### Concepts of Mathematics (21-127) — Feedback on Homework 8

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 $[a] = \mathrm{PreIm}_f(\{f(x)\})$

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