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1 = 0.999…?

February 20, 2011

There is a proof of this that runs like so

First let x = 0.999..., the … (triple) dots mean an infinite number of digits of 9s.

Then multiplying both sides by 10, we get 10x = 9.999...

Thus solving for x, we subtract the first equation from the second getting (10 - 1) x = 9.999... - 0.999...

From this we obtain 9x = 9. This implies x = 1. So the proof goes.

My first question is, do people accept this as a valid proof? Apparently lots of mathematicians do. I find it hard to accept this proof. If anything, even if I accept the proof, I can only conclude that the system of real numbers does indeed result in a contradiction! The number 0.999... can be subjected to equivocation. That is the problem. It is not obvious that what is meant here is that $0.999… = 3 (0.333…) = 3 (1/3)$. In which case there is not even need for a proof. It is a renaming exercise and no genius needed.

At any rate look at the proposed proof. I am skeptical in the technical sense due to the fact that we are not told what 0.999... means. It gives the impression that one starts off with an assumption, x = 0.999..., then one derives that x = 1. Philosophically as I understand real numbers, 0.999… is never the same as 1.0, the former always falls short from the latter. That is if it is taken as a value. In logical terms in the light of the proposed proof, this means one starts off with an assumption \phi and then from it one arrives at \neg \phi. Hence \phi \vdash \neg \phi, implying that \vdash \phi \to \neg \phi. However, IMHO, this is a contradiction.

If anything, the proof system allows for a contradiction and that is not cool.

OK, not convinced of my objection? Would you then accept that 0 = 0.000...?

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