Homework 8 was due on Thursday 3rd April 2014. I graded Q2,3 and the grader graded Q1,4,5. All questions are marked out of 6.
Question 1. Most errors in this question came from people having forgotten or misunderstood how congruence classes are defined.
Question 2. The biggest problem with this question was a lack of precision, and some extremely elaborate arithmetical acrobatics were performed when trying to reach a contradiction. The idea's this: if $m$ is composite then $m=rs$ for some $2 \le r,s < m$. But then $rs \equiv 0 \bmod m$. If $\mathbb{Z}_m$ were a field then there would exist $r{-1}$ such that $r^{-1}r \equiv 1 \bmod m$, meaning that $$s \equiv r^{-1}rs \equiv 1\cdot 0 \equiv 0 \bmod m$$ But this contradicts the fact that $2 \le s < m$. There were other nice proofs involving Bézout's identity.
Question 3. This question was algebraically ugly and somewhat difficult. Given $y \in \mathbb{R}$ you needed to find $x \in S$ such that $g(x)=y$. If $y=0$ you get $x=\frac{1}{2}$ and everything's fine. If $y \ne 0$, a few lines of algebra yield $$x = \frac{1-y \pm \sqrt{y^2+1}}{2y}$$ This certainly yields $g(x)=y$, but an important part of the question is that $\mathrm{Im}_f(S)=\mathbb{R}$, not just $\mathrm{Im}_f(\mathbb{R}) = \mathbb{R}$. That is, you need to prove that one of these two (corresponding to choice of $\pm$ sign) values of $x$ lies in $S$. This was where the majority of points were lost.
Question 4. Most people who applied the definitions correctly got this question right. There wasn't much thinking power required beyond piecing together definitions.
Question 5. The biggest error here was people not completing the question: you were asked to express equivalence classes in terms of preimages. Some people wrote $[a] = \mathrm{PreIm}_f(\{b\})$ or something similar, without saying what $b$ is. But $b$ has to depend on $a$ somehow, otherwise how can we distinguish it from any other equivalence class? In fact the correct answer to the last part of the question was $[a] = \mathrm{PreIm}_f(\{f(x)\})$