Is 0.9 recurring equal to 1?

  • Thread starter curleymatsuma
  • Start date
In summary: It's helpful to remember what we mean by "equal in all respects." 1 is an integer, while .999... is a real number, so they are not equal "in all respects."But we can define and prove a lot of things about real numbers, and in those contexts 1 and .999... turn out to be equal, in that any statement that is true about one of them is also true about the other.This is the sense in which we can say that 1 and .999... are the same number. They are distinct expressions for the same mathematical object.So, to answer the OP's question, yes, in mathematics, 0.9 recurring is accepted as being equal to
  • #1
curleymatsuma
2
0
Hi there, this is my first post here so I'm sorry if this is in the wrong place, asked before, standard newbie apologies :P

So I am well aware of the proof that 0.9 recurring =1

x = 0.99999999...

10 x = 9.99999999...

10x - x = 9.999999... - 0.99999...

9x = 9

x = 1 = 0.999999...

However my question is this. Is 0.9 recurring accepted in mathematics as being equal to 1, or are they considered distinct numbers?

I also understand that there are an infinite number of nines and in pretty much ANY situation theoretical or otherwise the difference is completely negligible. I would just like to clarify in my mind whether this number is thought of as being equal to 1 or not?

Thanks,
Matt
 
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  • #2
curleymatsuma said:
However my question is this. Is 0.9 recurring accepted in mathematics as being equal to 1

Yes. They're two different representations of the same number.
 
  • #3
gb7nash said:
Yes. They're two different representations of the same number.

Thanks! this has been bugging me a bit recently :P
 
  • #4
See the FAQ:

https://www.physicsforums.com/showthread.php?t=507002 [Broken]
 
Last edited by a moderator:
  • #5
curleymatsuma said:
Hi there, this is my first post here so I'm sorry if this is in the wrong place, asked before, standard newbie apologies :P

So I am well aware of the proof that 0.9 recurring =1

x = 0.99999999...

10 x = 9.99999999...

10x - x = 9.999999... - 0.99999...

9x = 9

x = 1 = 0.999999...

However my question is this. Is 0.9 recurring accepted in mathematics as being equal to 1, or are they considered distinct numbers?

I also understand that there are an infinite number of nines and in pretty much ANY situation theoretical or otherwise the difference is completely negligible. I would just like to clarify in my mind whether this number is thought of as being equal to 1 or not?

Thanks,
Matt


1 and .999... are two different representations of the same number, as others have noted.

But the "proof" above, which is often shown to people, is actually a fake proof. That's because without carefully defining the real numbers and proving some theorems about infinite series, you don't actually know for sure that you can multiply an infinitely-long decimal such as .999... by 10 term-by-term; and you don't know that you can subtract one infinitely-long decimal from another. It turns out that you can; but only after you prove you can.
 
  • #6
I think it's more correct to say that 0.9999999... approaches 1 as a limit as the approximation is extended. It seems clear that there will always be a non zero difference between 0.9999999... and 1.
 
  • #7
But 0.999... is the limit. So the difference between 0.999... and 1 is always 0.

Unless of course you mean 0.999... = 0.999...9, in which case there is definitely a non-zero difference. But this is irrelevant to the OP since 0.999... where the decimal expansion never terminates is equal to 1.
 
  • #8
SW VandeCarr said:
I think it's more correct to say that 0.9999999... approaches 1 as a limit as the approximation is extended. It seems clear that there will always be a non zero difference between 0.9999999... and 1.

This is not true. It makes no sense to say that 0.99999... "approaches" 1. It makes as much sense as saying that 2 approaches 3. There is no approaching going on. 0.9999... is just a number.
 
  • #9
SW VandeCarr said:
I think it's more correct to say that 0.9999999... approaches 1 as a limit as the approximation is extended. It seems clear that there will always be a non zero difference between 0.9999999... and 1.

Obviously, if you stop writing 9's and have a finite expansion 0.9999, it will never be 1. But you are writing 0.9999... ( with the dots ) , so you are implicitly talking about the "limiting decimal expansion"
 
  • #10
I'd suggest, possibly pendantic but perhaps also clarification of mathematics:

0.9999... and 1 are the same, and equal, value, but they are different numbers.

A number is a thing we have invented to represent a value. Therefore there may be a many-to-one correspondence, depending on what and why we invented the numbers.
 
  • #11
They have the same value and they are the same number. 1 and 0.999... are absolutely equal in all respects.
 
  • #12
micromass said:
This is not true. It makes no sense to say that 0.99999... "approaches" 1. It makes as much sense as saying that 2 approaches 3. There is no approaching going on. 0.9999... is just a number.

Can we consider the limit of the sum:

[itex]lim \sum_{n=1}^\infty (1 -(1/10)^{n}) = 1[/itex]

?
 
  • #13
SW VandeCarr said:
Can we consider the limit of the sum:

[itex]lim ( 1 - \sum_{n=1}^\infty (1/10)^{n}) = 1[/itex]

?

Limit with respect to which variable?
 
  • #14
micromass said:
Limit with respect to which variable?

I corrected the formula which was only up for about 1 minute. The variable is the exponent of ten ( a positive integer).

In other words I'm talking about an infinite series 9/10 + 9/100 + 9/1000.....
 
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  • #15
cmb said:
I'd suggest, possibly pendantic but perhaps also clarification of mathematics:

0.9999... and 1 are the same, and equal, value, but they are different numbers.

A number is a thing we have invented to represent a value. Therefore there may be a many-to-one correspondence, depending on what and why we invented the numbers.
It's more accurate to say that they are different "expressions", or different "strings of text", that represent the same number.
 
  • #16
SW VandeCarr said:
I corrected the formula which was only up for about 1 minute. The variable is the exponent of ten ( a positive integer).

In other words I'm talking about an infinite series 9/10 + 9/100 + 9/1000.....

Yes, that infinite series converges to a certain number. Namely, 1. Thus

[tex]\sum_{n=1}^{+\infty}{\frac{9}{10^n}}=1[/tex]
 
  • #17
SW VandeCarr said:
I corrected the formula which was only up for about 1 minute. The variable is the exponent of ten ( a positive integer).

In other words I'm talking about an infinite series 9/10 + 9/100 + 9/1000.....
That series would be written as [tex]\sum_{k=1}^\infty \frac{9}{10^k}=\lim_n\sum_{k=1}^n\frac{9}{10^k}[/tex] The sequence [itex]\langle\sum_{k=1}^n 9/10^k\rangle_{n=1}^\infty[/itex] of partial sums is convergent. Its limit is 1.

The string of text "0.99..." represents "the limit of the sequence 0.9+0.09+0.009+...". Since that limit is 1, the two strings of text "1" and "0.99..." represent the same number.

Edit: Lol...it's not the first time that Micromass said what I was going to say while I was typing it. :smile:
 
  • #18
micromass said:
They have the same value and they are the same number. 1 and 0.999... are absolutely equal in all respects.

I can assure you that I am pressing a different number button to type '0.999..' compared with '1' on my number pad.


Like I say, I'd not say it is necessarily more than pedantery, but my point is I would define numbers as human inventions to represent values, whereas values are values. This is a Big Deal for little kids when they start learning maths, and in some cases until someone points out that numbers are just things we've invented to describe the world around us, they often don't get arithmetic because they think there is something magical and God-given, thus unintelligible, about the numbers themselves.
 
  • #19
cmb said:
I can assure you that I am pressing a different number button to type '0.999..' compared with '1' on my number pad.


Like I say, I'd not say it is necessarily more than pedantery, but my point is I would define numbers as human inventions to represent values, whereas values are values. This is a Big Deal for little kids when they start learning maths, and in some cases until someone points out that numbers are just things we've invented to describe the world around us, they often don't get arithmetic because they think there is something magical and God-given, thus unintelligible, about the numbers themselves.

What you are calling numbers, the rest of us call numerals.

What you are calling values, the rest of us call numbers.

If your objections go beyond these syntactic substitutions, please advise. Otherwise, use standard terminology to avoid confusion.
 
  • #20
SteveL27 said:
Otherwise, use standard terminology to avoid confusion.

So, I guess what you're saying implies 'PIN' stands for 'Personal Idetification Numerals'? Never hear that one before!
 
  • #21
cmb said:
So, I guess what you're saying implies 'PIN' stands for 'Personal Idetification Numerals'? Never hear that one before!
This is a math forum.

If this were commerce forum where discussions of PINs were common and there were some ambiguity in terminology, yes, you'd be asked to use that industry's terminology.
 
  • #22
My copy of Chambers Dictionary of Science and Technology says;

Number (Maths.). An attribute of objects or labels obtained according to a law or rule of counting.

I don't see that this is describing a 'value' of a quantity or size. Sounds more like my definition.

Can we clarify what/whose definitions we're using here, then?
 
  • #23
micromass said:
Yes, that infinite series converges to a certain number. Namely, 1. Thus

[tex]\sum_{n=1}^{+\infty}{\frac{9}{10^n}}=1[/tex]

I agree, but infinity isn't a real number. Therefore for 9/10^∞ isn't real. Just like saying the function y = 1/x is never equal zero. Is is "at" infinity, but where is infinity? its not real, so zero is not part of the set of values for y.

In my mind comparing 0.9999... and 1 is apples to oranges: one is real, one isn't. The sum mention above approaches 1 as the number of terms approaches infinity, yes. But you won't EVER have that infinite term, so they won't ever be equal as long as both numbers are real. Once you hit that infinite term, 0.99... becomes non-real.

What solidified my issues was this proof

1/3 = 0.3333...

3(1/3) = 0.99999...
3/3 = 1
 
  • #24
wsabol said:
I agree, but infinity isn't a real number. Therefore for 9/10^∞ isn't real. Just like saying the function y = 1/x is never equal zero. Is is "at" infinity, but where is infinity? its not real, so zero is not part of the set of values for y.

In my mind comparing 0.9999... and 1 is apples to oranges: one is real, one isn't. The sum mention above approaches 1 as the number of terms approaches infinity, yes. But you won't EVER have that infinite term, so they won't ever be equal as long as both numbers are real. Once you hit that infinite term, 0.99... becomes non-real.

What solidified my issues was this proof

1/3 = 0.3333...

3(1/3) = 0.99999...
3/3 = 1

You don't have a term [itex]9/10^\infty[/itex]. Such a term simply does not occur. The notation [itex]\sum_{i=1}^{\infty}{9/10^n}[/itex] is well-defined without using infinity. Indeed, it can be defined as the limit of the partial sums. You don't need to work with infinity to have that.

0.9999... is a real number and it is equal to 1. 0.9999... is as real as 1.00000... There is no difference between recurring 9's and recurring 0's.
 
  • #25
micromass said:
You don't have a term [itex]9/10^\infty[/itex]. Such a term simply does not occur. The notation [itex]\sum_{i=1}^{\infty}{9/10^n}[/itex] is well-defined without using infinity. Indeed, it can be defined as the limit of the partial sums. You don't need to work with infinity to have that.

0.9999... is a real number and it is equal to 1. 0.9999... is as real as 1.00000... There is no difference between recurring 9's and recurring 0's.

I feel like you just made my point... The infinite term simply does not occur, therefore, is will always be 0.999999...9 which is not equal to one. The limit equaling one is exactly that, a limit.

0 is not a member of the set of values in the sequence 9/10^n for n[1,Inf). This sequence is always changing; the derivative is also nonzero. Therefore its sum will never settle at any number, let alone 1. The sum will approach 1, get closer and closer, but never get there, unless you are at Infinity, which doesn't exist.

I don't understand the break down in this logic.
 
  • #26
wsabol said:
I feel like you just made my point... The infinite term simply does not occur, therefore, is will always be 0.999999...9 which is not equal to one. The limit equaling one is exactly that, a limit.

0 is not a member of the set of values in the sequence 9/10^n. This sequence is always changing; the derivative is also nonzero. Therefore its sum will never settle at any number, let alone 1. The sum will approach 1, get closer and closer, but never get there, unless you are at Infinity, which doesn't exist.

I don't understand the break down in this logic.

You are correct that the sequence

[tex]0.9,~0.99,~0.999,...[/tex]

will never settle. It will always come closer and closer to 1. The point is that the very definition of 0.9999... is a limit, namely the limit of the series

[tex]\sum_{n=1}^{\infty}{\frac{9}{10^n}}[/tex]

this is the definition of 0.9999...
The above series will approach 1, and the limit of the above series will equal one. Thus 0.9999..., being defined as a limit, will equal 1.
 
  • #27
I get it. 0.9999... is just short hand for a limit. It isn't a real number in the traditional sense.
 
  • #28
wsabol said:
I get it. 0.9999... is just short hand for a limit. It isn't a real number in the traditional sense.

It is a real number in the traditional sense. Limits of sequences are real numbers!

We call a real number L the limit of [itex](x_n)_n[/itex] if

[tex]\forall \varepsilon>0:\exists n_0:\forall n>n_0:|x_n-L|<\varepsilon[/tex]

So you see that only real numbers can be limits, by definition.


0.9999... is simply notation for the real number defined as the limit of the series

[tex]\sum_{n=1}^{\infty}{\frac{9}{10^n}}[/tex]

This series has a limit which is a real number. We denote this real number by 0.9999...
But 1 is also a limit, thus 1=0.99999...
 
  • #29
cmb said:
My copy of Chambers Dictionary of Science and Technology says;
Number (Maths.). An attribute of objects or labels obtained according to a law or rule of counting.


I don't see that this is describing a 'value' of a quantity or size. Sounds more like my definition.

Can we clarify what/whose definitions we're using here, then?
That definition seems very imprecise to me. We are distinguishing between "number" and "numeral" here.

For example, we can write the number 4 in a variety of ways: 4, IV (in Roman numerals), 100 (in binary). These are different numerals that all represent the same number.

In the same way, 0.999... and 1 are different representations of the same number.

I can't believe we're having this same discussion again!
 
  • #30
micromass said:
It is a real number in the traditional sense. Limits of sequences are real numbers!

We call a real number L the limit of [itex](x_n)_n[/itex] if

[tex]\forall \varepsilon>0:\exists n_0:\forall n>n_0:|x_n-L|<\varepsilon[/tex]

So you see that only real numbers can be limits, by definition.0.9999... is simply notation for the real number defined as the limit of the series

[tex]\sum_{n=1}^{\infty}{\frac{9}{10^n}}[/tex]

This series has a limit which is a real number. We denote this real number by 0.9999...
But 1 is also a limit, thus 1=0.99999...

1 is a real whole number. 0.999... is a limit. That limit is equal to 1, not the real decimal number 0.9999...(as close as you can get to infinity without getting there, because the infinite term of the sequence ever happen)...9
 
  • #31
wsabol said:
1 is a real whole number. 0.999... is a limit. That limit is equal to 1, not the real decimal number 0.9999...(as close as you can get to infinity without getting there, because the infinite term of the sequence ever happen)...9

No, 0.99999... is a notation for a real number. The real number is defined by a limit.

Limits are real numbers.
 
  • #32
micromass said:
No, 0.99999... is a notation for a real number. The real number is defined by a limit.

Limits are real numbers.

Damn you got me. Ok.
 
  • #33
There is also the http://en.wikipedia.org/wiki/Infinitesimal" [Broken] approach.

There is a http://en.wikipedia.org/wiki/Hyperreal_number" [Broken] [itex]\epsilon[/itex] that is smaller than the smallest real number, so we can define the following: [itex] 1 - \epsilon = .999...[/itex].

This implies that [itex]1[/itex] and [itex]1 - \epsilon[/itex] (.999...) are different numbers in the hyperreal numbering system.
 
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  • #34
Matt Benesi said:
There is also the http://en.wikipedia.org/wiki/Infinitesimal" [Broken] approach.

There is a http://en.wikipedia.org/wiki/Hyperreal_number" [Broken] [itex]\epsilon[/itex] that is smaller than the smallest real number, so we can define the following: [itex] 1 - \epsilon = .999...[/itex].

This implies that [itex]1[/itex] and [itex]1 - \epsilon[/itex] (.999...) are different numbers in the hyperreal numbering system.


I fear you have not fully understood hyperreals. In the hyperreals, the definition [itex]1-\varepsilon=0.9999...[/itex] is not made. Furthermore, in the hyperreals, there is no such thing as the smallest real number.
 
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  • #35
This topic has come up so often here that we have an FAQ that addresses this concept: [thread]507001[/thread].

Thread closed.
 

1. Is 0.9 recurring equal to 1?

Yes, 0.9 recurring is equal to 1. This is because in decimal notation, 0.9 recurring represents an infinite decimal expansion of 0.9999... which is equivalent to the whole number 1.

2. How is it possible for 0.9 recurring to equal 1?

This concept can be explained using mathematical proofs and the concept of limits. Essentially, as the number of 9s in the decimal expansion of 0.9 recurring increases, the difference between 0.9 recurring and 1 becomes infinitesimally small, approaching a limit of 0. Therefore, for all practical purposes, 0.9 recurring can be considered equal to 1.

3. Can you provide an example to demonstrate that 0.9 recurring is equal to 1?

One way to demonstrate this is by using the geometric series formula: a/(1-r), where a is the first term and r is the common ratio. In this case, a=9 and r=1/10. Plugging these values into the formula, we get 9/(1-1/10) = 9/(9/10) = 10. This shows that 0.9 recurring is equivalent to the whole number 10, which is equal to 1.

4. Is 0.9 recurring equal to 1 a mathematical fact or just a convention?

This is a mathematical fact that can be proven using various mathematical concepts and principles. It is not simply a convention or assumption.

5. Are there any real-world applications or implications of 0.9 recurring being equal to 1?

Yes, this concept has real-world applications in fields such as mathematics, physics, and engineering. It is particularly useful in understanding and solving problems involving infinite series and limits. Additionally, the understanding of this concept can also help in accurately representing and interpreting data and measurements in various scientific studies and experiments.

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