What goes wrong here ...

Forum for the GRE subject test in mathematics.
mobius70
Posts: 19
Joined: Thu Jan 24, 2008 1:11 am

What goes wrong here ...

Postby mobius70 » Sat Aug 30, 2008 12:10 am

Hi

I was going through Cantor Second Diagonal Method. It pertains to proof that there are uncountable real numbers between 0 and 1.
Lets say x is a number such that ..
x = 0.a1 a2 a3 ... an .... (ak takes values between 0 and 9 both included)
I quite did not figure out why there is a condition on ak.
ak can't be 9 for all values greater than equal to n for some constt n.

A related question .. while i was trying to understand above mentioned theorem.

Lets say x= 0.199999999999999999.......ad infinitum
then 10x = 1.999999999999999.........ad infinitum
also 100x = 19.9999999999999999........ad infinitum

it implies 90x = 18
it implies x = 0.2

I CAN'T FIGURE THIS OUT..
SINCE 0.2 != 0.19999999999999.......ad infinitum

Plz help me understand this.

User avatar
lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am

Postby lime » Sat Aug 30, 2008 3:40 pm

What you're actually asking here is better to rephrase as
"Whether is 1=0.9999999... ?"
Googling on this would be possible to find myriad forum threads concerning this question. The answer does exist though.

YES. Both "1" and "0,9999999..." denote the same real number. Indeed, real numbers are something that exist independently of our attempts (like numeral systems, which as we see can be quite ambiguous ) to represent them.
The rigorous proof of this fact comes from axiomatic definition of real numbers. Either by Dedekind cuts or Cauchy sequences.

Also note, between any different real numbers there always is another one (continuous property). There is no such number between "1" and "0,99999999...". Then they represent the same number.

http://en.wikipedia.org/wiki/Talk:0.999... - here is a good link concerning this problem.

mobius70
Posts: 19
Joined: Thu Jan 24, 2008 1:11 am

Postby mobius70 » Mon Sep 01, 2008 1:41 am

Thanks lime.




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