Someone help me with the 0.9 repeating = 1 proof

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Jun 26, 2006
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Eeezee said: Someone help me with the 0.9 repeating = 1 proof

IF 0.9 with nines repeating comes from lim (x->1) x=1 and the function f(x)=x THEN you can prove using theorems in calculus that f(x) at x=1 APPROACHES AND BECOMES EQUAL TO 1 (basically the limit equals the function value and it is a continous function) FOR THE SPECIFIC FUNCTION, LIMIT AND INPUT VALUE. BUT THE LIMIT DOES NOT NECESSARILY EQUAL THE VALUE OF THE FUNCTION HENCE THERE IS NO PROOF THAT 0.9 WITH NINES REPEATING EQUALS 1. HERE IS A EXAMPLE:

f(x)= { x if x is not equal to one
50000000 if x is equal to one}

Another example:

lim (x->1) (x-1)/(x^2-1)=0.5

(x-1)/(x^2-1) at x=1 is INDETERMINATE. FUNCTION VALUE DOES NOT EQUAL THE LIMIT!
 

edcarman

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May 23, 2005
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Originally posted by: Lucent
(x-1)/(x^2-1) at x=1 is INDETERMINATE. FUNCTION VALUE DOES NOT EQUAL THE LIMIT!

This is one of the reasons why limits are needed in maths.

Look at the function y = x/sinx

Obviously if x = 0 then the function is undefined (0/0), but if you plot the function (e.g. in Excel between x = -3 and x = 3) you will see that it is definitely tending towards something as x gets closer to 0. What it does at x = 0 (for the moment) depends on the charting program (in Excel, you get a #Div/0 and y=0 on the chart - depending on how you've set it to plot absent values)

To find the value that y is approaching near x (Lim(x->0) x/sinx ) we take the derivative of the top and bottom functions to get:

(d/dx x)/(d/dx sinx) = 1/cosx

At x = 1, 1/cosx = 1

This allows us to conclude that lim(x->0) x/sinx = 1 (See links to L'hopital's rule below for explanation)

The function thus tends to 1 as x approaches 0.

Note: It's been a while since I studied this, so I am unsure whether it is valid to say
y(0) = 0/sin0 = 1 (I think not), but
lim(x->0) x/sinx = 1 is definitely valid

Cheers
Ed

L'Hopital's Rule - Wolfram MathWorld
L'Hopital's Rule - Wikipedia

And just so this is not entirely OT:
Proof that 0.999... = 1
Proof that 0.999... equals 1 - Wikipedia
Repeating Decimal - MathWorld

Edit: A general explanation that will make your brain hurt:
Decimal Expansion - MathWorld
 
Jun 26, 2006
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Originally posted by: edcarman

Note: It's been a while since I studied this, so I am unsure whether it is valid to say
y(0) = 0/sin0 = 1 (I think not), but
lim(x->0) x/sinx = 1 is definitely valid

To put it succiently, in your example, the function will NEVER, EVER, EVER be equal to the limit value, only get darn close to it. (Function is not continous) :)
 
Jun 26, 2006
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Thanks for the links edcarman. Looking at one page http://mathforum.org/dr.math/faq/faq.0.9999.html it says:

So in modern mathematics, the string of symbols 0.9999... = 1 is understood to mean "the infinite sum 9/10 + 9/100 + 9/1000 + ...". This in turn is shorthand for "the limit of the sequence of numbers

9/10,
9/10 + 9/100,
9/10 + 9/100 + 9/1000,
...."

So the function f(x)=x cannot represent this infinite sum. The infinite sum is represented by:

Sum 9/10^n
(n=1 -> Infinity)

Although the limit can be calculated to be one, I don't see anyway the sum can actually reach 1. If there are any mathmaticians that can demonstrate how this sum can actually equal 1, that would be awesome. But as I see it, the limit is equal to 1 but the sum will never actually be equal to 1 (1 will be a horizontal asymptote). Hence .999 ad nauseam does not equal 1 to my knowledge.
 

edcarman

Member
May 23, 2005
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Originally posted by: Lucent
To put it succiently, in your example, the function will NEVER, EVER, EVER be equal to the limit value, only get darn close to it. (Function is not continous) :)

Having given it more thought you're absolutely right - it won't equal ANYTHING at the limit value. This is one of those things that will tie your brain in knots if you try to visualise it physically - there's a 'hole' in the middle, but the 'hole has no size so you can't see it.

The problem (I think) is that people are confusing the limit of a function with an infinite sum:

lim(x?a) f(x) = b (a,b are any real numbers) does not necessarily mean that f(a) = b

whereas:

?(n = i to 8) fn(x) = g(x) (n integer) can have some exact value at x = a

Take for example:
?(n = 0 to 8) (-1)^n/(2n)! * x^2n = cosx

or more topically:
?(n = 1 to 8) 9/10^n = 1

I think the original proof using 1/3 (= 0.333...) + 2/3 (=0.666...) = 1 (=0.999...) is flawed as it requires you to make the assumption that 1/3 = 0.333... (you are essentially assuming what you want to prove - some discrete fraction = a repeating decimal)

The algebraic proof using x = 0.999... as shown above and in the first proof that I linked to is better as here your proof follows logically from your definition of x, without your definition of x having to rely on what you are about to prove.

Ed

PS. I am an engineer so my pure maths knowledge is not always 100% accurate - feel free to correct me (If possible provide some references so I can learn from my mistakes)

Edit: Oops you beat me...mustn't take so long to write my posts
 

edcarman

Member
May 23, 2005
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Originally posted by: Lucent
Although the limit can be calculated to be one, I don't see anyway the sum can actually reach 1. If there are any mathmaticians that can demonstrate how this sum can actually equal 1, that would be awesome. But as I see it, the limit is equal to 1 but the sum will never actually be equal to 1 (1 will be a horizontal asymptote). Hence .999 ad nauseam does not equal 1 to my knowledge.

I think this is the very crux of the whole problem. As an engineer I always try to look for physical interpretations of maths etc. and this is one of those cases where that approach fails.
I think it boils down to the definition of an infinite sum - it will converge (and be equal) to some discrete value, but will take infinitely many iterations.
Infinity is a concept that we as humans, I believe, can never trully visualise - our brains are just not wired to do it (similar to thinking about more than 3 spacial dimensions) - Try too hard to do it and you want to beat your head against the table.

 
Dec 30, 2004
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She's right not to believe it. 1/3 is not .333 repeating, no matter how far you go out. It will always be an infinite sequence of .3333's, but it will never be what it needs to be 1/3. We simply represent 1/3 in decimal form with a sequence of 3's because we can't do anything else. This is an example of the limit of our base 10 system. That is why we use a base 3 (1/3) or base 9 system (3/9) to represent a third of something, because it cannot be done in base 10.
 

imported_inspire

Senior member
Jun 29, 2006
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Originally posted by: TuxDave
Originally posted by: Eeezee
Originally posted by: Aluvus
Originally posted by: byosys
Originally posted by: Aluvus

The number she is describing is 1/infinity, which is identically zero.

1/infinity actually indeterminate, not zero.

Any real (and non-infinite) number divided by infinity is zero. Infinity divided by infinity is undefined.

What's zero times infinity? If it's zero, which I think it might be, then all of this falls apart.

Zero times infinity is also undefined. :p


The principle problem is that innfinity is not a number - it's a concept. The zero product property, though, still holds since you're basically dealing witha ring there.
 

imported_inspire

Senior member
Jun 29, 2006
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Originally posted by: Lucent
Eeezee said: Someone help me with the 0.9 repeating = 1 proof

IF 0.9 with nines repeating comes from lim (x->1) x=1 and the function f(x)=x THEN you can prove using theorems in calculus that f(x) at x=1 APPROACHES AND BECOMES EQUAL TO 1 (basically the limit equals the function value and it is a continous function) FOR THE SPECIFIC FUNCTION, LIMIT AND INPUT VALUE. BUT THE LIMIT DOES NOT NECESSARILY EQUAL THE VALUE OF THE FUNCTION HENCE THERE IS NO PROOF THAT 0.9 WITH NINES REPEATING EQUALS 1. HERE IS A EXAMPLE:

f(x)= { x if x is not equal to one
50000000 if x is equal to one}

Another example:

lim (x->1) (x-1)/(x^2-1)=0.5

(x-1)/(x^2-1) at x=1 is INDETERMINATE. FUNCTION VALUE DOES NOT EQUAL THE LIMIT!

That's exactly correct, because limits neglect removable discontinuities. I actually tried to use a limit to prove this at first, and then quickly realized that it wouldn't work for the vaery same reason.

 

Born2bwire

Diamond Member
Oct 28, 2005
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Originally posted by: edcarman
Originally posted by: Lucent
Although the limit can be calculated to be one, I don't see anyway the sum can actually reach 1. If there are any mathmaticians that can demonstrate how this sum can actually equal 1, that would be awesome. But as I see it, the limit is equal to 1 but the sum will never actually be equal to 1 (1 will be a horizontal asymptote). Hence .999 ad nauseam does not equal 1 to my knowledge.

I think this is the very crux of the whole problem. As an engineer I always try to look for physical interpretations of maths etc. and this is one of those cases where that approach fails.
I think it boils down to the definition of an infinite sum - it will converge (and be equal) to some discrete value, but will take infinitely many iterations.
Infinity is a concept that we as humans, I believe, can never trully visualise - our brains are just not wired to do it (similar to thinking about more than 3 spacial dimensions) - Try too hard to do it and you want to beat your head against the table.

Exactly. In the physical sense, infinity is not something we can attain or truly conceive. People like to try and think of the problem as being some physical entity, like writing out on a huge blackboard the number .9999... But then, there is always going to be an end to the 9's, and thus it must only approach 1. But in mathematics, infinity is a tangible entity. There is no reason why when we specify .999... that we can't say that there is an infinite number of nines. .999... is the INFINITE sum of 1/10^x, not the sum of 1/10^x to some incredibly large number. By definition, .999... is 1 because we are allowing it to be the sequence of an infinite number.
 

imported_inspire

Senior member
Jun 29, 2006
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Originally posted by: soccerballtux
She's right not to believe it. 1/3 is not .333 repeating, no matter how far you go out. It will always be an infinite sequence of .3333's, but it will never be what it needs to be 1/3. We simply represent 1/3 in decimal form with a sequence of 3's because we can't do anything else. This is an example of the limit of our base 10 system. That is why we use a base 3 (1/3) or base 9 system (3/9) to represent a third of something, because it cannot be done in base 10.

See edcarman's link to the wiki article - the order proof relies on dedekind's axioms of betweeness of the real line is nice - and slightly more geometric. It relies on induction, though, so I would conclude that the algebraic proofs are more elegant.

Base systems do not affect mathematical properties like this. Between every base systems, there is an isomorphism that preserves all algebraic properties.

1/3 = .333 repeating because we go all the way out - we don't stop. That's something that takes an abstract mathematical mind to fully comprehend. Engineers may not always get it - but mathematicians do. And I are one. =D
 

RossGr

Diamond Member
Jan 11, 2000
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Here is a proof which addresses the very point your gf is trying to make. It shows that no matter how small a number you add to .999.... you get a result GREATER then 1. For the final proof of equality relies on something called the Nested Interval Theorem. This theorem is proven in a University level Real Analysis course. So even though most will not know of it I present it without proof.
 

BrownTown

Diamond Member
Dec 1, 2005
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Originally posted by: inspire
Originally posted by: soccerballtux
She's right not to believe it. 1/3 is not .333 repeating, no matter how far you go out. It will always be an infinite sequence of .3333's, but it will never be what it needs to be 1/3. We simply represent 1/3 in decimal form with a sequence of 3's because we can't do anything else. This is an example of the limit of our base 10 system. That is why we use a base 3 (1/3) or base 9 system (3/9) to represent a third of something, because it cannot be done in base 10.

See edcarman's link to the wiki article - the order proof relies on dedekind's axioms of betweeness of the real line is nice - and slightly more geometric. It relies on induction, though, so I would conclude that the algebraic proofs are more elegant.

Base systems do not affect mathematical properties like this. Between every base systems, there is an isomorphism that preserves all algebraic properties.

1/3 = .333 repeating because we go all the way out - we don't stop. That's something that takes an abstract mathematical mind to fully comprehend. Engineers may not always get it - but mathematicians do. And I are one. =D

The problem with infinity is that alot of people think its a number like 1,2,3 etc.., but really it is just a word. However, I wouldn't say thats its all that hard to get the concept of infinity, the only problem is that math types put too much stress on the difference between something approaching a number as it goes to inifity and actually reaching it. Its not that engineers don't get this concept its that in the real world you just have to be "close enough", so you might as well think of .99999999999999991 as being equal to 1 even though its not since pretty much nothing will ever matter to those tolerences. For example if you have some vehicle that goes half the distance to the destination every second a mathmetician (at least in theory) will say its worthless since it never reaches the destiniation, wheras an engineer will say "it gets close enough, i'll just walk the last millimeter".

 

CSMR

Golden Member
Apr 24, 2004
1,376
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Originally posted by: Eeezee
I tried it two different ways

The pattern method
1/9 = 0.1 repeating
2/9 = 0.2 repeating
3/9 = 0.3 repeating
...
9/9 = 0.9 repeating = 1

And then I tried it this way
3/9 = 0.3 repeating
6/9 = 0.6 repeating
3/9 + 6/9 = 0.3 repeating + 0.6 repeating
9/9 = 0.9 repeating = 1

She still wouldn't believe me though! She keeps saying that since there is always something at the end of the string of numbers (despite being infinite) then it can be added together with something else to get 1 and is therefore not 1. I'm certain that you can't add anything to 0.9 repeating and get 1, but how can I prove this to her?
But 1/9=0.111... is no easier to see than 1=0.999...

Define 0.999...=sum from 1 to infinity of 9/(10^i)

(Forget about convergence for now which your friend might not understand.)

Let x=0.999...
ask what is 10x?
9.999...
What is 10x-9?
0.999...=x
So 10x-9=x, so x=1.
 

BrownTown

Diamond Member
Dec 1, 2005
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the point of the other one is that we all know that 1/9 is just a long series of ones, and 5/9 is a long series of fives, so of course 9/9 is also a long series of nines, but wait, 9/9 is also equal to 1, so .9999... must therefore equal one. Personally that seems like the easiet explanation to me.
 

CSMR

Golden Member
Apr 24, 2004
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Originally posted by: BrownTown
the point of the other one is that we all know that 1/9 is just a long series of ones
Because your calculator tells you so?
 

BrownTown

Diamond Member
Dec 1, 2005
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well, i would expect that most people would have that memmorized by that age, but if not your calculator, or long division can easily show you that is the case? Why does that matter for anything?
 

Eeezee

Diamond Member
Jul 23, 2005
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Originally posted by: byosys
Originally posted by: Aluvus

The number she is describing is 1/infinity, which is identically zero.

1/infinity actually indeterminate, not zero.

She says .0 repeating with a 1 on the end for however many 9s are in 0.9 repeating. I don't think she understands how there is no end to the 9s!

OK, then ask what number could exist between 0 and 0.0(rep)1. This is the only way I can think of, but if she dosn't really get the consept of repeating decimals and how they do not end, I doubt this approach will work.

I think that would confuse her more because what number is 0.0(rep)1? By definition that would have a finite number of zeros because it ends with the 1. Therefore one would simply add another 0 to the string to make a smaller number.
 

Eeezee

Diamond Member
Jul 23, 2005
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Originally posted by: soccerballtux
She's right not to believe it. 1/3 is not .333 repeating, no matter how far you go out. It will always be an infinite sequence of .3333's, but it will never be what it needs to be 1/3. We simply represent 1/3 in decimal form with a sequence of 3's because we can't do anything else. This is an example of the limit of our base 10 system. That is why we use a base 3 (1/3) or base 9 system (3/9) to represent a third of something, because it cannot be done in base 10.

With an infinite number of 3s, 1/3 is corretly represented by 0.3 repeating. The string of 3s can never end in order for 1/3 to equal 0.3 repeating.

It's a limit of our base 10 system, but it is still technically correct if you go out to an infinite number of 3s. There was your problem in the post; you made an assumption that you had to stop somewhere in the decimal line, but in actuality you can not stop anywhere. You must go on forever, into infinity.
 

Eeezee

Diamond Member
Jul 23, 2005
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Originally posted by: CSMR
Originally posted by: Eeezee
I tried it two different ways

The pattern method
1/9 = 0.1 repeating
2/9 = 0.2 repeating
3/9 = 0.3 repeating
...
9/9 = 0.9 repeating = 1

And then I tried it this way
3/9 = 0.3 repeating
6/9 = 0.6 repeating
3/9 + 6/9 = 0.3 repeating + 0.6 repeating
9/9 = 0.9 repeating = 1

She still wouldn't believe me though! She keeps saying that since there is always something at the end of the string of numbers (despite being infinite) then it can be added together with something else to get 1 and is therefore not 1. I'm certain that you can't add anything to 0.9 repeating and get 1, but how can I prove this to her?
But 1/9=0.111... is no easier to see than 1=0.999...

Define 0.999...=sum from 1 to infinity of 9/(10^i)

(Forget about convergence for now which your friend might not understand.)

Let x=0.999...
ask what is 10x?
9.999...
What is 10x-9?
0.999...=x
So 10x-9=x, so x=1.

I find that 1/9=0.111... is actually very easy to see. All you need to do is elementary long division to see that you get an infinite string of 1s after the decimal. The human brain can pick up the pattern and, indeed, the human brain is correct in predicting that the number is infinitely long.
 

Eeezee

Diamond Member
Jul 23, 2005
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Originally posted by: CSMR
Originally posted by: BrownTown
the point of the other one is that we all know that 1/9 is just a long series of ones
Because your calculator tells you so?

Or you can just calculate it yourself... 1/9 is an endless series of ones
 

Eeezee

Diamond Member
Jul 23, 2005
9,922
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Originally posted by: BrownTown
you can use google, or the search button on this forum to find many other proofs of this and choose the one you think works best. Personally I just consider it one of those silly little quirks in math that people think they are smart because they can repeat it. Kinda like that proof that 1=2.

The proof that 1=2 requires a fallacy step though, where you divide by zero. If you can divide by zero and get one, then you can prove that any number is equal to any other number :p
 

CSMR

Golden Member
Apr 24, 2004
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Originally posted by: EeezeeI find that 1/9=0.111... is actually very easy to see. All you need to do is elementary long division to see that you get an infinite string of 1s after the decimal. The human brain can pick up the pattern and, indeed, the human brain is correct in predicting that the number is infinitely long.
Then you have to justify long division which is very easy but still much harder than the original problem.
Originally posted by: BrownTown
well, i would expect that most people would have that memmorized by that age, but if not your calculator, or long division can easily show you that is the case? Why does that matter for anything?
Or you have to justify the infallibility of your maths books, which would involve adding an extra axiom to mathematics and possibly making it self-contradictory. Similarly with calculators.
 

BrownTown

Diamond Member
Dec 1, 2005
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You do realise this is why mathematicians annoy everyone. Needing all kinda pedantic proofs just to show something we all learned in 1st grade. I will take it as the word of God himself that long division works until show otherwise. And I'll be happy to believe that my calculator works flawlessly too. You go off and spend your life prooving the circuits are falliable whilst we go off and actually contribute to society...