Is 1 = 0.9999......

[josphII]

Originally posted by: MadRat

Originally posted by: silverpig

Originally posted by: MadRat
Hence, objective values cannot be used to define subjective ones.

What, pray tell, is an objective value, and what is a subjective value?

Null and Infinity do not exist in the natural universe, they are subjective definitions to represent fantasy.

An objective value can be expressed in a concrete sense.

While .333... is approximately 1/3 it is not equal to the latter. We use approximations based on objective values to solve problems close enough to an answer than it acceptable and within significant tolerances, but never do they truly represent the subjective value within the original equation. Substitution of the objective value for the subjective value is simply the key to approximation.

0.333... is equal to 1/3, it is not an approximation.

.33 or .333 would be an approximation, but 0.333... is exactly 1/3

 

[TuxDave]

Originally posted by: erikiksaz
I'm not sure if this applies fully, but as a store clerk, would you accept 99 cents if the item being bought is worth a dollar?

Nope... 99.9999... cents and nothing less. Hehehe...

 

[SilentRunning]

Originally posted by: RSMemphis
Heh, heh... It really helps to have taken calculus. I always thought "no", until I did the proof myself.

Proof I liked even more: In the space of the real numbers, you can project all the numbers from negative to positive infinity to the span of negative to positive one (or any other two numbers), because between every rational number, there is an infinite number of real numbers.

Talk about headache - but once understood, the coolest thing evar.

If you took calculus you missed the most important point. It is about approximations. You chose a dx value that is so small that your answer will be approximately equal to the actual value you seek.

 

[TuxDave]

Originally posted by: SilentRunning

Originally posted by: RSMemphis
Heh, heh... It really helps to have taken calculus. I always thought "no", until I did the proof myself.

Proof I liked even more: In the space of the real numbers, you can project all the numbers from negative to positive infinity to the span of negative to positive one (or any other two numbers), because between every rational number, there is an infinite number of real numbers.

Talk about headache - but once understood, the coolest thing evar.

If you took calculus you missed the most important point. It is about approximations. You chose a dx value that is so small that your answer will be approximately equal to the actual value you seek.

I thought once you choose a dx value infinitely small, the answer becomes the exact value you seek.

<edit> I bring as evidence, the derivative.

The creation of a derivative was picking two points infinitely close to each other, but our final answer should be the exact derivative, not an approximation.

 

[TheVrolok]

It does, this is absolute fact, you cannot debate it.

 

[SilentRunning]

In these courses, the student will have the opportunity

to explore the fundamental paradigm of Calculus: approximate and find the limit of increasingly more precise approximations;



Link

 

[SilentRunning]

Originally posted by: TuxDave

Originally posted by: SilentRunning

Originally posted by: RSMemphis
Heh, heh... It really helps to have taken calculus. I always thought "no", until I did the proof myself.

Proof I liked even more: In the space of the real numbers, you can project all the numbers from negative to positive infinity to the span of negative to positive one (or any other two numbers), because between every rational number, there is an infinite number of real numbers.

Talk about headache - but once understood, the coolest thing evar.

If you took calculus you missed the most important point. It is about approximations. You chose a dx value that is so small that your answer will be approximately equal to the actual value you seek.

I thought once you choose a dx value infinitely small, the answer becomes the exact value you seek.

<edit> I bring as evidence, the derivative.

The creation of a derivative was picking two points infinitely close to each other, but our final answer should be the exact derivative, not an approximation.

The bold statement is what this whole thread is about....wether you believe that statement or not.

 

[Sunner]

To me, 1 is 1, everything else is not 1, and "9.whateveryouwant" is part of "everything else".

 

[TuxDave]

Originally posted by: SilentRunning

Originally posted by: TuxDave

Originally posted by: SilentRunning

Originally posted by: RSMemphis
Heh, heh... It really helps to have taken calculus. I always thought "no", until I did the proof myself.

Proof I liked even more: In the space of the real numbers, you can project all the numbers from negative to positive infinity to the span of negative to positive one (or any other two numbers), because between every rational number, there is an infinite number of real numbers.

Talk about headache - but once understood, the coolest thing evar.

If you took calculus you missed the most important point. It is about approximations. You chose a dx value that is so small that your answer will be approximately equal to the actual value you seek.

I thought once you choose a dx value infinitely small, the answer becomes the exact value you seek.

<edit> I bring as evidence, the derivative.

The creation of a derivative was picking two points infinitely close to each other, but our final answer should be the exact derivative, not an approximation.

The bold statement is what this whole thread is about....wether you believe that statement or not.

Perhaps there will never be an answer, like two types of number theory. Almost like how there's several forms of geometry all based on one question.

Given a point and line, how many lines can go through the point and never intersect the line.

 

[Mucman]

I can't believe this was brought up again, and that there are still people who do not believe the proof

 

[silverpig]

Originally posted by: MadRat

Originally posted by: silverpig

Originally posted by: MadRat
Hence, objective values cannot be used to define subjective ones.

What, pray tell, is an objective value, and what is a subjective value?

Null and Infinity do not exist in the natural universe, they are subjective definitions to represent fantasy.

An objective value can be expressed in a concrete sense.

While .333... is approximately 1/3 it is not equal to the latter. We use approximations based on objective values to solve problems close enough to an answer than it acceptable and within significant tolerances, but never do they truly represent the subjective value within the original equation. Substitution of the objective value for the subjective value is simply the key to approximation.

0 = {}
1 = {0}
2 = {0,1}

etc...

The basis of the real number line is based upon the definition of zero.

1, 2, 4, 5, 98, etc do not exist in the natural universe either. You can have 2 apples but that does not equal 2. It represents twoness, but that isn't two.


Calculus doesn't allow you to find only approximations. Riemann sums are approximations. Integrals are the exact limits of those approximations. If you don't believe that, try graphing a semi circle of radius r and use calculus to integrate it's area. You should get pi/2 * r^2, the EXACT value for it's area.

 

[silverpig]

Originally posted by: SilentRunning
My point about e=mc^2 is that that is not an actual equation, It is an approximation of the actual equation.

And you seem to be missing the point

1/3 = 0.3333........ + 10^(-n) as n approaches infinity


to say that as n approaches infinity 10^(-n) = zero is wrong, it approaches zero.

Nothing you can say will change that fact. So 0.33333...... is an approximation of 1/3

E = mc^2 is an actual equation.

It is NOT an approximation in any sense.

If you are referring to the expanded version of the equation for E, then you mean that mc^2 is a special case SIMPLIFICATION for when v = 0. You aren't approximating anything there. It is only an approximation if the mass is moving and you choose to ignore the terms with v included.

 

[SilentRunning]

Originally posted by: silverpig
Calculus doesn't allow you to find only approximations. Riemann sums are approximations. Integrals are the exact limits of those approximations. If you don't believe that, try graphing a semi circle of radius r and use calculus to integrate it's area. You should get pi/2 * r^2, the EXACT value for it's area.

Limits for Dummies

If we refer to the polygon as an n-gon, where n is the number of sides, we can make some equivalent mathematical statements. (Each statement will get a bit more technical.)

As n gets larger, the n-gon gets closer to being the circle.
As n approaches infinity, the n-gon approaches the circle.
The limit of the n-gon, as n goes to infinity, is the circle!

The n-gon never really gets to be the circle, but it will get darn close! So close, in fact, that, for all practical purposes, it may as well be the circle. That's what limits are all about!

and

Let's look at the sequence whose nth term is given by n/(n+1). Recall, that we let n=1 to get the first term of the sequence, we let n=2 to get the second term of the sequence and so on.

What will this sequence look like?

1/2, 2/3, 3/4, 4/5, 5/6,... 10/11,... 99/100,... 99999/100000,...

What's happening to the terms of this sequence? Can you think of a number that these terms are getting closer and closer to? Yep! The terms are getting closer to 1! But, will they ever get to 1? Nope! So, we can say that these terms are approaching 1. Sounds like a limit! The limit is 1.

As n gets bigger and bigger, n/(n+1) gets closer and closer to 1...


Now who here wants me to believe the sky is green and the grass is red?

 

[MadRat]

.333... <> 1/3 but it is AN APPROXIMATION that is accepted as being one and the same.

 

[silverpig]

Again, that's perfectly fine, but that's NOT what is happening here.


Calculus uses limits to compute EXACT areas. Riemann sums are approximations. The integral of a function is the LIMIT of the riemann summation of the rectangles that approximate that area. The limit is the exact area, calculus allows us to find this exact area.

I don't see how you can construe that calculus is an approximation.

Secondly, 0.999... is the LIMIT of a sum. It is the number that sum approaches but never reaches. It is not the value of the sum at the nth term where you can pick whatever n is.

 

[silverpig]

Originally posted by: MadRat
.333... <> 1/3 but it is AN APPROXIMATION that is accepted as being one and the same.

It is not an approximation.

The sum S = (3/10 + 3/100 + 3/1000 + ...) is an approxmation, but this is not what 0.333... is.

 

[Kyteland]

Wow, what have I started? I subscribed to the thread when I created it and now my mailbox has >200 messages in it. It's almost worse than spam...

Anyway, the simplest explaination I heard that basically put this debate to rest at work was by using an accepted rule in number theory. Every two real numbers that are not equal have some number between them. If X>Y then X>(X+Y)/2>Y. You really can't debate this. It's true. If this rule breaks down then the numbers must be equal.

So the question this poses is can you name a number that is bigger than 0.99999..... but less than 1? No such number exists. That is because they are equal.

 

[silverpig]

Originally posted by: Kyteland
Wow, what have I started? I subscribed to the thread when I created it and now my mailbox has >200 messages in it. It's almost worse than spam...

Anyway, the simplest explaination I heard that basically put this debate to rest at work was by using an accepted rule in number theory. Every two real numbers that are not equal have some number between them. If X>Y then X>(X+Y)/2>Y. You really can't debate this. It's true. If this rule breaks down then the numbers must be equal.

So the question this poses is can you name a number that is bigger than 0.99999..... but less than 1? No such number exists. That is because they are equal.

Thank you.

I've posted this before, but somehow people insist that 0.00...01 is the difference between 0.999... and 1. Hence, they would argue that 1 + 0.999... = 1.999... and 1.999.../2 = 0.999...9945, which is ridiculous.

edit: typo

 

[silverpig]

Hot damn! I just realized something when I made that post.

Let's use the numbers 4 and 5 to show something here.

5 - 4 = 1
(4 + 5)/2 = 4.5
4.5 - 4 = 0.5 = 1/2 ie, half the difference


Now I'll use some people's logic here to try and repeat that.

1 - 0.999... = 0.000...01
(1 + 0.999)/2 = 0.999... (try it via long division)
0.999... - 0.999... = 0 (duh)

But this must be half the difference right? So half the difference is 0, so the difference must be 0.

If you assume the 1 - 0.999... = 0.000...01 statement is true, then you MUST agree that the result 2 lines down SHOULD be 0.000...005 right? Can anyone show how you would get this?

 

[Kyteland]

I've posted this before, but somehow people insist that 0.00...01 is the difference between 0.999... and 1. Hence, they would argue that 1 + 0.999... = 1.999... and 1.999.../2 = 0.999...995, which is ridiculous.

Sorry, I didn't know you posted it. I won't have time to read through all of the responses until Monday.

 

[silverpig]

I'm not complaining... just stating that even though it has been posted already, some still choose to believe that it doesn't make a difference.

 

[Haircut]

Originally posted by: Kyteland
Wow, what have I started? I subscribed to the thread when I created it and now my mailbox has >200 messages in it. It's almost worse than spam...

Anyway, the simplest explaination I heard that basically put this debate to rest at work was by using an accepted rule in number theory. Every two real numbers that are not equal have some number between them. If X>Y then X>(X+Y)/2>Y. You really can't debate this. It's true. If this rule breaks down then the numbers must be equal.

So the question this poses is can you name a number that is bigger than 0.99999..... but less than 1? No such number exists. That is because they are equal.

You can prove this as well.

Let x = 0.999... = 9 * Sum [i=1..oo] (0.1^i)
oo used to represent infinity here.
We can write x as two sums
x = 9 * Sum [i=1..n] (0.1^i) + 9 * Sum [i=n+1..oo] (0.1^i)

Now, let k(n) = 9 * Sum [i=n+1..oo] (0.1^i)
k(n) is arbitrary, but we know it must be > 0 for all natural n.
Also, k(n) = x*(0.1^n) so k(n) < 2*(0.1^n)

x = 9 * Sum [i=1..n] (0.1^i) + k(n)
This is a finite geometric series. Using the formula here
We get that x = 1- (0.1^n) + k(n)

Now, we can introduce a new variable, y(n) = 0.1^n, y(n) > 0 for all natural n

Lets calculate x+y, it is trivial to see that x+y(n) = 1+k(n)
We have that k(n) > 0 so, x+y(n)>1
From this x cannot be less than 1, as if it were we could simply take n, so that y(n) is smaller than or equal to the difference between x and 1 and then x+y <=1, which contracticts the statement above.

Thus x >= 1

Let us now work out x-y

Thus x-y = 1 - 2*(0.1^n) +k(n)
From above we have that k(n) < 2*(0.1^n)

Therefore x-y < 1. Using a similar argument to before we can see that from this x cannot be greater than 1

x <= 1

We now have that x >= 1 and x <= 1, meaning that x=1.

QED

 

[RSMemphis]

Originally posted by: SilentRunning

Originally posted by: RSMemphis
Heh, heh... It really helps to have taken calculus. I always thought "no", until I did the proof myself.

Proof I liked even more: In the space of the real numbers, you can project all the numbers from negative to positive infinity to the span of negative to positive one (or any other two numbers), because between every rational number, there is an infinite number of real numbers.
Talk about headache - but once understood, the coolest thing evar.

If you took calculus you missed the most important point. It is about approximations. You chose a dx value that is so small that your answer will be approximately equal to the actual value you seek.

Dude, what's up your butt??? I was just saying that I found that concept hard to believe myself until I learned enough to properly grasp the whole aspect of the problem.

Also, calculus is not about decimal representations, 0.333333..... is just an easier way for a calculator to express 1/3.
Multiply that by 3, but I am disgressing.

Geez, who knew I had to wear a flame resistant suit in this thread.

 

[josphII]

Originally posted by: silverpig

Originally posted by: MadRat
.333... <> 1/3 but it is AN APPROXIMATION that is accepted as being one and the same.

It is not an approximation.

The sum S = (3/10 + 3/100 + 3/1000 + ...) is an approxmation, but this is not what 0.333... is.

the sum you stated above is EXACTLY what 0.333... is, they are both exactly 1/3.

 

[Skyclad1uhm1]

Originally posted by: Krk3561

Originally posted by: Kyteland
We're having a debate at work. Is 1=0.99999..... repeating. I say that this holds but one of my coworkers claims that multiplication breaks down for an infinitely repeating number.

x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1.

What do you think?

9 x .99999999999999999999999999999999999999999999999999999999999999999999999999 does not equal 1. Any calculator (except for a very accurate computer) you put that in will give you 9 because it cant process that many digits, so it will round up

The point is that infinity cannot be defined.

x = 0.999...
10x= 9.999...
10x - x =9.999... - 0.999...
9x = 9

This requires you to see that infinity = infinity, that there is no infinity -1 or infinity +1. It's an abstract number. (infinity - infinity can basically be anything)

9.999... (infinity -1 decimals) - 0.999... (infinity decimals) is hard to imagine.

Try this:
Start walking in a circle. As you keep starting the same road again it is infinite. Now walk 10 feet less. Does that mean the circle is less infinite? 'infinite' does not point at the distance covered, but instead it points at the circle.