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>if you shift the decimal point by an infinite number of places, then there are still an infinite number of 9s to the right

>0.9bar7 is a completely nonsensical number

For the same reason that 0.9...7 isn't a meaningful number, you cannot move the decimal 'an infinite number of times' and then after this, look at what number you have left and see it still has infinite 9s left. It's like you're trying to perform transfinite induction on the set of numbers generated by moving the decimal point. You can only move the point a countable number of times, so there is no sense in which the property can still be true after infinity many times.



> For the same reason that 0.9...7 isn't a meaningful number, you cannot move the decimal 'an infinite number of times' and then after this, look at what number you have left and see it still has infinite 9s left.

Yes you absolutely can, for exactly the same reason. 0.9bar7 is nonsensical precisely because you can move the decimal to the right an infinite number of times, and still have an infinite number of 9s before the 7.

> You can only move the point a countable number of times

Not true, the implicit definition of “...”, the very statement that there are an “infinite” number of 9s, means exactly the opposite of what you claim, it means you can move the decimal an infinite number of times.


In your proof you say

>And here is the logical (induction) step: if you shift the decimal point by an infinite number of places, then there are still an infinite number of 9s to the right

This is not how induction works. The induction shows that you can shift the decimal point any finite number of steps to the right and there will still be infinite 9's after it. If you want to show something is still true after infinity steps, you require transfinite induction, but this doesn't make sense because the '...' decimal representation only represents a countably infinte number of 9's.

This is the same reason 0.9bar7 doesn't make sense - because decimal representations only have countably many digits.


>because decimal representations only have countably many digits

This should rather say that it's because decimal representations have digits indexed by the natural numbers I guess, rather than by any larger countable ordinal




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