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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
‘I can’t be having with this.’ - Esmeralda Weatherwax
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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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