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I am a software engineer without a math degree, so I am planning to learn something today.

Take this bijection between the naturals and reals. (This is a valid bijection, no?)

...03020 => 0.02030... 
...11111 => 0.11111... 
...51413 => 0.31415...
.
.
.

Walking along the diagonal, we can find a real number not listed. However, that would also find a natural number not listed as well, correct?

brandon_busby
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  • You need to elaborate how the above describes a bijection between the naturals and the naturals. – copper.hat Sep 18 '22 at 21:02
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    Your "natural number" not listed will have infinitely many digits. – Conifold Sep 18 '22 at 21:06
  • That was supposed to say "naturals and reals". Fixed. Is more elaboration still needed? – brandon_busby Sep 18 '22 at 21:06
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    $\ldots 03020$ is not a natural number unless all but finitely many of the digits in $\ldots$ are $0$. – Robert Israel Sep 18 '22 at 21:07
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    This is a very frequently asked question, it was asked just a few days ago and the farthest back I can find goes back to 2011: https://math.stackexchange.com/questions/35107/why-doesnt-cantors-diagonal-argument-also-apply-to-natural-numbers – Qiaochu Yuan Sep 18 '22 at 21:09
  • OK, I see. Natural numbers cannot have infinitely many non-zero digits. What is the why? Why do the reals get this special treatment, but not the natural numbers? – brandon_busby Sep 18 '22 at 21:10
  • The set of real numbers whose expansion is zero after some point is a subset of the rationals and so countable. See @RobertIsrael's comment. – copper.hat Sep 18 '22 at 21:11
  • "What is the why?" The reals simply… are bigger. You might as well ask why a teacup has the special treatment that it can hold tea, while a plate does not. – Patrick Stevens Sep 18 '22 at 21:12
  • It is not special treatment. Natural numbers come from counting, talking about their digits came much later and is an incidental effect of their positional representation. The same concerns real numbers. They reflect geometric measurement, continuum, infinite divisibility, continuous motion, etc. It so happens that representing that positionally requires infinity of digits. You can extend natural numbers to transfinite ordinals, and get a bijection between some of those and reals, but we do not use them for counting, except in abstract mathematics. – Conifold Sep 19 '22 at 03:17

2 Answers2

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All natural numbers have finitely many digits.

Adam Rubinson
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Forget diagonals, the point of being an bijection is that you can go both ways. If your map is a bijection, it has an inverse that sends reals to naturals. But where would this inverse send the famous real number $1/3 = 0.33333333...$?

Vincent
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  • This is similar to my example of $\pi/10$, no?

    ...51413 => 0.31415...

     – brandon_busby Sep 18 '22 at 21:17
  • Yes, but what is this number on the left? – Vincent Sep 18 '22 at 21:19
  • Let me ask a more specific question. How many digits does the natural number on the left in your example have? – Vincent Sep 18 '22 at 21:20
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    I'm not sure that this is so much about bijections as it is about my misunderstanding of natural numbers. – brandon_busby Sep 18 '22 at 21:21
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    Aaah, I now see your comment: 'why can reals have infinitely many digits and naturals can not?' What is important to realize is that neither can have infinitely many digits before the decimal point – Vincent Sep 18 '22 at 21:21
  • Natural numbers are 1, 2, 3, 4, 5, 6 etc. For each natural number $n$ you can count from 1 to $n$ in finite time. The number of digits of $n$ is even smaller than $n$ itself (because there are multiple numbers with the same number of digits, e.g. 14 and 15 both have two digits) so it will certainly be finite. – Vincent Sep 18 '22 at 21:23
  • Every natural number has the property that you write it down in full. It may take some time, like in the case of google plex, but in principle there is no need to resort to thing like '...' – Vincent Sep 18 '22 at 21:24
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    Thanks for pointing that out! It is nice to realize that reals don't have special treatment. Any particular real or integer must be bounded. I'm sure the are not the right words, but the difference then is that the "granuarity" of reals is "unbounded". – brandon_busby Sep 18 '22 at 21:26
  • Yes I think your last comment gets it exactly right. – Vincent Sep 18 '22 at 21:27
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    (Minor nitpick on my last comment: the notion that both reals and naturals are bounded, but reals, unlike naturals, have unbounded granularity does explain why your bijection is not a bijection, but it does not by itself explain why reals are uncountable. Confusingly enough the rational numbers, which also have unbounded granularity in the same way as the reals can be brought into bijection with the natural numbers, just not in the way of the original post. But maybe that is a subject for a separate question) – Vincent Sep 18 '22 at 21:29