1. ## Homework help!!!

I have a midterm tomorrow and can't spend all night trying to figure out what is going on with these problems. PLEASE help!!!! I understand congruence classes but am just having a problem putting together exactly what is being asked here. Please clarify.

Fix a positive integer m. In this exercise, you will study which elements of Z(m) have inverses. Recall that we write Z(m) for the set of congruence classes modulo m, and that there are m of these: [0],[1],[2],.....,[m-1]. Here [a] = b element of Z such that b is congruent a modm. Now given [a] element of Z(m), a congruence class [b] is called the multiplicative inverse of [a] if [a][b]=1. If such a class [b] exists then [a] is called invertible.

a) Show that if gcd (a,m) =1, then [a] is invertible. (Hint: begin by using the pypothesis to find x,y element of Z for which ax + my=1)

b) Show that if [a] is invertible, then gcd(a,m)=1 (Hint: If [a] is invertible show first that there exists b element of Z such that ab is congruent (mod m). Then prove that if gcd (a,m) is not equal to 1 then a contradiciton is reached. )

c) Find all of the invertible elements of Z(15), using parts (a) and (b). Find inverses for each of the elements you wrote down, by inspection.

I NEED HELP ON a) and b) the most!!!! THANKS IN ADVANCE

2. ## ???

No help out there I guess

Let $m\geq 2$ be a positive integer. Define $\mathbb{Z}_m^{\text{x}}$ to be the congruences classes $[x]_m$ such that $\gcd(x,m)=1$. Prove that $\mathbb{Z}_m^{\text{x}}$ is a group under multiplication (not addition). Thus if you have the equation $[a][x]=[1]$ and $\gcd(a,m)=1$ then by the property of groups this equation is solvable for $[x]$.