MATH questions

It seems most people will agree that 0.333... = 1/3. Since 0.999... is 3 x 0.333... , it is equal to 3 x 1/3 = 1.

Yes, I know this is not a formal proof. But this was the explanation that finally had me convinced.

I'm not sure what you are saying here.

I am not disproving anything. 0.999..=1 is only true by definition in the real number system because infinitesimals are not allowed and repeating decimals are interpreted as the limit of an infinite series. I was arguing that the popular proofs such as using 1/3=0.333.. only try to cleverly hide this fact.

You're attacking proofs by referncing things that don't exist in the number system that they're taking place in. You explicitly said that the x=.999..., 10x=9.999..., proof is flawed because of the existence of infinitesimals.

Essentially you're doing the equivalent of telling someone who says that square roots of negative numbers don't exist in the real number system that they're wrong because i exists in the complex number system.

The statement 0.999...=1 is practically equivalent to saying that infinitesimals do not exist in the reals.

It is pointless to "prove" that 0.999...=1 if you assume that infinitesimals do not exist in your proof.

A better analogy: here is the proof that the square root of -1 does not exist (in the reals)

sqrt(-1)

=sqrt(2*-1/2)

=sqrt(2) *sqrt(-1/2)

But sqrt(-1/2) doesn't exist so neither does sqrt(-1).

Obviously this isn't really showing anything.

It's not an infinitesimal you are talking about, it is zero.

In other words, by starting out with the assumption that 1 - 0.999... is something infinitesimal number other than zero and then basing the subsequent counterexample to the proof on that all you've done is construct a tautology.

I'd ask you this:  What infinitesimal number can you add to 0.999... that gives you a number still less than one?  (Hint, it's half if the "infinitesimal" number that is the difference between 0.999... and 1)

Arguing that a proof will not work without assuming the result is not a tautology but a valid criticism.

Just help me understand...are you saying that since it is "only true by definition", that you have created a different and equally useful system of mathematics where this is NOT true by definition?

http://en.wikipedia.org/wiki/Hyperreal_number

the sum of all the natural numbers is -1/12

http://www.wolframalpha.com/input/i=sum+x%3D1+to+infinity+9%2F%2810%5Ex%29

consider that sum. It converges to 1 meaning if you added every value with x replaced from 1 to infinity it equals 1. the moment you stop adding you have a finite value near 1 but not 1, you have to keep adding forever

I was not aware of this "Leibniz bridge to Abraham Robinson" alternate approach to calculus that does not make use of limits.

Is it "equally useful"?

The link isn't working for me.

But anyway convergence is an infinite limit of a series, not the explicit infinite sum. 

DO NOT WANT TO WASTE MY TIME ON THIS STUFF!!!!!!!

Too late.

This is the longest I have seen a thread in these forums last without extreme trolling and name-calling...(knock on wood).

You've aragued that the proof won't work based on the assumption that 1 - 0.999... is a non zero number (i.e. infinitesimal).  This is the same as assuming that 0.999... is not equal to one.  You are starting out with your conclusion as an assumption.

If this fact were true, you could tell me what that infinatesimal number was (aside from zero).  You could also halve it, add it to .999... and arrive at a number that was not 1.

You are saying that there is indeed a counterexample that meeds the conditions above but you can't provide it.

This is kind of twisting it, don't you think?

I am not disproving that 0.99.. =1, and I am not assuming that 1-0.999.. isn't zero. I am not arguing for a side but against the validity of the 9x=1 proof.

I am saying that the latter was simply assumed in the proof to prove the former, which makes the entire proof a tautology

weeeee