fixed comments

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diff --git a/Problem Set 2.page b/Problem Set 2.page
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@@ -109,7 +109,7 @@ Proofs:
 
 Here are some issues with the above solutions.  Feel free to make corrections, even if they weren't the original poster.  However, if you do decide to redo a solution, please don't delete the old one, so that people can see where the old idea went wrong and how it was fixed.
 
-- The map from $\mathbb{Z} \times \mathbb{Z}$ to $\mathbb{Q}$ sending a pair $(a,b)$ to a ratio $\frac{a}{b}$ is right in principle, but not exactly right.  One minor problem with it is that the map is not defined when $b = 0$.  Another more serious issue is that multiple pairs can map to the same rational.  For example the pairs $(2,1)$ and $(4,2)$ each map to the rational number $2$.  So, while it's true that every rational number is a ratio of some pair, it's not true that there's a \emph{unique} such pair.  To make it unique, you might demand that the pair be in lowest terms (and there's also an issue with negative numbers that you'd have to incorporate into the idea of lowest terms) but then you would have a bijection between some \emph{subset} of $\mathbb{Z} \times \mathbb{Z}$ and $\mathbb{Q}$.  So a little bit more is needed to make this work.
+- The map from $\mathbb{Z} \times \mathbb{Z}$ to $\mathbb{Q}$ sending a pair $(a,b)$ to a ratio $\frac{a}{b}$ is right in principle, but not exactly right.  One minor problem with it is that the map is not defined when $b = 0$.  Another more serious issue is that multiple pairs can map to the same rational.  For example the pairs $(2,1)$ and $(4,2)$ each map to the rational number $2$.  So, while it's true that every rational number is a ratio of some pair, it's not true that there's a \emph{unique} such pair.  To make it unique, you might demand that the pair be in lowest terms (and there's also an issue with negative numbers that you'd have to incorporate into the idea of lowest terms) but then you would have a bijection between some of $\mathbb{Q}$ with some subset of $\mathbb{Z} \times \mathbb{Z}$ rather than all of $\mathbb{Z} \times \mathbb{Z}$.  So a little bit more is needed to make this work.
 
 - The bijection from $[0,1]$ to $\mathbb{R}$ doesn't work because the function given would send $0$ to $- \infty$ and $1$ to $+ \infty$.  Somehow there are two extra points you have to sneak in.  It might be helpful to consider the bijection from $\{0,1,2,3, \dots \}$ and $\mathbb{N} = \{1,2,3,\dots \}$ given by sending $n$ to $n+1$.  How could you modify this idea to put $[0,1)$ in bijection with $(0,1)$?