Someone help me with the 0.9 repeating = 1 proof

[silverpig]

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!

1. This is number theory, not calculus
2. We have a bounded monotonic series here, not a piecewise defined function
3. 0.9r = 9 * lim(n->inf) Sum(x=1..n) 1/10^x

In other words, the quantity 0.9r we are talking about is DEFINED to be the limit of that series. That limit can be easily calculated as it is a geometric series and is found to be 1.

 

[silverpig]

Oh and for the OP:

I suggest you explain it like this:

0.9
0.1
one 9, one 1

0.99
0.01
two 9s, one 0, one 1

0.999
0.001
three 9s, two 0s, one 1

0.999999999...
0.000000000...
infinity 9s, infinity - one 0s.

infinity - 1 = infinity

 

[Howard]

There's a formula for it, but heck if I can find it now.

Anyway, 0.0...1 doesn't exist.

 

[RossGr]

Originally posted by: Howard
There's a formula for it, but heck if I can find it now.

Anyway, 0.0...1 doesn't exist.

Sure it exists, you just wrote it down! But it AINT A NUMBER!

 

[smack Down]

Originally posted by: byosys
Ask her what number could possibly exist between .9 repeating and 1. If you can't fit a number between two other numbers (.9 repeating and 1.0), the numbers must be equal.

That doesn't prove equality. What number comes between pi and the nearest rational number. Bet you can't name it. Does that mean Pi equals a rational number?

 

[silverpig]

Originally posted by: smack Down

Originally posted by: byosys
Ask her what number could possibly exist between .9 repeating and 1. If you can't fit a number between two other numbers (.9 repeating and 1.0), the numbers must be equal.

That doesn't prove equality. What number comes between pi and the nearest rational number. Bet you can't name it. Does that mean Pi equals a rational number?

There is no nearest rational number. And yes that does prove equality.

Two numbers are equal if their difference is 0.

 

[zugzoog]

Originally posted by: Eeezee

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.

This is not your proof, you have made the assumption 1/9 = .11111...... exactly, let's test this assumption.

back to basic division. Sorry for the bad (or total lack of) formatting.

1/9 = 0.1 with 0.1 remainder.
= 0.11 with 0.01 remainder
= 0.111 with 0.001 remainder
= 0.1111 with 0.0001 remainder
= 0.11111 with 0.00001 remainder
= 0.111111 with 0.000001 remainder
rinse and repeat an infinite number of times
1/9 = 0.1111.... with an infinitesimally small remainder divided by 9.

Therefore 1/9 = .11111...... is an approximation.

The assumption does not hold up, this is not the proof.

Running throught the calculation again (I am calling the infinitesimally small addition to the number (to make it exact) "remainder", I am not sure of the mathematical nomenclature for such a concept);

1/9 = .111... + remainder/9
+
1/9 = .111... + remainder/9
+
.
repeat 7 times
-------------------
9/9 = .999... + remainder

therefore
9/9 = 1



footnote. If you say that the infinitesimally small remainder can be ignored then you are making the assumption that an infinitesimally small amount can be ignored to prove that an infinitesimally small difference = 0. This is circular reasoning.

 

[Rangoric]

Originally posted by: silverpig

Originally posted by: smack Down

Originally posted by: byosys
Ask her what number could possibly exist between .9 repeating and 1. If you can't fit a number between two other numbers (.9 repeating and 1.0), the numbers must be equal.

That doesn't prove equality. What number comes between pi and the nearest rational number. Bet you can't name it. Does that mean Pi equals a rational number?

There is no nearest rational number. And yes that does prove equality.

Two numbers are equal if their difference is 0.

But the difference between 1 and 0.9r is 1/Infinity (or 0.00...1 which is an invalid decimal representation, but you get the idea )

And since 1 > 0, 1/Infinity has a value. Not a value you can show via a decimal, but a value nonetheless. While in useful mathmatics, it has a value not that unlike 0, and has a limit of 0 (when turned into a function), that only means that effectivly it can be represented by 0.

Due to the infinitly small difference, it is generally accepted that 1 = 0.9r. It also allows the mathmatical and science fields to continue on their way.

I just can't wait for the day for that assumption to bite some engineer or scientist in the ass.
(Note: I accept 0.9r = 1, but I don't believe it. Noone I've ever met has been able to prove it to me to my satisfaction)

 

[RossGr]

Originally posted by: Rangoric

Originally posted by: silverpig

Originally posted by: smack Down

Originally posted by: byosys
Ask her what number could possibly exist between .9 repeating and 1. If you can't fit a number between two other numbers (.9 repeating and 1.0), the numbers must be equal.

That doesn't prove equality. What number comes between pi and the nearest rational number. Bet you can't name it. Does that mean Pi equals a rational number?

There is no nearest rational number. And yes that does prove equality.

Two numbers are equal if their difference is 0.

But the difference between 1 and 0.9r is 1/Infinity (or 0.00...1 which is an invalid decimal representation, but you get the idea )

And since 1 > 0, 1/Infinity has a value. Not a value you can show via a decimal, but a value nonetheless. While in useful mathmatics, it has a value not that unlike 0, and has a limit of 0 (when turned into a function), that only means that effectivly it can be represented by 0.

Due to the infinitly small difference, it is generally accepted that 1 = 0.9r. It also allows the mathmatical and science fields to continue on their way.

I just can't wait for the day for that assumption to bite some engineer or scientist in the ass.
(Note: I accept 0.9r = 1, but I don't believe it. Noone I've ever met has been able to prove it to me to my satisfaction)

This is just wrong!

First of all 1/infinity must be defined, because infinity is NOT a number. So when mathematicians DEFINE 1/infinity to be zero, then it is zero. There can be no argument, it is a definition. Oh yeah and in the extented real numbers that is the normal definition. Why? because it works. 1 = .999... is not an approximation it is a result of the construction of the real number system, that is the way it works. Many people, including You seem to have trouble grasping infinity, you treat it just like it was some large number, it is NOT a large number it is greater then ANY number and THAT is the definition. So when you say things like:
"And since 1 > 0, 1/Infinity has a value. Not a value you can show via a decimal, but a value nonetheless. It shows that you do not have your head wrapped around infinity. For there to be a difference between 1 and .999... the nines would have to stop. They do not stop so there is no difference.

 

[Rangoric]

Originally posted by: RossGr
This is just wrong!

First of all 1/infinity must be defined, because infinity is NOT a number. So when mathematicians DEFINE 1/infinity to be zero, then it is zero.

Except is it a construct of the real number system. Real Number system not meaning "The number system we use every day". And quite frankly, if you accept 1/Infinity = 0, then it is a defined number. Learn your arguement before talking next time.

http://mathforum.org/library/drmath/view/62486.html
http://www.everything2.com/index.pl?node_id=152045
http://www.physicsforums.com/archive/index.php/t-511.html

By definition 1/Infinity has no value and saying 1/Infinity = 0 is mathmatically wrong as it is the limit as x approuches infinity for 1/X that equals 0. As such a definition says that the value of that fraction approuches 0, not equals 0.

Its only when you deal with a specific number set that you get 1/Infinity defined to be 0. However this definition also equates 1/+Infinity and 1/-Infinity to be equal.

|1/Infinity| is the smallest number larger then 0.

 

[DrPizza]

Originally posted by: Rangoric

Originally posted by: RossGr
This is just wrong!

First of all 1/infinity must be defined, because infinity is NOT a number. So when mathematicians DEFINE 1/infinity to be zero, then it is zero.

Except is it a construct of the real number system. Real Number system not meaning "The number system we use every day". And quite frankly, if you accept 1/Infinity = 0, then it is a defined number. Learn your arguement before talking next time.

http://mathforum.org/library/drmath/view/62486.html
http://www.everything2.com/index.pl?node_id=152045
http://www.physicsforums.com/archive/index.php/t-511.html

By definition 1/Infinity has no value and saying 1/Infinity = 0 is mathmatically wrong as it is the limit as x approuches infinity for 1/X that equals 0. As such a definition says that the value of that fraction approuches 0, not equals 0.

Its only when you deal with a specific number set that you get 1/Infinity defined to be 0. However this definition also equates 1/+Infinity and 1/-Infinity to be equal.

|1/Infinity| is the smallest number larger then 0.

You're getting closer... yes, "1 over infinity" is frequently seen as a limit in calculus, and it evaluates to 0. Exactly 0. So, if we're looking at the value of 1/x as x approaches infinity, the limit = 0. (1/x "approaches" 0, but the limit *IS* zero.) I always have to give my calculus students counter-intuitive limits to show them why we can't simply "plug in" infinity (or plug in zero). That's because when doing limit problems, it's inappropriate to say "plug in" infinity, but unfortunately that's exactly what's often done out of laziness. Off the top of my head, here's an example showing this difference (although my example is one for plugging in 0)

What is the sqrt(0 + sqrt(0 +sqrt(0 + sqrt(0 +...
(infinitely many nested square roots)
the answer is 0.

However, what is the limit as x approaches 0 (from the right, since we're not interested in imaginary numbers) of sqrt(x + sqrt(x +sqrt(x + sqrt(x +...
Nope, it's not 0. The correct answer is 1.
(I can explain why when I'm not on vacation... the bass await my lure as we speak)

1/infinity is meaningless, except when we're discussing limits.

Now, I don't know if it's been defined somewhere as 1/infinity = 0; I didn't think it was. But this really doesn't make sense, as explained in one of your links above.

So, recapping: the limit of 1/x as x approaches infinity IS EXACTLY ZERO... it's not the smallest number greater than zero.

 

[silverpig]

Originally posted by: zugzoog

Originally posted by: Eeezee

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.

This is not your proof, you have made the assumption 1/9 = .11111...... exactly, let's test this assumption.

back to basic division. Sorry for the bad (or total lack of) formatting.

1/9 = 0.1 with 0.1 remainder.
= 0.11 with 0.01 remainder
= 0.111 with 0.001 remainder
= 0.1111 with 0.0001 remainder
= 0.11111 with 0.00001 remainder
= 0.111111 with 0.000001 remainder
rinse and repeat an infinite number of times
1/9 = 0.1111.... with an infinitesimally small remainder divided by 9.

Therefore 1/9 = .11111...... is an approximation.

The assumption does not hold up, this is not the proof.

Running throught the calculation again (I am calling the infinitesimally small addition to the number (to make it exact) "remainder", I am not sure of the mathematical nomenclature for such a concept);

1/9 = .111... + remainder/9
+
1/9 = .111... + remainder/9
+
.
repeat 7 times
-------------------
9/9 = .999... + remainder

therefore
9/9 = 1



footnote. If you say that the infinitesimally small remainder can be ignored then you are making the assumption that an infinitesimally small amount can be ignored to prove that an infinitesimally small difference = 0. This is circular reasoning.

Sorry if I'm not convinced by your mathematical arguments when you say that 1/9 is 0.1 with 0.1 remainder. Where I come from 0.1 with 0.1 remainder is 0.2. You just said that 1/9 = 1/5.

The rest of your argument just simply does not follow as you start off by saying that 1/9 = 0.2.

And 1/9 does equal 0.111... EXACTLY. Here's why:

0.111... is defined to be the limit of the sum of 1/10^n as n goes from 1 to infinity. In other words, 0.111... is the limit (which you will never actually reach) of that division you were trying to do. 1/9 is also this limit (again, which you will never actually reach). Or, if you like, 0.111... is the shorthand representation of a decimal point with an infinite number of 1s after it. We can't ever write it, but it exists. All are equivalent.

 

[RossGr]

Rangoric,
I am just working out of my old Real Analysis text by Rudin. The fact is, infinity must be defined before it is useful in mathematics. According to Rudin, (see the posts by Integral in your last link.) His definition of the extended Reals is "customary".


If you wish I can post a scan of the page in Rudin where this definition occurs. (have to be tomorrow, cuz I am currently at work.)

I still maintain that you have not got a good concept of a mathematical infinty. As I said before, it is not just some large number. What you say is exactly correct for any very large number. There is a subtle difference you are sweeping under the rug.

1/(any extremly large number you can name) is very close to zero, you can make it arbitrarly close to zero. Just specify how close you want to come to zero and you can find an extremly large number, say x, such that 1/x is closer to zero then your specification. OTOH you cannot specify a number closer to zero then 1/oo. The key phrase here is specify. This is a "feature" of the Real Number System. When you are working in the Real Number System You must be able to specify the number. This is where infinity is different and why it is not a normal Real number, it must be carefully defined, then, since it is a defintion, you must define how it interacts with the rest of the real numbers.


Edit: When I say Real Number, what do you think?
I find your link to PhysicsForums some what amusing as I am Integral @ PhysicsForums.com

 

[Aluvus]

Originally posted by: silverpig
Sorry if I'm not convinced by your mathematical arguments when you say that 1/9 is 0.1 with 0.1 remainder. Where I come from 0.1 with 0.1 remainder is 0.2. You just said that 1/9 = 1/5.

The rest of your argument just simply does not follow as you start off by saying that 1/9 = 0.2.

Then where you come from does not handle remainders properly.

A remainder means that you arrive a particular answer, and then have some more left over that still needs to be divided by the divisor.

0.1r0.1 is not the same expression as 0.1 + 0.1. Rather, in this context it would be 0.1 + 0.1/9. Note that the remainder is still divided by the divisor.

And that is factually true. 0.1 + 0.1/9 = 9*0.1/9 + 0.1/9 = (9*0.1 + 0.1)/9 = 1/9

 

[imported_inspire]

Originally posted by: silverpig

Originally posted by: smack Down

Originally posted by: byosys
Ask her what number could possibly exist between .9 repeating and 1. If you can't fit a number between two other numbers (.9 repeating and 1.0), the numbers must be equal.

That doesn't prove equality. What number comes between pi and the nearest rational number. Bet you can't name it. Does that mean Pi equals a rational number?

There is no nearest rational number. And yes that does prove equality.

Two numbers are equal if their difference is 0.

... having trouble wrapping my brain around this... Aren't the rationals countable? So wouldn't a nearest rational exist? Honestly, it's been awhile, help me out. Nevermind, I got it. Same concepts, new names.

Just a curiosity:
Read Pages 18-21

 

[kevinthenerd]

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.

No, zero times infinity is zero. However, 0/0 and inf/inf is neither mathematically meaningful nor meaningless. Ask her what 1/3 is, and then ask her to multiply that by 3. You'll end up with

3/3=0.9999...
therefore 1=0.9999...

My favorite proof, however, is still belons to Aluvus:

Algebraic: (Edited by kevinthenerd.)

Let A = 0.999 (repeating)

Let B = 10 * A = 9.999 (repeating)

B - A = (10 * A) - A = A * (10 - 1) = 9 * A
Note that no assumption about A is made yet, so B-A MUST equal 9*A, regardless of the value of A, given that B is defined as 10*A.
B - A = 9.999 (rep) - 0.999 (rep) = 9

Therefore,
B - A = 9 * A (given by definition)
B - A = 9 (given by the proof above)
9 * A = 9 (induced by both the definition and proof)
A = 1
and since A=0.999... (rep)
0.999 (repeating) = 1
[/b]

<-- calculus nerd who loves infinities

 

[kevinthenerd]

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.

lim(x->inf) (1/x) = 0

since lim(x->inf) x = inf
then inf = lim(x->inf)

therefore (1/inf) = (1 / [lim(x->inf) x] ) = lim (x->inf) (1/x) = 0

 

[kevinthenerd]

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.

Taken as an infinite sum of numbers (not as a set of numbers approaching an infinite length), 1/3=0.3333 repeating. That's the difference here. Once you define that the string is repeating, nothing can come after it but another repetition.

All ratios of whole numbers will yield a number with a decimal that either truncates or repeats. (There are no transcendental ratios of whole numbers.) My favorite example is when people define pi as 22/7, not realizing that pi is transcendental.

22/7 = 3.142857 142857 142857 142857 142857 ... (repeating)

Another proof, although quite a bit less formal than I'd like it to be, might go like this:

1 = 0.9999... + (1/inf)

 

[kevinthenerd]

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?

Do the long division. It's easy to see the pattern that results. You can prove that the pattern will never change because its inputs never change. First, when you divide 1.0 by 9 you get 0.1 with 1 remaining to carry to the tenths decimal place. Then, when you divide 0.1 by 9 you get 0.01, which carries another 1 over to the hundredths, and so forth. THIS is the proof that (1/9)=0.111 (repeating), so (9/9) must equal 0.999 repeating = 1

 

[imported_inspire]

I thought about this last night (not really trying to, but since I'm a math nerd...) and I got it. Of course, an algebraic definition equality is a=b iff a-b=0. Which, is equivalent to saying that if the difference between two numbers is zero, if and only if they are equal.

So here's my simple arguement, and the way I think you should explain it to your girlfriend. This is the conversation I had with my wife last night:

Me: So, have you heard that .999 repeating an infinite number of times is equal to 1?
Wife: No, that doesn't sound right.
Me: Sure it is. Can we agree that two numbers are equal if the difference between them is zero?
Wife: Yeah....
Me: Okay, so what's the diferrence betwen .999 repeating and 1?
Wife: .000 repeating with a 1 on the end.
Me: On the end of what?
Wife: The zeros.
Me: The infinitely repeating zeros?
Wife: Yes.
Me: So, let me get this straight, you want to put a 1 on the 'end' of an infinite amount of zeros to the right of a decimal point?
Wife: Yes.
Me: Okay, let me know when you get there.

Point is that there is no end to infinity. Duh.

 

[kevinthenerd]

Originally posted by: inspire
I thought about this last night (not really trying to, but since I'm a math nerd...) and I got it. Of course, an algebraic definition equality is a=b iff a-b=0. Which, is equivalent to saying that if the difference between two numbers is zero, if and only if they are equal.

So here's my simple arguement, and the way I think you should explain it to your girlfriend. This is the conversation I had with my wife last night:

Me: So, have you heard that .999 repeating an infinite number of times is equal to 1?
Wife: No, that doesn't sound right.
Me: Sure it is. Can we agree that two numbers are equal if the difference between them is zero?
Wife: Yeah....
Me: Okay, so what's the diferrence betwen .999 repeating and 1?
Wife: .000 repeating with a 1 on the end.
Me: On the end of what?
Wife: The zeros.
Me: The infinitely repeating zeros?
Wife: Yes.
Me: So, let me get this straight, you want to put a 1 on the 'end' of an infinite amount of zeros to the right of a decimal point?
Wife: Yes.
Me: Okay, let me know when you get there.

Point is that there is no end to infinity. Duh.

As mentioned earlier, the best way to talk about the "1 on the end" is to say that, once there's a 1 on the end, there's a countable number of 0's between the decimal point and the 1, meaning it's not an infinite string. You can't have, say, an infinite string of 0's followed by another infinite string of 1's... and although it sounds crazy, it's just as logical as saying there's anything else after an infinite string of anything.

 

[imported_inspire]

Originally posted by: kevinthenerd

Originally posted by: inspire
I thought about this last night (not really trying to, but since I'm a math nerd...) and I got it. Of course, an algebraic definition equality is a=b iff a-b=0. Which, is equivalent to saying that if the difference between two numbers is zero, if and only if they are equal.

So here's my simple arguement, and the way I think you should explain it to your girlfriend. This is the conversation I had with my wife last night:

Me: So, have you heard that .999 repeating an infinite number of times is equal to 1?
Wife: No, that doesn't sound right.
Me: Sure it is. Can we agree that two numbers are equal if the difference between them is zero?
Wife: Yeah....
Me: Okay, so what's the diferrence betwen .999 repeating and 1?
Wife: .000 repeating with a 1 on the end.
Me: On the end of what?
Wife: The zeros.
Me: The infinitely repeating zeros?
Wife: Yes.
Me: So, let me get this straight, you want to put a 1 on the 'end' of an infinite amount of zeros to the right of a decimal point?
Wife: Yes.
Me: Okay, let me know when you get there.

Point is that there is no end to infinity. Duh.

As mentioned earlier, the best way to talk about the "1 on the end" is to say that, once there's a 1 on the end, there's a countable number of 0's between the decimal point and the 1, meaning it's not an infinite string. You can't have, say, an infinite string of 0's followed by another infinite string of 1's... and although it sounds crazy, it's just as logical as saying there's anything else after an infinite string of anything.

I see what you're getting at, and I apologize if I repated what you had already said. However, having a countable number of 0's between the decimal point and the 1 at the end does not mean that it's not an infinite string. Being countable (in the mathematical sense) does not imply being finite.

 

[kevinthenerd]

Originally posted by: inspire

Originally posted by: kevinthenerd

Originally posted by: inspire
I thought about this last night (not really trying to, but since I'm a math nerd...) and I got it. Of course, an algebraic definition equality is a=b iff a-b=0. Which, is equivalent to saying that if the difference between two numbers is zero, if and only if they are equal.

So here's my simple arguement, and the way I think you should explain it to your girlfriend. This is the conversation I had with my wife last night:

Me: So, have you heard that .999 repeating an infinite number of times is equal to 1?
Wife: No, that doesn't sound right.
Me: Sure it is. Can we agree that two numbers are equal if the difference between them is zero?
Wife: Yeah....
Me: Okay, so what's the diferrence betwen .999 repeating and 1?
Wife: .000 repeating with a 1 on the end.
Me: On the end of what?
Wife: The zeros.
Me: The infinitely repeating zeros?
Wife: Yes.
Me: So, let me get this straight, you want to put a 1 on the 'end' of an infinite amount of zeros to the right of a decimal point?
Wife: Yes.
Me: Okay, let me know when you get there.

Point is that there is no end to infinity. Duh.

As mentioned earlier, the best way to talk about the "1 on the end" is to say that, once there's a 1 on the end, there's a countable number of 0's between the decimal point and the 1, meaning it's not an infinite string. You can't have, say, an infinite string of 0's followed by another infinite string of 1's... and although it sounds crazy, it's just as logical as saying there's anything else after an infinite string of anything.

I see what you're getting at, and I apologize if I repated what you had already said. However, having a countable number of 0's between the decimal point and the 1 at the end does not mean that it's not an infinite string. Being countable (in the mathematical sense) does not imply being finite.

I doubt a math n00b would agree, but you're right, although I'll have to say it's a matter of semantics of which I've never been fond. I meant to say that, as long as you can count them and assign a finite number once you're done counting (i.e., countable in a practical human sense), then it's not an infinite string. Provided you do say that there's a zero at the end of an infinite line of zeroes (which I still say is impossible), that zero is in a decimal place of a truly null significance and does not affect the value of the number whatsoever. If it did affect the number, it would have a significance of 10^-n where n is a finite number of decimal places away from the decimal point.

 

[Eeezee]

Originally posted by: inspire
I thought about this last night (not really trying to, but since I'm a math nerd...) and I got it. Of course, an algebraic definition equality is a=b iff a-b=0. Which, is equivalent to saying that if the difference between two numbers is zero, if and only if they are equal.

So here's my simple arguement, and the way I think you should explain it to your girlfriend. This is the conversation I had with my wife last night:

Me: So, have you heard that .999 repeating an infinite number of times is equal to 1?
Wife: No, that doesn't sound right.
Me: Sure it is. Can we agree that two numbers are equal if the difference between them is zero?
Wife: Yeah....
Me: Okay, so what's the diferrence betwen .999 repeating and 1?
Wife: .000 repeating with a 1 on the end.
Me: On the end of what?
Wife: The zeros.
Me: The infinitely repeating zeros?
Wife: Yes.
Me: So, let me get this straight, you want to put a 1 on the 'end' of an infinite amount of zeros to the right of a decimal point?
Wife: Yes.
Me: Okay, let me know when you get there.

Point is that there is no end to infinity. Duh.

My girlfriend made the same hypothesis that your wife made. How do we overcome the misunderstanding of infinity? By definition there can't be a 1 on the end, there are infinite zeros! Once you put a 1 on the end, the 0s become finite, and adding that number to 0.999... results in a number that is greater than 1 (0.999 repeating + 0.01 = 1.00999999...), so no matter where you put your 1 you're going to result in the sum of these two numbers being greater than one. If you can find a non-zero number whose sum with 0.9 repeating equals 1, then you've proved that 0.9 repeating is not equal to 1.

I like it. That will be my next approach. I keep forgetting to bring it up lol

 

[Eeezee]

Originally posted by: inspire

Originally posted by: kevinthenerd

Originally posted by: inspire
I thought about this last night (not really trying to, but since I'm a math nerd...) and I got it. Of course, an algebraic definition equality is a=b iff a-b=0. Which, is equivalent to saying that if the difference between two numbers is zero, if and only if they are equal.

So here's my simple arguement, and the way I think you should explain it to your girlfriend. This is the conversation I had with my wife last night:

Me: So, have you heard that .999 repeating an infinite number of times is equal to 1?
Wife: No, that doesn't sound right.
Me: Sure it is. Can we agree that two numbers are equal if the difference between them is zero?
Wife: Yeah....
Me: Okay, so what's the diferrence betwen .999 repeating and 1?
Wife: .000 repeating with a 1 on the end.
Me: On the end of what?
Wife: The zeros.
Me: The infinitely repeating zeros?
Wife: Yes.
Me: So, let me get this straight, you want to put a 1 on the 'end' of an infinite amount of zeros to the right of a decimal point?
Wife: Yes.
Me: Okay, let me know when you get there.

Point is that there is no end to infinity. Duh.

As mentioned earlier, the best way to talk about the "1 on the end" is to say that, once there's a 1 on the end, there's a countable number of 0's between the decimal point and the 1, meaning it's not an infinite string. You can't have, say, an infinite string of 0's followed by another infinite string of 1's... and although it sounds crazy, it's just as logical as saying there's anything else after an infinite string of anything.

I see what you're getting at, and I apologize if I repated what you had already said. However, having a countable number of 0's between the decimal point and the 1 at the end does not mean that it's not an infinite string. Being countable (in the mathematical sense) does not imply being finite.

I believe that depends on how strict a definition you're using for infinite. If you want to assume that very large, basically uncountable numbers are infinite, then yes. The difference is insignificant from any point of view except this proof. If there are a near-infinite number of 9s following the decimal point, then by a mere technicality the string is infinitely long by human standards just because we don't have the time required to count all of those numbers (let's assume that a supercomputer spends all of its time counting these numbers when you divide 1 by 3; its workload would never end).

The strictest definition for infinite would be one in which the number can never be counted. The definition of 1/3 requires the strictest definition for infinite because the string repeats forever. You have an infinite number of remainders to account for, and the result is an infinitely repeating string of numbers. This is not a near-infinite number or a practically infinite, but 100% infinite without approximation.

A near-infinite number of 0s after the decimal point followed by a 1 maintains significance for this problem because in the end the string is finite even if the 1 has almost no significance. The point is that it has some significance.

Ultimately the infinity that we're talking about for this proof must be too large for approximations. It has to be the final form of infinity, one without any limits whatsoever.