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diff --git a/Problem Set 2.page b/Problem Set 2.page
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@@ -99,7 +99,7 @@ Proofs:
 
 -$\mathbb{Q}=\mathbf{N}$ by combining the bijections from $\mathbf{N}$ to $\mathbf{Z}$ and from $\mathbf{N}$ to $\mathbb{N} \times \mathbb{N}$. This provides a bijection from $\mathbf{N}$ to $\mathbb{Z} \times \mathbb{Z}$, and since every element of $\mathbb{Q}$ can be represented as the ratio of the two components of an ordered pair of integers, we have a bijection from $\mathbb{Z} \times \mathbb{Z}$ to $\mathbb{Q}$.
 
--Josh explained Cantor's proof of the uncountability of the real numbers on the 28th; Wikipedia provides a good description thereof: [external](http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument). In a nutshell, you assume you can have an ordered list of the reals (i.e., a bijection to the naturals), then construct a real number not on that list by having its nth digit be different from the nth digit of the nth number on the list.
+-Josh explained Cantor's proof of the uncountability of the real numbers on the 28th; [Wikipedia](http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument) provides a good description thereof. In a nutshell, you assume you can have an ordered list of the reals (i.e., a bijection to the naturals), then construct a real number not on that list by having its nth digit be different from the nth digit of the nth number on the list.
 
 -$(0,1)=\mathbf{R}$ and $[0,1]=\mathbf{R}$ under the same bijection: $n \mapsto \tan\left (\pi n - \pi/2 \right)$.