And it's only a small step from here to Q, which kinda looks like ZxZ using (numerator, denominator)
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Better yet, algebraic numbers, i.e. roots of non-zero polynomials with integer coefficients, also form a countable set, of which ℤ and ℚ are subsets.pic.twitter.com/xmsvzc6oqu
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As amazing as this mathematical phenomenon is, neither Billy Gibbons nor Dusty Hill of ZZ Top ever won the Fields medal.pic.twitter.com/Da7iBKdMfv
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How about: for (x, y) ∈ ℤ x ℤ, compute 2^x ⋅ 3^y. This maps to a unique ℚ by fundamental theorem of arithmetic, and any subset of ℚ is countable. Argument extends to any ℤ^n.
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And this is also the proof of Q is countable...
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And Julio Le Parc from the Op'Art movements knew about it! https://www.geogebra.org/m/cM7jFTUw#material/t9ZpaQN5 …pic.twitter.com/r6KROFLZn2
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set of integers, my guy
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can you visualize for us Z^3 please?
@fermatslibrary -
Unroll the blue curve of the picture into a straight line (which will represent Z^2 in a single axis), and draw another axis with Z, which will give a representation of Z^2×Z. Now create the blue curve as before! Someone should make a gif showing this.
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