Question for finitists -- 0.999... = 1?

[Jehanne]

Warning: the following post contains a contiguous slice that is a PART of a larger post:

I'm not sure if your proof is valid...

It was valid in discrete math class when I was in University.

You were taught naive set theory, as was I. Little to no introduction to ZFC.

[rocinantexyz]

You were taught naive set theory, as was I.  Little to no introduction to ZFC.

How did you get my transcripts (eta: to figure out what I was and was not taught)?

[Jehanne]

You were taught naive set theory, as was I.  Little to no introduction to ZFC.

How did you get my transcripts?

Because, you were taught from Kenneth H. Rosen's book, and no, I am not clairvoyant.

[rocinantexyz]

Because, you were taught from Kenneth H. Rosen's book, and no, I am not clairvoyant.

That cover says it was written by: László Lovász, József Pelikán, and Katalin Vesztergombi. Why should I believe you instead of the cover?

[Jehanne]

Because, you were taught from Kenneth H. Rosen's book, and no, I am not clairvoyant.

That cover says it was written by: László Lovász, József Pelikán, and Katalin Vesztergombi. Why should I believe you instead of the cover?

Ha! I told you that I was not clairvoyant, and that proves it, doesn't it?!

[polymath257]

Step 1: Prove the expression .999.... makes sense.

Otherwise, you could argue as follows:

x=1+2+4+8+16+...

2x=2+4+8+16+32+...

Hence,

x=1+2x

so

-x=1

x=-1

In particular,

1+2+4+8+... <0.

Not so, since x = 0.999... has an upper bound as the decimal expansion continues without limit, but x = 1+2+4... diverges.

And *that* is the point. A discussion about convergence and what, precisely, a decimal expansion means in terms of limits.

Once the definition is given, the equality is automatic. No algebraic manipulations required. Just the fact that 1/10^n --> 0 as n-->infty.

[rocinantexyz]

Ha!  I told you that I was not clairvoyant, and that proves it, doesn't it?!

Colloquially it might prove something; I'm not sure.

[polymath257]

Not so, since x = 0.999... has an upper bound as the decimal expansion continues without limit, but x = 1+2+4... diverges.

True.  In your case, though, you are performing the same trick - it is just that the additional digit you shifted in is an infinitesimal instead of a large value.

I'm not sure if your proof is valid, but of course one can get as close to correct as you like.

The proof is valid *if* you first demonstrate convergence.

But if you know the definition of convergence, the equality is trivial without the algebraic manipulations.

[polymath257]

Warning: the following post contains a contiguous slice that is a  PART of a larger post:
It was valid in discrete math class when I was in University.

You were taught naive set theory, as was I.  Little to no introduction to ZFC.

ZFC vs naive ST isn't so much the issue here as clarity about what an infinite decimal expansion means. Which means being a bit more precise about what it means to be a real number (although, in this case, we are doing convergence in the rationals).

[polymath257]

Step 1: Prove the expression .999.... makes sense.

Otherwise, you could argue as follows:

x=1+2+4+8+16+...

2x=2+4+8+16+32+...

Hence,

x=1+2x

so

-x=1

x=-1

In particular,

1+2+4+8+... <0.

Not so, since x = 0.999... has an upper bound as the decimal expansion continues without limit, but x = 1+2+4... diverges.

But it is a perfectly good proof in the 2-adics, where the sum 1+2+4+... does, in fact, converge to -1.