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June 26th, 2018, 09:54 AM   #1
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An Anomaly of Decimal Representation

An Anomaly of Decimal Representation of Real Numbers

1/3=.333333333......... therefore
3(1/3)=1=.999999...............
The resolution is that 1/3 $\displaystyle \neq$ .33333.......
Proof:
If 1/3 = .333333..... then the open interval (0,1/3) has a largest member, which violates a fundamental axiom of the real numbers.

The situation is analogous to:
.00000......0001, with an endless number of 0's.
If this were equal to 0, the open interval (0,1/3) would have a smallest member.
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June 26th, 2018, 11:03 AM   #2
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Quote:
Originally Posted by zylo View Post
An Anomaly of Decimal Representation of Real Numbers

1/3=.333333333......... therefore
3(1/3)=1=.999999...............
The resolution is that 1/3 $\displaystyle \neq$ .33333.......
Proof:
If 1/3 = .333333..... then the open interval (0,1/3) has a largest member, which violates a fundamental axiom of the real numbers.

The situation is analogous to:
.00000......0001, with an endless number of 0's.
If this were equal to 0, the open interval (0,1/3) would have a smallest member.
Sigh. Again? You start with true statements then jump to false non-sequiturs. Yes, 1/3= 0.33333.... Yes 3(1/3)= 1= 0.999999....

There is no "resolution" needed. Both of those statements are true. But the statement making up your "proof" is incorrect. "If 1/3= 0.3333..." (and it is) it does not follow that the open interval (0, 1/3) would have a largest member. Are you thinking that 0.3333... would be in that interval? "If 1/3= 0.3333..." then obviously 0.3333... could NOT be in that interval.

On the other hand ".00000.....0001 with an endless number of 0's" simply is meaningless. You can't put a "1" at the end of an endless number of 0's!
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June 26th, 2018, 11:19 AM   #3
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Quote:
Originally Posted by zylo View Post

The situation is analogous to:
.00000......0001, with an endless number of 0's.
Zylo, this shows that you are confused about what decimal representation is.

First, I hope you agree that there is no end to the sequence of positive integers 1, 2, 3, 4, 5, 6, ...

Now in a decimal representation, to the right of the decimal point there is one digit for each positive integer. So there can be no "last" decimal digit, for exactly the same reason there's no last positive integer.
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June 26th, 2018, 01:09 PM   #4
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$\dfrac{1}{3} = \displaystyle \sum_{j=1}^{\infty} \dfrac{3}{10^j} \implies$

$3 * \dfrac{1}{3} = 3 * \displaystyle \sum_{j=1}^{\infty} \dfrac{3}{10^j} \implies$

$1 = \displaystyle \sum_{j=1}^{\infty} \dfrac{9}{10^j} \implies$

$1 + \displaystyle \sum_{j=1}^{\infty} \dfrac{0}{10^j} = \displaystyle \sum_{j=1}^{\infty} \dfrac{9}{10^j}.$

There is NOTHING to resolve. Line 4 follows from line 1 by simple logic.

This of course does not represent a proof that 1/3 does equal 0.333.... But there is no contradiction
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June 26th, 2018, 02:00 PM   #5
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Quote:
Originally Posted by JeffM1 View Post
Line 4 follows from line 1 by simple logic.
Actually line 3 does not follow from line 2 by simple logic. Rather, term-by-term multiplication of a convergent infinite series by a constant is a theorem that must be proved. It does not follow from the axioms for real numbers without first, a careful definition of the real numbers, and second, a careful definition of the limit of a convergent infinite series.
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June 26th, 2018, 10:38 PM   #6
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Quote:
Originally Posted by Maschke View Post
Zylo, this shows that you are confused about what decimal representation is.
Critically, a decimal representation of a number is not a number. It is a representation of it. In fact, it is a representation of an convergent infinite series. You can then use JeffM1's derivation (or even just the definition of convergence) to prove that $1 = 0.999\ldots$. Once you have that proof, there is no need for a "resolution" because there is no "anomaly".

You are treating the decimal representation of a number as a mathematical object. It isn't. It's a representation of a mathematical object.



Ceci n'est pas un pipe - "this is not a pipe" it's a picture of a pipe. It's a representation of a pipe. You cannot use it as a pipe.



A similar message here: the representation of the view is not the view. Although (and now I'm straying off the real point) in this case it's a little more complicated, because the painting in the picture is not a painting - it's a representation of a painting. Something that is brought into greater relief in The Two Mysteries:



So the pipe is not a pipe, it's a picture of a pipe. And the picture of a pipe is not a picture of a pipe.
Except that the pipe is not a pipe, but a representation of a pipe. But the representation of a pipe (the picture within the picture) is a representation of a pipe and also a representation of a representation of a pipe.
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June 27th, 2018, 12:00 AM   #7
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So we have a function
$$\{0,1,...,9\}^\mathbb{N}\rightarrow [0,1]$$
which sends $(\alpha_n)_n$ to $\sum_{n=1}^{+\infty} \frac{\alpha_n}{10^n}$.

This map is surjective, but not injective, so what is the big deal with this??
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June 27th, 2018, 09:07 AM   #8
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For the sake of continuity, please rename this thread to "Zylo doesn't understand real numbers #219"
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June 27th, 2018, 10:07 AM   #9
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1/3 is a limit point of the open set (0,1/3). As such, no member of (0,1/3), including .3333333........, can equal 1/3.
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June 27th, 2018, 10:27 AM   #10
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Quote:
Originally Posted by zylo View Post
1/3 is a limit point of the open set (0,1/3). As such, no member of (0,1/3), including .3333333........, can equal 1/3.
It is right that 1/3 is a limit point of the open set (0,1/3).
It is right that no member of (0,1/3) can equal 1/3 (although this doesn't follow from 1/3 being a limit point).
It is not right that 0.3333... is an element of (0,1/3).
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