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off topic: 0.999 repeating Equals 1, proof inside!

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Frosty90
Wed Dec 09 2009, 07:13AM
Frosty90 Registered Member #1617 Joined: Fri Aug 01 2008, 07:31AM
Location: Adelaide, South Australia
Posts: 139
 		There is another simple way to show this, and you can also show that 0.333...=1/3 with the same technique:

let n=0.999...

then 10n=9.999...

then 10n-n=9n=9.999... - 0.999...=9

then n=9/9=1=0.999...

QED

likewise:

let m=0.333..

then 10m=3.333...

then 10m-m=9m=3.333...-0.333=3

then m=3/9=1/3

so no assumptions need to be made, only that the ellipsis means you have an unlimited repeating pattern after the decimal point to play with!

Cheers,
Jesse

Edit: Sorry I didnt see that Dr Slack had already written this.
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Andyman
Thu Feb 11 2010, 10:13PM
Andyman Registered Member #1083 Joined: Mon Oct 29 2007, 06:16PM
Location: Upland, California
Posts: 256
 		Sorry for gravedigging this topic, but i just wanted to share another method for proving this.
In a geometric series, we have the first n terms defined as
19fd6ad9c868fccaec943f840355cd43
If n is infinity and r is less than 1, then r^n = 0 and we get the sum as a/(1-r)
We can say .9999999 is the same as .9 + .09 + .009..... so a=0.9=9/10 and r=1/10

The sum of each term is therefore (9/10)/(1-1/10) = (9/10)/(9/10) = 1

Of course this is assuming that a fraction raised to infinity equals exactly zero...
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Frosty90
Sat Feb 13 2010, 06:22AM
Frosty90 Registered Member #1617 Joined: Fri Aug 01 2008, 07:31AM
Location: Adelaide, South Australia
Posts: 139
Of course this is assuming that a fraction raised to infinity equals exactly zero...

It is "in the limit".

Jesse
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Steve Conner
Sat Feb 13 2010, 10:17AM
Steve Conner Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
 		Engineer's proof:

0.999... is within a few percent of 1.
If we just round it up to 1, nobody will ever notice the difference.
Therefore 0.999... = 1.
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WaveRider
Sun Feb 14 2010, 02:31PM
WaveRider Registered Member #29 Joined: Fri Feb 03 2006, 09:00AM
Location: Hasselt, Belgium
Posts: 500
 		A simple proof of this is based on evaluating the limit of

y = lim x->inf (x / (1+x))

By induction, you can show that 0.999999... can be made arbitrarily close to 1..

Cheers!
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Steve Conner
Sun Feb 14 2010, 06:37PM
Steve Conner Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
 		Granted, but my proof is better because it works for anything bigger than 0.95 smile
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