Instigator / Pro
TNBinc's avatar
7
TNBinc
1534
rating
5
debates
80.0%
won
Topic
#2237

Is 0.9999 with the nine repeating equal to 1

Status
Finished

The debate is finished. The distribution of the voting points and the winner are presented below.

Winner & statistics
Better arguments
3
0
Better sources
2
0
Better legibility
1
1
Better conduct
1
1

After 1 vote and with 5 points ahead, the winner is...

TNBinc
Tags
Math
Education
Parameters
Publication date
Last updated date
Type
Standard
Number of rounds
3
Time for argument
Two days
Max argument characters
10,000
Voting period
One week
Point system
Multiple criterions
Voting system
Open
Contender / Con
seldiora's avatar
2
seldiora
1417
rating
158
debates
32.59%
won
Description

repeating: an action done over and over again
equal: same in numerical amount/value

Round 1
TNBinc's avatar
TNBinc
Pro
#1
Thanks, seldiora.

In this argument, when I mention 0.999 repeating, it is basically the 9 repeating which is 0.9999999999999999999999999999999999999999999999999999999...


Burden of Proof

I must prove 0.999 repeating is equal to 1 better than seldiora in order to win this debate. My opponent wins if he does the opposite better.


My round one argument is based on three proofs:

  • Algebraic equation proof
  • Denominator of 9s proof
  • Geometric series
Proof 1: Algebraic equation proof

Let x equal 0.999 repeating
This means:
x = 0.999 repeating
By multiplying both sides by 10, I get:
10x = 9.999 repeating
9.999 repeating is equal to 9 + 0.999, so I can express the equation like this:
10x = 9 + 0.999 repeating
Recall from the first line x is 0.999 repeating. I can substitute x for this.
10x = 9 + x
I will subtract x from both sides.
9x = 9
I will divide by 9 on both sides.
9/9 x = 9/9
9/9 is equal to 1
x = 1

By using transitive property (a=b and b=c so a=c), I can say 1 = 0.999 repeating since x = 0.999 repeating and x = 1 in this equation. 

Proof 2: Denominator of 9s

In the book Competition Math For Middle School by J. Batterson, the author says on page 164, where he talks about repeating decimals, "The repeating block of a fraction whose denominator is 9, 99, 999, etc. in the denominator will be the numerator. Use leading zeroes where necessary."

This means if we have a fraction whose denominator is composed of nines as digits (9, 99, 999, etc.), the numerator will keep repeating with leading zeroes if the number of digits in the denominator is not equal to the numerator.

He also gave various examples after the quoted text

5/9 = 0.5 with "5" repeating (0.555555...)
24/99 = 0.24 with the "24" repeating (0.2424242424...)
817/999 = 0.817 with the "817" repeating (0.817817817...)
1/999 = 001/999 = 0.001 with the "001" repeating since the # of digits in the numerator must be equal to the number of digits in the denominator. (0.001001001...)

Using this logic, 999/999 should be equal to 0.999 repeating (0.999999999...), or 99/99 = 0.999 repeating.

However, it is known that n/n = 1 for any number n. In this case, 999/999 = 1

Using transitive property (a=b and b=c so a=c), I can deduce 0.999 repeating = 1.

Proof 3: Infinite geometric series with sum.

According to Purplemath
0.999 repeating can be expressed as an infinite geometric series.

0.999 repeating = 9/10 + 9/100 + 9/1000 + ...
This is equal to:
0.999 repeating = 9/10^1 + 9/10^2 + 9/10^3 + ...

In a geometric sequence, there are two parts: "a", or the first term, and "r" or a common ratio.
In this example, a is 9/10 and r is 1/10.

Recall that the sum of an infinite geometric series is a/1-r.

This means I can express the sum as (9/10)/(1-1/10) or (9/10) * (1)/(1-1/10).

This is equal to (9/10)* (1)/(9/10). Notice that (1)/(9/10) is equivalent to 10/9

This is, using the statement above, 9/10 * 10/9. This is 9*10/10*9 = 90/90 = 1.

Therefore, 0.999 repeating is equal to 1.


I look forwars to CON's argument.

Sources used
1: https://i.imgur.com/ydKGqb3.png (referenced as "this image" in Proof 1)
2: Competition Math For Middle School by J. Batterson (used in Proof 2)
3: https://www.purplemath.com/modules/howcan1.htm (referenced as "PurpleMath" in Proof 3)
seldiora's avatar
seldiora
Con
#2
My opponent has failed to address all types of numbers, including non-real numbers. If we look at the Wikipedia page on this math statement, it says

Some proofs that 0.999... = 1 rely on the Archimedean property of the real numbers: that there are no nonzero infinitesimals. Specifically, the difference 1 − 0.999... must be smaller than any positive rational number, so it must be an infinitesimal; but since the reals do not contain nonzero infinitesimals, the difference is therefore zero, and therefore the two values are the same.
However, there are mathematically coherent ordered algebraic structures, including various alternatives to the real numbers, which are non-Archimedean. Non-standard analysis provides a number system with a full array of infinitesimals (and their inverses).[49] A. H. Lightstone developed a decimal expansion for hyperreal numbers in (0, 1)∗.[50] Lightstone shows how to associate to each number a sequence of digits,
{\displaystyle 0.d_{1}d_{2}d_{3}\dots ;\dots d_{\infty -1}d_{\infty }d_{\infty +1}\dots ,}
indexed by the hypernatural numbers. While he does not directly discuss 0.999..., he shows the real number 1⁄3 is represented by 0.333...;...333... which is a consequence of the transfer principle. As a consequence the number 0.999...;...999... = 1. With this type of decimal representation, not every expansion represents a number. In particular "0.333...;...000..." and "0.999...;...000..." do not correspond to any number.
The standard definition of the number 0.999... is the limit of the sequence 0.9, 0.99, 0.999, ... A different definition involves what Terry Tao refers to as ultralimit, i.e., the equivalence class [(0.9, 0.99, 0.999, ...)] of this sequence in the ultrapower construction, which is a number that falls short of 1 by an infinitesimal amount. More generally, the hyperreal number uH=0.999...;...999000..., with last digit 9 at infinite hypernatural rank H, satisfies a strict inequality uH < 1. Accordingly, an alternative interpretation for "zero followed by infinitely many 9s" could be
{\displaystyle {\underset {H}{0.\underbrace {999\ldots } }}\;=1\;-\;{\frac {1}{10^{H}}}.}[51]
All such interpretations of "0.999..." are infinitely close to 1. Ian Stewart characterizes this interpretation as an "entirely reasonable" way to rigorously justify the intuition that "there's a little bit missing" from 1 in 0.999....[52] Along with Katz & Katz, Robert Ely also questions the assumption that students' ideas about 0.999... < 1 are erroneous intuitions about the real numbers, interpreting them rather as nonstandard intuitions that could be valuable in the learning of calculus.[53][54] Jose Benardete in his book Infinity: An essay in metaphysics argues that some natural pre-mathematical intuitions cannot be expressed if one is limited to an overly restrictive number system:
The intelligibility of the continuum has been found–many times over–to require that the domain of real numbers be enlarged to include infinitesimals. This enlarged domain may be styled the domain of continuum numbers. It will now be evident that .9999... does not equal 1 but falls infinitesimally short of it. I think that .9999... should indeed be admitted as a number ... though not as a real number.
In layman terms, this means if we assume there is a smallest number X where X>0 but there is no number such that X>Y>0, then X is the difference between 1 and 0.999... As all my opponent's formulas rely on the idea that 0.999... is a real number, he fails the idea when applied to hyperreal numbers.

Round 2
TNBinc's avatar
TNBinc
Pro
#3
From round 1:

Proof 1: Algebraic equation proof
Con has not said anything wrong about this proof, so he has dropped it.

Proof 2: Denominator of 9s
Con has not said anything wrong about this proof, so he has dropped it.

Proof 3: Infinite geometric sequence with sum
Con has not said anything wrong about this proof, so he has dropped it.

Rebuttals:

My opponent has failed to address all types of numbers, including non-real numbers. If we look at the Wikipedia page on this math statement, it says
Right, but these proofs and also these numbers do not have to do anything with non-real numbers, which deal with the imaginary unit i.

If I am not wrong, you are trying to describe that 0.999 repeating and 1 are not the same, but there is no numerical difference. However, for two equal values, their difference is zero, which means it has no numerical difference. Therefore, 0.999 repeating and 1 are the same.


Suppose I have a square like this image. How much of the area is on the left side?

The answer would be one half. However, if your argument would apply to this, according to you, it would not be half because there is a small border between the left and right side. However, recall that a line is only 1-D and has no width, only length. This means there is nothing on the line, therefore it is indeed half. 

The line in the square is like the area between 0.999 repeating and 1. It does not have any width, therefore, it does not exist. Previously, I have said that two values with a difference of zero are equal. This means that again, 0.999 repeating and 1 are equal.

In layman terms, this means if we assume there is a smallest number X where X>0 but there is no number such that X>Y>0, then X is the difference between 1 and 0.999... As all my opponent's formulas rely on the idea that 0.999... is a real number, he fails the idea when applied to hyperreal numbers.
However, like in the line example, there is no difference between 0.999 and 1, therefore they are the same number.

According to brilliant.org 0.999 is equal to 1 because "Since no number exists between 0.999… and 1, it must be that they are the same."

What my opponent has said, fails to realize my Proof 2 in round 1. When you say this, you say a repeating decimal is not real, but a fraction equivalent to the same repeating decimal is?

I look forward to CON's round two.



seldiora's avatar
seldiora
Con
#4
my opponent has not addressed infinitesimals in calculus being essential, a function when going with X to infinity could be "equivalent" to 0.999... but never reach 1. That's why his argument is wrong. His proof only put 0.999... as a real number, and doesn't consider the possibility that it might not be the same interpretation in every circumstance.
Round 3
TNBinc's avatar
TNBinc
Pro
#5
My Rebuttals:

a function when going with X to infinity could be "equivalent" to 0.999... but never reach 1
I see your argument.

Brilliant.org has a wiki page for this argument.

They have listed some common rebuttals and refuted them. Here is one of the rebuttals on that website, which I believe is your claim.

Rebuttal: 0.999… only tends to 1. It is not equal to 1. We only have an approximation.
Reply: It is true that for the sequence a_n = 0.999, it is the limit as n→∞ that is equal to 1. However, since 0.999… is defined to be that limit, it is defined to equal 1.
Without using this definition for infinitely repeating decimals, there would be many numbers, such as 1/3 = .333..., that we wouldn't be able to write out as decimals since there exists no finite sum of tenths, hundredths, thousandths, etc. that exactly equals 1/3
Seldiora says
 doesn't consider the possibility that it might not be the same interpretation in every circumstance.
What do you mean by this? A number is defined as the exact same value in every equation and every time it is used. You can't say 3 is equal to 4 or any other number. 


Argument

Brilliant.org has also listed many other claims to which they refuted.
Rebuttal: 0.999... and 1 are not equal because they're not the same decimal. With the exception of trailing 0's, any two decimals that are written differently are different numbers.
Reply: 0.999...=1 is another case when two decimals that are written differently are, in fact, the same number. The fact that there are two different ways to write 1 as a decimal is a result of the role that infinite sums have in defining what non-terminating decimals mean.
The decimal system is just a shorthand for writing a number as a sum of the powers of 10, each scaled by an integer between 0 and 9 inclusive. For example, 0.123 means 1/10 + 2/100 + 3/1000. Similarly, 0.999... means 9/10 + 9/100 + 9/1000 + ... and the value of this infinite sum is equal to 1.
Infinite sums are possible:
Rebuttal: Infinite sums don't make any sense. It's not possible to add up infinitely many things, so any infinite sum is only an approximate value, not a real value.
Reply: Without using infinite sums, there would be many numbers, such as 1/3 = .333..., that we wouldn't be able to write out as decimals. The definition of such an infinite sum is rigorous, but strange in that the sum is defined to equal the limit approached as many terms are added together, whenever this limit exists.
Not all infinite sums of fractions can be evaluated. For example, 1/2 + 2/3 + 3/4 + 4/5 + ... could not have a real numerical value. But the definition of an infinite sum includes the restriction that an infinite sum only has a well-defined valuation when, as we add up terms of the series, the total sum zooms in towards one, specific value. In the case of 3/10 + 3/100 + 3/1000 + ..., 1/3 is the value being approached. There would be no other way to define 1/3 as a decimal otherwise since 1/3 is not equal to any finite sum of tenths, hundredths, thousandths, etc.
In the case of 9/10 + 9/100 + 9/1000 + ..., 1 is the value being approached as more and more of the terms are added together.
For a more complete explanation of infinite sums, check out the Infinite Sums wiki page.

Brilliant.org also says, about the limits,
This argument is valid in that the perspective that we're adding "term by term" is how we are evaluating the limit.

Conclusion

Con has not rebutted any of my proofs or said that there is anything wrong with them; this means that he is acknowledging, to his perspective, that everything I said in those proofs is correct.

Con's first argument does not make sense, because two numbers are not equal if there is a nonzero difference between those numbers. This means there is a number between those two quantities. There is no number between 0.999... and 1 because all the digits are at max (9) and there are an infinite number of digits. This means 0.999... and 1 have a difference of 0, which means they are equal since equal numbers have a difference of 0. There is no "little bit missing" as Con says in his argument.

Con cannot say that there is a number that is equal to 0.999... but not equal to 1 because according to the previous statement, there is no number between them which means it is a zero difference. Hence, they are equal.

If Con says that in his example of X and Y that there is no number Y, this means that there is no difference between X and 0, which means X=0, which invalidates his argument.

I used three proofs in describing how 0.999... is equal to 1.



I hope CON had a good time debating with me in this debate.

Vote Pro!!!
seldiora's avatar
seldiora
Con
#6
my opponent misses my point which is that there is an infinitesimal that is equal to 1-0.999..., equal to 1-1/10^h, given above. He has not negated this possibility.

The crux of 0.999... and the problem of it, is that the infinite sums of 0.9, 0.09, 0.009... only add up to the limit of 1, and hence is not equal. It will never equal 1, not even if X equals infinity. It is just an incredibly small value which no value is smaller. 1/10^h with H= infinity approaches zero and is not zero. This is why 0.999 =/=1.