January 16th, 2019, 12:10 PM  #11  
Senior Member Joined: Jun 2014 From: USA Posts: 479 Thanks: 36  Quote:
Quote:
$$1$$ or as $$0.\overline{9}$$ Therefore, one is a counterexample because it has two decimal representations. Any number of the form $\frac{n}{10^m}$, where $n$ is an integer and $m$ is a natural number, will have two decimal expansions (except $0$, which only has one). The rest of the reals will have only one decimal expansion. Last edited by skipjack; January 16th, 2019 at 02:59 PM.  
January 16th, 2019, 12:40 PM  #12  
Senior Member Joined: Oct 2009 Posts: 733 Thanks: 247  Quote:
You should prove however that $1$ cannot have more than $1$ expansion. How do you know there isn't some weirdo expansion that happens to equal $1$? Or do you DEFINE real numbers as such, by their decimal expansions?  
January 16th, 2019, 12:42 PM  #13 
Senior Member Joined: Dec 2015 From: iPhone Posts: 387 Thanks: 61 
$\displaystyle 0. \overline{9}$ converges to $\displaystyle 1$ and not equal to $\displaystyle 1$

January 16th, 2019, 12:50 PM  #14 
Senior Member Joined: Oct 2009 Posts: 733 Thanks: 247  
January 16th, 2019, 01:08 PM  #15 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 
You are talking about applications of the decimal definition. 1 has no expansion. A decimal number can have an unlimited number of digits, but doesn't have to. 33 and 333333.... are both valid (different) decimal numbers. Defining the natural numbers by decimal digits is trivial: 1, 2, 3, ... Application of a decimal point to all the natural numbers gives (defines) all the real numbers in [0,1). But that's been gone over many times in previous posts. .9 and 1 are not the same thing. But considering them the same is part of the practical application of the decimal definition of the real numbers. EDIT One way to define all the real numbers in [0,1) is to list the natural numbers, put a period after them, and read them in reverse. 1. Is .1 2. Is .2 . 10. Is .01 . Last edited by zylo; January 16th, 2019 at 01:20 PM. 
January 16th, 2019, 01:18 PM  #16 
Senior Member Joined: Aug 2012 Posts: 2,157 Thanks: 631  
January 16th, 2019, 05:56 PM  #17 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,042 Thanks: 815 Math Focus: Wibbly wobbly timeywimey stuff. 
Same old arguments as many times before... Dan 
January 16th, 2019, 06:08 PM  #18 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,600 Thanks: 2588 Math Focus: Mainly analysis and algebra  
January 17th, 2019, 04:56 AM  #19 
Senior Member Joined: Dec 2015 From: iPhone Posts: 387 Thanks: 61 
For each statement , a proof is needed. Show a proof that $\displaystyle 0. \overline{9} =1$ .(or disprove it) 
January 17th, 2019, 06:17 AM  #20  
Senior Member Joined: Jun 2014 From: USA Posts: 479 Thanks: 36  Quote:
https://math.stackexchange.com/quest...edirect=1&lq=1 I say if $0.\overline{9} < 1$, then there must be a real number $x$ such that: $$0.\overline{9} < x < 1$$ In that case, what is the decimal representation of $x$? The OP says it must be unique.  

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