4

(7q+2)^(1/2)=x

Answer 5

Am I headed in the right direction

Answer 6

You really want to stick with integers. So a square root is probably not the best idea.

Answer 7

When you are working with a small modulus, like 7, its a little easier to just check each element and see if its a solution to the equation. For example, 1 isn't a solution, because 1^2=1 and thats not congruent to 2 mod 7.

However, 3 is a solution, because 3^2 is 9, and 9 is congruent to 2 mod 7.

Answer 8

Ah I see. so how could you get "all" of them. Wouldn't it be infinite?

Answer 9

So you can come up with all the solutions, just by trying out 0, 1, 2, 3, 4, 5, and 6. Checking to see which will square to 2 mod 7

Answer 10

So Z_7 means only up to 6?

Answer 11

Why wouldn't 7 be included?

Answer 12

Just to join in, once you have one square root, you have both. Since \((-3)^2=3^2\), those would be your two square roots of 2 mod 7. (Usually -3 is written as 4 mod 7 though)

Answer 13

I see but the z_7 table doesn't show any negative integers.

Answer 14

http://demonstrations.wolfram.com/ModularArithmeticTables/

Answer 15

In modular arithmetic, you really only care about the remainder thats left after division by the modulus. So in our case (mod 7):\[\ldots \equiv -14\equiv -7\equiv 0\equiv 7\equiv 14\equiv \ldots \pmod{7}\]All these integers represent the same congruence class, because the remainder they leave when divided by 7 is 0. So by using 0 in the above calculation, we have taken care of all the multiples of 7 (which is why 7 isnt in the list). Similarly:\[\ldots \equiv -13\equiv -6\equiv 1\equiv 8\equiv 15\equiv \ldots \pmod{7}\]are all the same because they leave a remainder of 1 when divided by 7.

Answer 16

When you divide by 7, the only possible remainders you can get are 0, 1, 2, 3, 4, 5, and 6. So by using those numbers, we are checking all the possible solutions.

Answer 17

Awesome! I totally understand now. So the table is really only illustrating the remainder.

Answer 18

Excellent explanation @joemath314159

Answer 19

@KingGeorge Thanks :) although to really really really understand whats going on, we would have to talk about equivalence relations, the addition/multiplication of congruence classes being well-defined, etc. The little details are important! lol.

Answer 20

Yeah, I'm actually working on equivalence relations but the modulus part of it all seems to confuse me. Thanks for the help. As usual I really appreciate it.