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Is $...000n$ (i.e a natural number with an infinite number of zeros on the left) a natural number?

If it is a natural number, can't we use something like Cantor's diagonalization method to prove that $\mathbb{N}$ is uncountable (I know that it is false, and that $\mathbb{N}$ is countable)?

Thanks in advance

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4 Answers 4

up vote 2 down vote accepted

To answer the other question: Cantor diagonalization cannot be used in this case: if you try using it for "most" choices, the digit you'll have will be $0$. So infinitely many times, you'll need to pick a digit $a \neq 0$, but this would create an expression with infinitely many non-zero digits, which is not a number.

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It's a little odd to write a number that way, but it's clear. However, every natural number written that way has a "leftward tail" of zeros, so that the diagonalization argument won't actually work: the new number you create won't be a natural.

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Yes.

Or in more detail, $\ldots 000n$ is (shorthand for) an infinite string of digits which notates a decimal. And that decimal does represent a natural number.

Note that $n$ and $\ldots 000n$ are two notations for the same decimal. A decimal has digits in every place, and our convention is that if we only write a string of finitely many digits, then all of the remaining places are filled with zeroes.

With standard interpretations, only the decimals whose left end is repeated zeroes actually notate real numbers.

There are some extended interpretations that allow more decimals to represent real numbers. In 10's complement, negative numbers can be represented by decimals whose left end is repeated 9's: e.g. $\ldots 9999 = -1$. Also, there are extensions that allow you to interpret decimals recurring to the left to be rational numbers: e.g. $\ldots010101 = -\frac{1}{99}$. However, these extensions are not in common use.

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$0's$ on the left don't affect the value of a number, so that $0....0n=n$

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