Homework 04

Read: Continue with 2.3 and skim through Appendix C: Definitions in Mathematics

Turn in: 2.35(b,d,e), 2.36(f,g,i,j,k,l), 2.37, 2.39, 2.40, 2.41

Remarks

• Notation on 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 $B$.
• A comment about 2.35. Please make sure to explain why each of the axioms is true. You can refer to what we proved in earlier problems (without reproving them) AND you can also use 2.29, but please make it clear when you are using them.

• A comment about 2.37 and 2.41. (Theorem 2.37 is our first theorem to prove—yay!) 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 $e_1=e_2$.

  To continue the proof, you will need to use that axiom 2 is valid for both $e_1$ and $e_2$, and then manipulate things to arrive at $e_1=e_2$. Similarly, on 2.41, you might consider starting as follows.

  Let $G$ be a group, and let $g\in G$. Assume that there exists $g’,g’’ \in G$ such that $gg’ = g’g = e$ and $gg’’ = g’’ g = e$. We will show that $g’=g’’$.

• 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 $h(gx)=h(gy)$; however, $h(gx)=(gy)h$ might not be true…do you see the difference? Now, to solve 2.39, choose something clever for $h$, so that you arrive at $x=y$.