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/sci/ - Science & Math

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5784871 No.5784871 [Reply] [Original]

is this proof enough for 1=.999999....

x=.999999.....

10x=9.999999.....

10x=9+x

9x=9

x=1=.999999....

>> No.5784888

Yes.

>http://www.youtube.com/watch?v=G_gUE74YVos

>> No.5784894

The proof is correct, but it depends on arithmetic properties of infinite decimal expansions that most people who dispute the original claim won't be familiar with; therefore, it won't convince most people who don't already know it.

>> No.5784896

Prove that 10x=9.999999...

>> No.5784921

>>5784888
thanks for the youtube link

>> No.5784923

At breakfast my alphabets cereal in milk proved it.
The cereal did the subtraction, 1-0.999999=0
It came up with 0.0000000. . .
Or maybe it was cheerios.

>> No.5784942
File: 106 KB, 953x613, 0.999 = 1.jpg [View same] [iqdb] [saucenao] [google]
5784942

>> No.5784973

>1-0.999...=0.000...1
>0.000...1=0 because you never reach 1
>thus 0.999...=1 as in the expression 1-x one is the only value for x to result in 0

>> No.5784977

>>5784973
This girl knows whats up.

>> No.5785055

>>5784871
nope, lot of holes in this demonstration. First you need to prove that your x is a real number and not +-infinity or a non-defined expression ( if you don't do that you can prove that x=1-1+1-1+1-1+1-....=1/2 ) but that's not everything, you would need to prove that arithmetical operation behave as you writed it with this infinite decimal sequence etc...

This is a nice proof and it's a correct proof but not a complete proof. see the blue part of >>5784942 if you want a complete proof.

>> No.5785078

How can it be proof, when the result is wrong?

>> No.5785710

0.99999...<a<1

There is no a, so 0.99999...=1