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Question for finitists -- 0.999... = 1?
#1
Question for finitists -- 0.999... = 1?
And, ultra-finitists, too!



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#2
RE: Question for finitists -- 0.999... = 1?
At work.

Oh...... 'Maths'....

*Brain fritz*
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#3
RE: Question for finitists -- 0.999... = 1?
I once had a real, professional mathematician corner me at a party and spend 20 minutes explaining to me why the interval from 0 to 1 is not the same as the interval from 1 to 2.

Because of the resultant trauma, I am unable to watch that video.

Boru
‘But it does me no injury for my neighbour to say there are twenty gods or no gods. It neither picks my pocket nor breaks my leg.’ - Thomas Jefferson
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#4
RE: Question for finitists -- 0.999... = 1?
0.000 ... 001 = 0

1 / ∞
"The first principle is that you must not fool yourself — and you are the easiest person to fool." - Richard P. Feynman
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#5
RE: Question for finitists -- 0.999... = 1?
0.999=0.999
1=1
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#6
RE: Question for finitists -- 0.999... = 1?
Define:

x = 0.999...

Multiply both sides by 10

10x = 9.999...

Isolate integer part

10x = 9 + 0.999...

By definition of x

10x = 9 + x

Subtract x from both sides

9x =9

Divide both sides by 9

x = 1
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#7
RE: Question for finitists -- 0.999... = 1?
(November 26, 2022 at 12:21 pm)LinuxGal Wrote: Define:

x = 0.999...

Multiply both sides by 10

10x = 9.999...

Isolate integer part

10x = 9 + 0.999...

By definition of x

10x = 9 + x

Subtract x from both sides

9x =9

Divide both sides by 9

x =  1

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.
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#8
RE: Question for finitists -- 0.999... = 1?
(November 26, 2022 at 5:17 pm)polymath257 Wrote: 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.
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#9
RE: Question for finitists -- 0.999... = 1?
(November 26, 2022 at 5:54 pm)LinuxGal Wrote: 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.
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#10
RE: Question for finitists -- 0.999... = 1?
Warning: the following post contains a contiguous slice that is a PART of a larger post:
(November 26, 2022 at 6:04 pm)HappySkeptic Wrote: I'm not sure if your proof is valid...
It was valid in discrete math class when I was in University.
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