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Is $2^{|\mathbb{N}|} = |\mathbb{R}|$? If so, how?

I was reading the Wiki page on the , and it says "Moreover, $\mathbb{R}$ has the same number of elements as the power set of $\mathbb{N}$", but I don't see how this is true?

I feel like it has something to do with binary, but I'm not too sure how it works? Do I have to show a map of all reals can be done in binary? I'm just very confused, and any advice would be appreciated!

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    What on earth does "a map of all reals can be done in binary" mean?2012-10-08
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    Have you tried searching this site for the answer?2012-10-08
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    See also: http://math.stackexchange.com/questions/553526/the-set-of-real-numbers2016-08-09
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    and see this http://math.stackexchange.com/questions/1885700/cardinality-of-power-set-of-mathbb-n-is-equal-to-cardinality-of-mathbb-r/1885782 . Basically you can map $\{0,1\}^N$ to [0,1] via mapping $(x_1,x_2, x_3,..... )$ to $\sum x_i/2^i$. The sum is a decimal expansion but written in binary. So the set of such sums is all reals between [0.1]. It's a simple matter to show $|[0,1]| = |\mathbb R|$. So we have $|\{0,1\}^N| = 2^N = |[0,1]| = |\mathbb R|$2016-08-09

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