Prove .999 repeating = 1

Homework Statement:

Prove .999 repeating = 1

Relevant Equations:

.999 repeated = 1
Let there be a number .999 repeated m times where m is a natural number, which is the smallest number less than 1.

Since there is a number .999 + .0000 with a 9 in the m+1 position, it that follows the number of the form .9999, with an infinite amount of 9's, is bigger. Since infinity doesn't end, there is no finite number of the form .9999 with x 9's, x being a natural number, that is less than 1.

Since the number of the form .999 with 9's repeating x amount of times, x being a natural number, is not less than 1, but it's not greater than 1 either, it must be 1.

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Infrared
Gold Member
That isn't really a proof. Try working from whatever definition of a decimal expansion you're using, which should be something like ##0.\overline{9}=0.9+0.09+\ldots## or ##0.\overline{9}=\lim_{n\to\infty}0.9...9## (where there are ##n## nines on the RHS). If you know about geometric series (and even if not!), you should be able to evaluate this.

That isn't really a proof. Try working from whatever definition of a decimal expansion you're using, which should be something like ##0.\overline{9}=0.9+0.09+\ldots## or ##0.\overline{9}=\lim_{n\to\infty}0.9...9## (where there are ##n## nines on the RHS). If you know about geometric series (and even if not!), you should be able to evaluate this.
I dont see how this can be done without assuming that there is a thing such as infinity. Like, I don't know what infinity is.

Infrared
Gold Member
You mentioned in your other thread that you are reading Apostol's calculus book. I'm sure it has a good account of limits. You don't need a concept of "infinity" to define limits.

You mentioned in your other thread that you are reading Apostol's calculus book. I'm sure it has a good account of limits. You don't need a concept of "infinity" to define limits.
It does, I read it before, but I'm trying to do this from memory. I'm working on this right now, and I'll report back tomorrow. Hopefully I don't have to go back and read the book again.

ehild
Homework Helper
Homework Statement:: Prove .999 repeating = 1
Relevant Equations:: .999 repeated = 1

Let there be a number .999 repeated m times where m is a natural number, which is the smallest number less than 1.

Since there is a number .999 + .0000 with a 9 in the m+1 position, it that follows the number of the form .9999, with an infinite amount of 9's, is bigger. Since infinity doesn't end, there is no finite number of the form .9999 with x 9's, x being a natural number, that is less than 1.

Since the number of the form .999 with 9's repeating x amount of times, x being a natural number, is not less than 1, but it's not greater than 1 either, it must be 1.
Your way of thinking is correct, but I want to show you how to find the fraction that is equal to an infinitely repeating decimal number.
Let be N= 0. abcabcabc.... 3 dgits repeating here. Multiply it by 10^3 , and subtract : N
1000N -N =abc.abcabc... -0.abcabc = abc.0
999N= abc, so N= abc/999
You can do the same with any length of repeating period.

atyy, jack action, Adesh and 2 others
Thanks. So x = .99..... n times where n approaches infinity
10x = 9.99999999 n times

10x-x = 9
9x=9

x=1 = .999...... n times

I guess that's simple enough but I really like proof by contradiction and wish I could have done it that way : ( Re-reading my OP it wasn't really that coherent anyways. Just gotta keep on grinding and keep doing problems/reading and understanding and hopefully I can get this math stuff down. Hope to be a regular in this community from here on.

Delta2
PeroK
Homework Helper
Gold Member
Thanks. So x = .99..... n times where n approaches infinity
10x = 9.99999999 n times

10x-x = 9
9x=9

x=1 = .999...... n times

I guess that's simple enough but I really like proof by contradiction and wish I could have done it that way : ( Re-reading my OP it wasn't really that coherent anyways. Just gotta keep on grinding and keep doing problems/reading and understanding and hopefully I can get this math stuff down. Hope to be a regular in this community from here on.
If you want a formal proof, you first need a formal definition of what ##0.999 \dots## actually means mathematically.

If you want a formal proof, you first need a formal definition of what ##0.999 \dots## actually means mathematically.
And wouldn't that require defining infinity as something that actually exists? I don't see how you could do this without circular logic.

Infrared
Gold Member
No, see my post 2. If you know about limits, you can define ##0.999...## as the infinite sum ##\sum_{n=1}^\infty \frac{9}{10^n}.##

dRic2 and jack action
PeroK
Homework Helper
Gold Member
And wouldn't that require defining infinity as something that actually exists? I don't see how you could do this without circular logic.
If you can't define ##0.999 \dots##, then you can't do anything with it.

Have you studied infinite sequences and series?

No, see my post 2. If you know about limits, you can define ##0.999...## as the infinite sum ##\sum_{n=1}^\infty \frac{9}{10^n}.##
You can define .999 repeating infinite may times, by defining it as an infinite sum? That still requires infinity to be defined

Infrared
Gold Member
No it doesn't. The statement "##\lim_{n\to\infty}a_n=a##" means: for every ##\varepsilon>0##, there exists a natural number ##N## such that if ##n>N##, then ##|a_n-a|<\varepsilon.## I don't need a concept of infinity.

And the infinite sum ##\sum_{n=1}^\infty\frac{9}{10^n}## just means ##\lim_{m\to\infty}\sum_{n=1}^m\frac{9}{10^n}.##

JD_PM, dRic2, jbriggs444 and 2 others
PeroK
Homework Helper
Gold Member
You can define .999 repeating infinite may times, by defining it as an infinite sum? That still requires infinity to be defined
If you look at the definition of a infinite sequence or series, it does not require a definition of "infinity". A sequence, for example, is a function on the natural numbers ##\mathbb{N}##.

PS @Infrared is doing all the work for you here. If you are not prepared to go to your calculus book and learn the formal definition of limits, sequences and series, then we can't help you very much.

rxh140630
Gotcha. I guess I'm going to go back to the book to really make sure I understand everything. Thanks infrared.

Infrared
ehild
Homework Helper
Thanks. So x = .99..... n times where n approaches infinity
Do not say "n times". it is 0.9999999 ... extending forever.
When you learnt about decimal numbers, you used the term "infinite".pi, for example, when written out, is an infinite decimal number. When the decimal number consists of repeating units, repeating forever, it is a rational number, and can be written as a fraction. Otherwise, when the decimal number does not terminate or does not have a forever repeating section, it is an irrational number.
https://en.wikipedia.org/wiki/Repeating_decimal

Delta2
etotheipi
Gold Member
2019 Award
Informally, you can think of the "=" sign as a game. Let's say you give me two numbers ##a## and ##b##, and we want to check whether they're equal. If and only if I cannot find another number between ##a## and ##b##, then ##a## and ##b## are equal.

In the case of ##a = 9.\dot{9}## and ##b=10##, whatever guess I make for a number closer to ##10## you can always refute by writing down a bunch more ##9##'s. So since I can't find a number between them, they're equal.

It's not a proof, but a little bit of intuition

Delta2
FactChecker
Gold Member
Suppose there is a definite difference between .99999... and 1. How much is the difference? You can logically get to a contradiction long before you have to use infinite '9's. So a "proof by contradiction" does not need to use infinity; it just needs to use enough '9's to get a contradiction. The availability of enough '9's is guaranteed by the infinity.

WWGD
jbriggs444
Homework Helper
2019 Award
No it doesn't. The statement "##\lim_{n\to\infty}a_n=a##" means: for every ##\varepsilon>0##, there exists a natural number ##N## such that if ##n>N##, then ##|a_n-a|<\varepsilon.## I don't need a concept of infinity.

And the infinite sum ##\sum_{n=1}^\infty\frac{9}{10^n}## just means ##\lim_{m\to\infty}\sum_{n=1}^m\frac{9}{10^n}.##
Note that the ##\infty## that appears in that ##\lim_{m \to \infty}## syntax does not literally mean "infinity". It is a syntactic placeholder. Instead of reading it as "as m approaches infinity", you could do as my calculus instructor suggested and read it as "as m increases without bound".

Then all you need is the idea that there is this set of things called natural numbers and that they have an order. The question of whether the set has infinitely many members does not arise in the definition of the limit.

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Infrared
The open interval (0, 1), i.e. the interval from 0 to 1 not including the endpoints, and the closed interval [0, 1], i.e. the interval from 0 to 1 including the endpoints, by definition have equal diameter.

Although the closed interval includes the 2 endpoints that are not included in the open interval, the 2 endpoints, like all the other points on the number line, by definition have a length of 0, so their inclusion or exclusion does not affect the diameter.

We can thus say that by definition 0.999... = 1, just as the half-open interval (0, 1] by definition has a diameter of 1.

jbriggs444
Homework Helper
2019 Award
The open interval (0, 1), i.e. the interval from 0 to 1 not including the endpoints, and the closed interval [0, 1], i.e. the interval from 0 to 1 including the endpoints, by definition have equal diameter.

Although the closed interval includes the 2 endpoints that are not included in the open interval, the 2 endpoints, like all the other points on the number line, by definition have a length of 0, so their inclusion or exclusion does not affect the diameter.

We can thus say that by definition 0.999... = 1, just as the half-open interval (0, 1] by definition has a diameter of 1.
The usual definitions for decimal notation do not mention anything about "diameter".

Instead, one might start by defining the real numbers as equivalence classes of Cauchy sequences of rationals and decimal strings as identifying specific Cauchy sequences that may be taken as exemplars of equivalence classes. Then a decimal string is defined to denote the real number of which its sequence is an exemplar. Two decimal strings are "equal" if they are both exemplars of the same equivalence class.

There are other ways to proceed. But when arguing that something is true "by definition", it is good to be able to state the definition that one is using.

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dRic2
Gold Member
I like @Infrared approch, but I was wondering if this could be considered a prof:

1) ##0. \bar 9 = 3 * 0. \bar 3##
2) ##0. \bar 3 = \frac 1 3##
3) ##0. \bar 9 = 3 * 0. \bar 3= 3 * \frac 1 3 = 1##

The second step is kind of a definition, but it's pretty intuitive. I don't think this is a valid prof, but It was the first thing that came ti my mind.

PeroK
Homework Helper
Gold Member
I like @Infrared approch, but I was wondering if this could be considered a prof:

1) ##0. \bar 9 = 3 * 0. \bar 3##
2) ##0. \bar 3 = \frac 1 3##
3) ##0. \bar 9 = 3 * 0. \bar 3= 3 * \frac 1 3 = 1##

The second step is kind of a definition, but it's pretty intuitive. I don't think this is a valid prof, but It was the first thing that came ti my mind.
There are a number of good informal proofs like this. Another is ##1 - 0. \bar 9 = 0.\bar 0 = 0##.

But, really, you need the definition as an infinite sum for a formal proof

jack action and dRic2
The usual definitions for decimal notation do not mention anything about "diameter".

Instead, one might start by defining the real numbers as equivalence classes of Cauchy sequences of rationals and decimal strings as identifying specific Cauchy sequences that may be taken as exemplars of equivalence classes. Then a decimal string is defined to denote the real number of which its sequence is an exemplar. Two decimal strings are "equal" if they are both exemplars of the same equivalence class.
From: https://en.wikipedia.org/wiki/Interval_(mathematics)

Bounded intervals are bounded sets, in the sense that their diameter (which is equal to the absolute difference between the endpoints) is finite. The diameter may be called the length, width, measure, range, or size of the interval.​

Perhaps the term 'diameter' is something you prefer to reserve for the distance of a maximum-length chord across (the Greek prefix 'dia' means across) a circle or sphere; however, a distance across an interval may be viewed in a manner like that of 'across a bridge', i.e. spanning a distance between 2 points, and hence diameter is a legitimate descriptor of an interval distance.
There are other ways to proceed. But when arguing that something is true "by definition", it is good to be able to state the definition that one is using.
I note that you did not disagree with the contentions regarding the equivalences being in accordance with definitions. There are many workable definitions. One such is: $$(\text{Min.}~ x:\forall(n>0)[\frac 1 {9^n}<x]=1)⇒(x=1)~ Def.$$(equivalent to the Archimedean Property, from Euclid 's Elements Book V definition 4, from Eudoxus' Axiom). I do not prefer such a definition over preservation of the infinitesimal. I think that in ordinary English it is absurd to say that 1 is the least number than which something is infinitesimally less and then to call that something consequently equal to 1; however, we can say that it is so because we define it to be so, and then we can decide elsewhere which to embrace: inconsistency or incompleteness.

jbriggs444
Homework Helper
2019 Award
From: https://en.wikipedia.org/wiki/Interval_(mathematics)

Bounded intervals are bounded sets, in the sense that their diameter (which is equal to the absolute difference between the endpoints) is finite. The diameter may be called the length, width, measure, range, or size of the interval.​

Perhaps the term 'diameter' is something you prefer to reserve for the distance of a maximum-length chord across (the Greek prefix 'dia' means across) a circle or sphere; however, a distance across an interval may be viewed in a manner like that of 'across a bridge', i.e. spanning a distance between 2 points, and hence diameter is a legitimate descriptor of an interval distance.

I note that you did not disagree with the contentions regarding the equivalences being in accordance with definitions. There are many workable definitions. One such is: $$(\text{Min.} x:\forall(n>0)[\frac 1 9^n<x]=1)⇒(x=1) Def.$$(equivalent to the Archimedean Property, from Euclid 's Elements Book V definition 4, from Eudoxus' Axiom). I do not prefer such a definition over preservation of the infinitesimal. I think that in ordinary English it is absurd to say that 1 is the least number than which something is infinitesimally less and then to call that something consequently equal to 1; however, we can say that it is so because we define it to be so, and then we can decide elsewhere which to embrace: inconsistency or incompleteness.
You need a definition of the meaning of a decimal string before you can pontificate on what it is or is not equal to. No amount of careful reasoning about Archimedian properties, measure theory or the like can gain any traction without that.