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"It’s larger than the rational numbers in the sense that it is a strict superset" Ok, did I miss the explanation of that? Or is it something in "part 2" which
by perl4ever 7y ago
"It’s larger than the rational numbers in the sense that it is a strict superset"
Ok, did I miss the explanation of that? Or is it something in "part 2" which I didn't see a link to?
- shawnz 7y agoClearly every single rational number is "useful", plus others that are not rational
- perl4ever 7y agoSo I guess what I lack is an understanding of why that doesn't affect cardinality.
- shawnz 7y agoThere is a 1-to-1 mapping between the sets, so by definition that means they have the same cardinality. Let me give a more common example. Consider these two sets: N, the set of non-negative integers, and Z, the set of all integers. Clearly, everything in N is also in Z, and then some. But N and Z still have the same cardinality, because there are 1-to-1 mappings between the two sets. Here is one example of such a mapping: N | Z --------- 0 -> 0 1 -> 1 2 -> -1 3 -> 2 4 -> -2 5 -> 3 6 -> -3 ..etc. The formula for this mapping would be floor(n/2)*(-1)^(n%2). Clearly everything in the left set has exactly one corresponding item in the right set and vice versa, so they must be the same "size", even though the right set contains every item in the left set and then some.