Is 0.999 Repeating 1? Debate & Opinions

[HallsofIvy]

It is a real number, but it will never be complete. It is not 1. One is be, but be is not .99... You will always be chasing your tail.

All you are telling us here is that you do not know what a real number is. Didn't you read the previous responses in this thread? There are a variety of (equivalent) definitions of real number. They all give the result that 0.999... is exactly the same as 1.00...

 

[jcsd]

TENYEARS, there is always another real number (infact there's always an infinite number of real numbers between any two real numbers, the proof is trivial) between any two real numbers, what's inbetween 1 and 0.9999...?

 

[robert Ihnot]

This question is a lot more important than is getting credit for here. Cantor was very concerned with this in in drawing up a list of rational numbers without duplicates. Remember that 1/2 + 1/4 + 1/8 +++ = 1. So that using base 2, which seems the simplest from a theoretical standpoint, then .111111111...=1.

 

[MiGUi]

What you suggest, TENYEARS, is wrong because you think in "0.99..." and "1" as two different numbers, but they are not. There are two different forms to write the same. You may not discuss that "8/4" and "2" are the same thing, and with "1" and "0.99..." occurs the same, because if not, if that two numbers were different, then aritmethic would not work properly.

Writting 0.99... may let you think (wrong) that the <i>next</i> real number is 1. But that thing is not correct, because in the real field, between two numbers there are infinity different numbers.


Bye
MiGUi

 

[TENYEARS]

I was going to change the post right after I posted, but did not. Throw out the word real. .99999... is not equal to 1. Case point end of discussion. It will have relavance to you some day. Apparently just not now.

 

[ahrkron]

I'm sorry, TENYEARS, but if you want to argue for a system in which 0.999... is not equal to 1.000..., you would need to do so in the TD forum, since their equality is a solid piece of math; it is not a matter that can "become relevant" at some point for someone, but an identity that follows from definitions and formal logic.

 

[HallsofIvy]

No, ahrkon, TENYEARS does not want to argue anything, He just likes to hear himself babble. I would say that I suspect TENYEARS is his age, but I don't want other ten year olds complaining that I am dissing them.

 

[jcsd]

TENYEARS takes ome advice: when people have nothing to say of any import, or indeed meaning, they usually keep quiet.

 

[TENYEARS]

Because ideas are accepted as fact by the masses, does it make it so? What defines logic but the logic before it which is based upon the agreement of a base worked forward. Because we have decided to accept something as such, is it such? It is such because you were taught so is it not.

It is an accepted value, I have accepted it as such a value in caluclations, it is rounded and creates one, but is it one? No it tries to be but never quite reaches. The issue can be avoided in large calculations by using factionary math, but when rounding must occur, it is what it is.

You can sweep it under the carpet, but that is why so many do not understand. Because you do not question and you accept. It is one by whos authority? This authority is certainly no greater than mine or yours unless you make it so by not seeing.

 

[jcsd]

Because ideas are accepted as fact by the masses, does it make it so? What defines logic but the logic before it which is based upon the agreement of a base worked forward. Because we have decided to accept something as such, is it such? It is such because you were taught so is it not.

It is an accepted value, I have accepted it as such a value in caluclations, it is rounded and creates one, but is it one? No it tries to be but never quite reaches. The issue can be avoided in large calculations by using factionary math, but when rounding must occur, it is what it is.

You can sweep it under the carpet, but that is why so many do not understand. Because you do not question and you accept. It is one by whos authority? This authority is certainly no greater than mine or yours unless you make it so by not seeing.

Meaningless waffle.

I don't need to accept anyone's authority on it I've seen many proofs. There's no need to appeal to authority when it can be shown with logical arguments, if you throw logic out the window you shouldn't expect to get answers of any menaing.

It's not about 'rounding up', it's basic maths.

 

[matt grime]

Because ideas are accepted as fact by the masses, does it make it so? What defines logic but the logic before it which is based upon the agreement of a base worked forward. Because we have decided to accept something as such, is it such? It is such because you were taught so is it not.

It is an accepted value, I have accepted it as such a value in caluclations, it is rounded and creates one, but is it one? No it tries to be but never quite reaches. The issue can be avoided in large calculations by using factionary math, but when rounding must occur, it is what it is.

You can sweep it under the carpet, but that is why so many do not understand. Because you do not question and you accept. It is one by whos authority? This authority is certainly no greater than mine or yours unless you make it so by not seeing.

to show you're not a crank explain from first principles the construction of the real numbers in your favourite way and explain how it is that 0.999... and 1 represent different real numbers (base 10 expansions).

if as i suspect you've no idea what the definitions of the real numbers are, then this could be an interesting exercise for you. HINT they are not decimal expansions.

 

[quartodeciman]

I am interested why TENYEARS insists:

"Throw out the word real".

 

[ahrkron]

TENYEARS,

It is nice that you express your feelings about the meaning of authority, and the need for people to question what they are taught. However, they have nothing todo whatsoever with the issue at hand. The reason is that, in math, it all starts with an agreed upon set of definitions. There is no dogma as to how "things should be". Once one understands that, it is clear that all conclusions are based on the definitions used.

In this particular case, the fact that 0.999...=1.000... is an inevitable consequence of the definitions used for the objects called "real numbers". You are free to use or not use that representational tool (the "real numbers"), but you cannot change the fact that, given the definitions of the system, the stated property follows.

A slightly different matter is the application of math concepts to physical science, but also in that case there is not much room for people to "express their freedom" or "question authority", since in that case the only possible adequacy test is given by experiment; again, this is so only because the agreed-upon definition of physical science.

If you want to use different definitions and methods, you are certainly free to do so, but then you need to realize (and you also need to be clear about it, so others don't get confused) that your statements do not refer to "math", "physics", or "real numbers", but other, distinct, disciplines and objects.

 

[Zorodius]

... infact there's always an infinite number of real numbers between any two real numbers, the proof is trivial ...

If it wouldn't be a bother, could I see that proof?

 

[Hurkyl]

Suppose a < b.

Then, a + a < a + b (add a to both sides)
Also, a + b < b + b (add b to both sides)
Divide both inequalities by 2 to get:

a < (a + b)/2 < b

In this way, given any two distinct numbers in an ordered field, such as the real numbers, we can show there is another number between those two.

Similarly,
a < (2a + b)/3 < b
a < (3a + b)/4 < b
a < (4a + b)/5 < b

and so on.


Or, if you prefer, you can structure the proof like this:

There must be a c such that a < c < b,
then, there must be a d such that a < d < c (< b),
then there must be an e such that a < e < d (< b),
and so on.

 

[HallsofIvy]

Let a, b be any two numbers, a< b. Then, adding a to both sides, a+b< 2b so (a+b)/2< b. Also, adding b to both sides of a< b, 2a< a+ b so a< (a+b)/2. That is, (a+b)/2 is a real number strictly between a and b. Repeat to show that (3a+b)/4 is a real number strictly between a and (a+b)/2 and use induction to show that ((2ⁿ-1)a+ b)/2ⁿ) is a real number strictly between a and ((2^n-1-1)+b)/2^n-1. Since the inequality is strict, these are all distinct real numbers. That is, there exist an infinite number of distinct real numbers strictly between a and b.

Although it is a bit harder, the proof can be extended to show that between any two real numbers there is a rational number and between any two real numbers there is an irrational number.

 

[Integral]

Because ideas are accepted as fact by the masses, does it make it so? What defines logic but the logic before it which is based upon the agreement of a base worked forward. Because we have decided to accept something as such, is it such? It is such because you were taught so is it not.

It is an accepted value, I have accepted it as such a value in caluclations, it is rounded and creates one, but is it one? No it tries to be but never quite reaches. The issue can be avoided in large calculations by using factionary math, but when rounding must occur, it is what it is.

You can sweep it under the carpet, but that is why so many do not understand. Because you do not question and you accept. It is one by whos authority? This authority is certainly no greater than mine or yours unless you make it so by not seeing.

This is a pretty funny argument! According to all the evidence I have seen (several web polls) The masses believe that 1<> .999... , so by your own argument it seems that you should believe the opposite. Yet you follow the masses like a blind sheep. Sound logic and knowledge of the Real Numbers lead to a conclusion contrary to the believe of the Masses. Perhaps rather then being a blind sheep you should educate yourself and learn to draw conclusions based on knowledge and logic rather then intuition.

 

[TENYEARS]

What proof you would like to come up with is irrelevant. .99999999... <> 1.
You would have to project it into infinity and it would approch 1 but never be 1. You may accept it as such, but it is not correct. 1/3 * 3/1 will yeild one, but .333333333333... * 3 will not yeild one. We accept it as one, but it is not one. One is infinite and complete. The approximation is not. Like I said, I do not expect you to understand. I am done with expectation. If you are trying to project infinite probability, you will always stop short with .999999... you must stop somewhere to project, if not you will always be running.

 

[ahrkron]

What proof you would like to come up with is irrelevant. .99999999... <> 1.

As I said, you should be explicit in the fact that you are not talking math any more. If proof is irrelevant, you are definitely talking about something other than math.

You would have to project it into infinity and it would approch 1 but never be 1.

This sounds like philosophy or pseudoscience.

1/3 * 3/1 will yeild one, but .333333333333... * 3 will not yeild one.

So, in the language you are using, 1/3 <> 0.3333...?

This is starting to go in circles.

 

[master_coda]

.999… not equal to 1? That’s kiddy stuff, just watch me argue that .3333… is not equal to 1/3

Wow, and I thought that was just a joke.

 

[Integral]

What proof you would like to come up with is irrelevant. .99999999... <> 1.
You would have to project it into infinity and it would approch 1 but never be 1. You may accept it as such, but it is not correct. 1/3 * 3/1 will yeild one, but .333333333333... * 3 will not yeild one. We accept it as one, but it is not one. One is infinite and complete. The approximation is not. Like I said, I do not expect you to understand. I am done with expectation. If you are trying to project infinite probability, you will always stop short with .999999... you must stop somewhere to project, if not you will always be running.

But .999... IS an infinite number of nines, what can you possibly mean by stop short?

Perhaps it is your concept of infinity that is flawed, that along with your lack of knowledge of the Real Numbers leads you to false conclusions.

 

[TENYEARS]

1/3 * 3/1 If I mulitple the across the top I have 1 * 3 = 3, if I multiple across the bottom I have 3 * 1 = 3 If I then divide I have 1.

If I take the number 1 divide it by 3 I have an infinite series of .33333. If I multiple that * 3 I have .9999999 in an infinite series. It is accepted as 1, but it is not 1.

By stopping short I mean if you take the .99999... in an infinite series and attempt a calculation, unless you were satisified with that degree, you could could never complete the calculation. In a projection of infinite proability this would be significant.

The processor would have to do a conversion of .999... series and equate that to 1 in order for proper calculations to occur otherwise the result could potentially be flawed.

The reason we .99999 series is accepted as 1 is because the assumption is it was taken from 1/3 of a whole which added together(.3333... 3333... 3333...) creates a whole object or 1 this would preclude one knowing that the past calculation was indeed done in this fashion because if not the projection could potentially go billions deep and not be repetative.

 

[Integral]

The reason we .99999 series is accepted as 1 is because the assumption is it was taken from 1/3 of a whole which added together(.3333... 3333... 3333...)

The DEMONSTRATION you give is not a complete or stand alone proof. There are many ways of proving this FACT without doing that calculation. Once again your limited knowldege of the field is leading you to bad assumptions.

Edit: completed a word, the out fell through the keyboard in "withOUT"

 

[Integral]

By stopping short I mean if you take the .99999... in an infinite series and attempt a calculation, unless you were satisfied with that degree, you could could never complete the calculation. In a projection of infinite proability this would be significant

Infinite geometric series CAN be summed EXACTLY. The formula for this is well known and it is NOT an approximation. http://home.comcast.net/~rossgr1/Math/one.PDF link shows a detailed use of the this formula. It is followed by a separate proof which you may be able to follow.

 

[TENYEARS]

My view on this will not change under any condition, but the initiating post was good. To test the accepted as if one has never seen it before is what creates breakthoughs.

 

[BoulderHead]

Willfulness?

 

[Integral]

My view on this will not change under any condition, but the initiating post was good. To test the accepted as if one has never seen it before is what creates breakthoughs.

Nonsense. Breakthroughs can only come when the current theories are well understood. The state of the sciences has progressed to the point where you MUST understand what is known before you can hope to add anything to the state of knowledge.

Your blind adherence to faulty concepts spells doom for any hope of your contributing anything of meaning.

 

[Hurkyl]

My view on this will not change under any condition,

Then I see no point in continuing this discussion. Do your preaching in some other thread.

 

[ahrkron]

I think we are done here.