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January 16th, 2019, 11:10 AM   #11
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Quote:
 Originally Posted by zylo Maschke. Please give "Counter example" {whatever that means} JeffM1. Please give your definition of a real number and a representative member, other than "x" or "a." EDIT I wavered between "represented by" and "defined by". But what is it you are representing? A Dedekind cut? How do you define a Dedekind cut without numbers, and then what decimal do you assign to it?
Quote:
 Originally Posted by JeffM1 A representative real number is 1. According to you it has a unique decimal representation. Prove it. You are the one trying to demonstrate something. I am under no obligation to do anything. I accept that you may define things as you please, but then what you have to say has no relevance to anyone else.
In decimal, we can express the number one as:

$$1$$
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 01:59 PM.

January 16th, 2019, 11:40 AM   #12
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Quote:
 Originally Posted by zylo 1 is a real number by definition: it is a unique decimal sequence. "No relevance to anyone else?" You mean to you, don't you
I completely agree that $1$ has a decimal expansion.
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, 11:42 AM #13 Senior Member   Joined: Dec 2015 From: somewhere Posts: 514 Thanks: 80 $\displaystyle 0. \overline{9}$ converges to $\displaystyle 1$ and not equal to $\displaystyle 1$
January 16th, 2019, 11:50 AM   #14
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Quote:
 Originally Posted by idontknow $\displaystyle 0. \overline{9}$ converges to $\displaystyle 1$ and not equal to $\displaystyle 1$
That's like saying the number 2 converges to 1. Does it? What does it mean for a number to converge? Can you prove the number 2 doesn't converge to the number 1?

 January 16th, 2019, 12:08 PM #15 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 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 12:20 PM.
January 16th, 2019, 12:18 PM   #16
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Quote:
 Originally Posted by zylo 1 is a real number by definition: it is a unique decimal sequence.
Really? You can't think of another, entirely different decimal that evaluates to 1?

 January 16th, 2019, 04:56 PM #17 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,162 Thanks: 879 Math Focus: Wibbly wobbly timey-wimey stuff. Same old arguments as many times before... -Dan
January 16th, 2019, 05:08 PM   #18
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Quote:
 Originally Posted by zylo Convergence in the conventional sense has nothing to do with this.
Nope. Convergence has everything to do with the value of a decimal representation.

 January 17th, 2019, 03:56 AM #19 Senior Member   Joined: Dec 2015 From: somewhere Posts: 514 Thanks: 80 For each statement , a proof is needed. Show a proof that $\displaystyle 0. \overline{9} =1$ .(or disprove it)
January 17th, 2019, 05:17 AM   #20
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Quote:
 Originally Posted by idontknow For each statement , a proof is needed. Show a proof that $\displaystyle 0. \overline{9} =1$ .(or disprove it)
Any one of these work for you?

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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