Let g:A à B and f: B à C

Let g:A à B and f: B à C. Prove that f is surjective if f o g is surjective.

Proof (By Ben Mueller):

To start this proof, let’s assume that f o g is surjective. From that, we want to prove that f must also be surjective.

We know that f o g represents a mapping from A à C as f(g(a)) = c, a being in the set A and c in C. Now, since f o g is surjective, there is a g(a) such that f(g(a)) = c for all c in C. Also, since g maps from the set A to the set B, we know that g(a) = b, b being in set B.

This completes our proof because then f(g(a)) = f(b) = c, which implies that f is also surjective (there is a b such that f(b) = c for all c)./(Proof complete)

Let p be a prime integer. Prove that if [a][b] = 0 in Zp, then either [a] = 0 or [b] = 0.

Proof (Written by Ben Mueller):

Since [a][b]=0, we know that the product of a and b must be divisible by p (prime integer). Recall Euclids Lemma which states if p divides ab and p is prime, then either p divides a or p divides b. From Euclids Lemma, we can state that p divides a or p divides b. Now, if p divides a, then [a] = 0 and if p divdes b, then [b] = 0./(Proof complete)

Application Problem

Find the multiplicative inverse for the element [5] in Z11.

Solution:

The identity element with respect to multiplication in Z11 is [1].

We want to find an element [x] in Z11 such that [5][x] = [1].

The following elements are equal to [1] in Z11: [1],[12],[23],[34],[45],[56],[67].

Now, [5][9] = [45] = [1] in Z11. Thus, the multiplicative inverse for the element [5] in Z11 is [9].