Read: Continue with Section 2.3, and Appendix A: Elements of Style for Proofs
Turn in: 2.35(e), 2.36(d–i), 2.37, 2.39
- In 2.36(g), the set $\mathbb{Q}\backslash \{0\}$ is the set of all rational numbers excluding 0. In general, if $A$ and $B$ are sets, then $A \backslash B$ is the set set of all elements of $A$ that are not also in $B$.
- About 2.37: a good way to show that there is only one element with a certain property is to assume you have two elements with that property and then show that they are the same. So here is one way you might start your proof of 2.37.
Let $G$ be a group. Assume that $e_1$ and $e_2$ are identity elements. We will show that $e_1=e_2$. (Now keep going using that both $e_1$ and $e_2$ satisfy the equation given in Axiom 2 of a group.)
- A comment about 2.39. This theorem (and many others) can be proved “multiplying” both sides of an equation by the same thing. However, since the operation may not be commutative, if you multiply one side of the equation by something on the left, then you need to do the same on the other side too. For example, if you know that $gx=gy$, then for any $h\in G$ it is also true that $hgx=hgy$; however, $hgx=gyh$ might not be true…do you see the difference? Now, to solve 2.39, choose something clever for $h$, so that you arrive at $x=y$.
Extra practice: work through some of the problems we are skipping