1 = 0.99... (RossGr)

[RossGr]

Originally posted by: Pudgygiant
My parents (1 with a masters in math and one with an associates in math) and my math professor (phd) are all saying that .3... and .6... are just approximations for 1/3 and 2/3, respectively.

You'd expect better, it must be a bit disappointing.

 

[sao123]

sao your list if undefined numbers is bunk as most of the numbers you put in there to start of with are undefinded. And to make maters worse 5 and 7 are actually defined.

Unless you are nonmathematically inclinde,,,indeterminate is the mathematicians way of saying undefinable or definable as simultaneous multiple conflicting values

I suggest you do some research on the indeterminate forms of numbers. These forms are all recognized & accepted by mathmaticians.

Perhaps you would like to check here...
BTW... 5 & 7 are not defined. Perhaps Dr. Math will answer your question...

 

[Dinominant]

So why can't you just add 1/oo ? I think that is an irrational number, but (sqrt2)^2=2, so wouldn't 0.9...+(1/oo)=1
you couldn't add 2/oo because that would produce 0.0...2 (yes I know that n/oo=1/oo but that is because we are working with oo and anything is game... right??)

My notation is probably very bad, but you get the idea. Wouldn't finding a loophole and exploiting it produce anything you want in math (since you can do it in the court)? Maybe all we are doing is exploiting a loophole when saying 1=0.9...

Keep in mind that I am still in high school and I don't know everything - go easy on me.

 

[Lynx516]

Ross Gr argue your case taht the difference bettweeen 0.999... and 1 is 1 * 10^(-N) where N tends to infinity.

By the way dont argue that much because this comes from someone who is a Senior Wrangle. If you dont know what that is it means he got a first (highest degree you can get) from Cambridge Univesrcity in the UK. And that is one of the hardest maths courses in the world.

 

[RossGr]

Originally posted by: Lynx516
Ross Gr argue your case taht the difference bettweeen 0.999... and 1 is 1 * 10^(-N) where N tends to infinity.

By the way dont argue that much because this comes from someone who is a Senior Wrangle. If you dont know what that is it means he got a first (highest degree you can get) from Cambridge Univesrcity in the UK. And that is one of the hardest maths courses in the world.

When did I state that? Did you examine my proof?

In my proof I make the following claims,
1 - 10^-N < .999....
and
.999.... + 10^-N > 1

For All N
I do not use the concept of a limit in my proof.

By the way it is not clear to me how you take a limit on a fixed number. Are you implying that .999... is not a fixed number?

 

[Lynx516]

sorry I miss typed.
it should read argue that it isnt. As you said may people woudl argue. And I am not limiting .9999 I am limiting the difference bettween it and 1 by first taking an aproximation then improving on it. Hence a limiting process.

 

[RossGr]

Originally posted by: Lynx516
sorry I miss typed.
it should read argue that it isnt. As you said may people woudl argue. And I am not limiting .9999 I am limiting the difference bettween it and 1 by first taking an aproximation then improving on it. Hence a limiting process.

I did not mean to imply that these people who argue this have rational arguments!

If you have lots of time read through this thread.
to see some of the objections to this concept.

 

[sao123]

I am still working on that equation, though i may have it soon....

But here is food for thought.

1...In order to prove that two numbers a and b are different, you must be able to explicitly define another number, c, that is larger than a and smaller than b.
(notation: E = summation, 00 = infinity, 00i = infinity with subscript i)

Let a = .999999.... or 9 * E(.1^n) as n->00
Let b = 1
I must find C that fits a < c < b. Possible or no?

Let me denote now that:
Let a = .999999.... or 9 * E(.1^n) as n->00a ( I will denote infinity here as infinity subscript a)

Since it has been proven that there are multiple levels of infinity (actually an infinite number of levels of infinity.)
Then I could choose any higher level of infinity ( lets say i chose infinity subscript c [00c] & infinity subscript d [00d] ) where 00d > 00c > 00a.

Then I have just defined a c which meets criteria 1? That being:
Let c = .999999....99999.... or 9 * E(.1^n) as n->00c
1 > 9 * E(.1^n) as n->00d > 9 * E(.1^n) as n->00c > 9 * E(.1^n) as n->00a

Thoughts plz.

 

[ShawnD1]

You guys are using flawed math to prove something a 2 year old can figure out.
1 != 0.999999 because 1 is a limit and 0.9 repeating is your number. As you know, a limit is NOT the same as the number approaching that limit.
Using the stupid explanation that a number is the same as its limit, I can also say that 1/y = 0 where Y approaches infinity. Can you ever divide by a number to somehow get 0? No. With the above equation, the number 0 is not an actual value, IT IS A LIMIT; the value does not actually reach 0 but it comes very close. Let's assume it was 0 though, then I could write 1 = 0 x y. Then I can write 1 = 0. Oh right on! By using incorrect math I have just proven that 1 = 0, infact, I have just proven that every number = 0.

Another thing, as said before, is that you are not using the proper number system. If we take 1/3 then add 2/3, we get 3/3 which is 1. By using a base 3 system we have got the right answer. You can't just use a screwy system and turn 1/3 and 2/3 into estimations then add estimations.



incorrected calculation deleted.

 

[RossGr]

Originally posted by: ShawnD1
You guys are using flawed math to prove something a 2 year old can figure out.
1 != 0.999999 because 1 is a limit and 0.9 repeating is your number. As you know, a limit is NOT the same as the number approaching that limit.
Using the stupid explanation that a number is the same as its limit, I can also say that 1/y = 0 where Y approaches infinity. Can you ever divide by a number to somehow get 0? No. With the above equation, the number 0 is not an actual value, IT IS A LIMIT. Let's assume it was though, then I could write 1 = 0 x y. Then I can write 1 = 0. Oh right on! By being an idiot I have just proven that 1 = 0, infact, I have just proven that every number = 0.

Another thing, as said before, is that you are not using the proper number system. If we take 1/3 then add 2/3, we get 3/3 which is 1. By using a base 3 system we have got the right answer. You can't just use a screwy system and turn 1/3 and 2/3 into estimations then add estimations.




Another thing I have to say. If you want to prove that .999999 = 1, you cannot use your calculator. Calculators work on limits so if you put in 10/.99999999999 it won't bother spitting out an answer very close to 10, it'll just say 10.
To prove 0.999999 != 1, just do it long hand. You'll see that you end up with 10.01001001001001 and the 001 just keeps repeating.

You should attempt to understand my proof before speaking about thay which you know so little.

My proof is vaild, the fact that 1=.99... is a fundamental fact of the construction of the Real Number line. You need to study a bit beyond a first year calculus course to understand this.
Edit;
Both 1 and .999.... are fixed numbers. Will you please explain how to take limits of fixed numbers?

 

[RossGr]

Originally posted by: sao123
I am still working on that equation, though i may have it soon....

But here is food for thought.

1...In order to prove that two numbers a and b are different, you must be able to explicitly define another number, c, that is larger than a and smaller than b.
(notation: E = summation, 00 = infinity, 00i = infinity with subscript i)

Let a = .999999.... or 9 * E(.1^n) as n->00
Let b = 1
I must find C that fits a < c < b. Possible or no?

Let me denote now that:
Let a = .999999.... or 9 * E(.1^n) as n->00a ( I will denote infinity here as infinity subscript a)

Since it has been proven that there are multiple levels of infinity (actually an infinite number of levels of infinity.)
Then I could choose any higher level of infinity ( lets say i chose infinity subscript c [00c] & infinity subscript d [00d] ) where 00d > 00c > 00a.

Then I have just defined a c which meets criteria 1? That being:
Let c = .999999....99999.... or 9 * E(.1^n) as n->00c
1 > 9 * E(.1^n) as n->00d > 9 * E(.1^n) as n->00c > 9 * E(.1^n) as n->00a

Thoughts plz.

The real numbers have cardinality of the continum, that is the infinity that we must use when dealing with them. We are simply not free to choose a sets cardinality, it is an inherient proberty of the set.

 

[uart]

Originally posted by: DrPizza

Originally posted by: uart

Originally posted by: nrg99
what about 1=0?

Sorry, the closest I can get to that is 0^0 = 1. Is that whacky enough for you?

0^0 is undefined. Simple explanation: x^12 divided by x^9 = x^3.... subtract the coefficients. x^12 divided by x^12 = x^ (12-12) = x^0 = 1. However, if x=0, then the initial fraction involves dividing by 0. Thus, 0^12 divided by 0^12 does NOT equal 1. It is division by 0 which is undefined. The entire rule for "x^0 = 1" includes "x != 0"

0^0 as a limit is indeterminate (and depending on what the limit is of, 1 is a possible answer).

Yes I know that strictly speaking 0^0 is undefined, I was just trying to be controversial.

x^0 = 1 and is well defined for all x other than 0.

0^x = 0 for all x>0 and is Inf (or undefined) for all x<0. I can remember a Maths Lecturer once told me that some authors "define 0^0" as being 1 for convenience, but that this was a little controversial and not universally accepted.

BTW, Both Windows calculator and Maple7 return 0^0=1.

 

[ShawnD1]

Originally posted by: RossGr


You should attempt to understand my proof before speaking about thay which you know so little.

My proof is vaild, the fact that 1=.99... is a fundamental fact of the construction of the Real Number line. You need to study a bit beyond a first year calculus course to understand this.
Edit;
Both 1 and .999.... are fixed numbers. Will you please explain how to take limits of fixed numbers?

That's where the disagreement is. You (and the people who agree with you) think that 0.9999 repeating is a fixed number. What the people like me are saying is that it is not a fixed number. To get a fixed number, it has to end or be written as a fraction or a root or an infinite series (like sine, cosine and e) or something else where you get an exact value. How do you get the exact value of 0.9 repeating? It gets closer to what you are trying to say as you add more digits but it doesn't reach the value you are trying to give it. Sine and cosine do the exact same thing but we can still write them down as equations. You just have to realize that the number is not the same as the number it is reaching, that's all. Remember in math when you would get a question like "what is the domain of x" when you have something like (x + 3)/(x - 5)? You could say that x is absolutely anything EXCEPT 5. You can say x is infinitesimally smaller or greater than 5.
1 is not the same as .99999 repeating because .9999 repeating can't be used in any sort of calculation. 1 can be used in any calculation but .9999 can't be used since it has no real defined answer. By real, I don't mean the number system, I mean the ability to write it down properly. By real I mean the difference between writing square root 3 and writing a decimal estimate for square root 3. The decimal value for root 3 doesn't end so I can't use it in any calculation to get an exact answer. There is a huge difference between writing 3 x root 3 and writing 5.1961524........ just as there is a huge difference in writing 0.9999999....... and writing 1. For all practical purposes they are the same but mathematically speaking, they are different.

To prove that 0.9999 = 1, somebody has to be able to write it properly and get an exact value for it. Is there any way to write it as an infinite series perhaps?

 

[DrPizza]

BTW, just a little clarification for people about limits

.999999 repeating = .9 +.09 +.009 +.0009 + . . .
or 9/10 +9/100 +9/1000 +9/10000 + . . .

to find the sum,

the sum IS the limit. It's not "the sum approaches the limit"

the limit is "what number is that sum approaching as n approaches infinity"
This does NOT mean that the sum is approaching a number.
The sum IS the number.

Also, as implied by a couple of people above, infinity is NOT a number, it's not "the highest number", it's a concept.

 

[uart]

Originally posted by: DrPizza
BTW, just a little clarification for people about limits

.999999 repeating = .9 +.09 +.009 +.0009 + . . .
or 9/10 +9/100 +9/1000 +9/10000 + . . .

to find the sum,

the sum IS the limit. It's not "the sum approaches the limit"

the limit is "what number is that sum approaching as n approaches infinity"
This does NOT mean that the sum is approaching a number.
The sum IS the number.

Also, as implied by a couple of people above, infinity is NOT a number, it's not "the highest number", it's a concept.

Yes I'd also noticed that misconsception that many people seem to be having here about what a limit is.

Thanks for posting that clarification Dr P, interestingly I was just thinking about posting almost exactly the same thing.

 

[RossGr]

, I don't mean the number system, I mean the ability to write it down properly. By real I mean the difference between writing square root 3 and writing a decimal estimate for square root 3. The decimal value for root 3 doesn't end so I can't use it in any calculation to get an exact answer. There is a huge difference between writing 3 x root 3 and writing 5.1961524........ just as there is a huge difference in writing 0.9999999....... and writing 1. For all practical purposes they are the same but mathematically speaking, they are different.

This of course is your belief. Your believies have nothing to do with mathematics. I am sorry you are simply wrong, MATHEMATICALLY speaking they are the same point on the number line.

Once again I will request you read the proofs linked in my sig. If you cannot take the time to do that please refrain from posting in this thread.

 

[uart]

, I don't mean the number system, I mean the ability to write it down properly. By real I mean the difference between writing square root 3 and writing a decimal estimate for square root 3. The decimal value for root 3 doesn't end so I can't use it in any calculation to get an exact answer. There is a huge difference between writing 3 x root 3 and writing 5.1961524........ just as there is a huge difference in writing 0.9999999....... and writing 1. For all practical purposes they are the same but mathematically speaking, they are different.

That?s a stupid argument Shawn. The ellipsis notation (...) you use makes NO sense for an irrational number like sqrt(27). When you write sqrt(27)=5.1961524... the ?...? has no real meaning, other than to indicate that there are more decimal places which are not shown. By the very nature of an irrational one can never specify all the decimal places, either explicitly or via a repeating relation. However when you write 1/3 = 0.3333(3)repeated then it has a very definite meaning. There is no comparison between the case of writing a rational as a repeated decimal and that of writing a decimal approximation of an irrational, these are completely different things.

 

[OokiiNeko]

n/m

 

[NittanyLAncer]

I'm not sure how many of you have actually read the proof, it's very slick, but very wrong.



<argue
1/3 != .3333(repeating) You should have learned that by the first time through a calc class. But that's not a point that should be argued, as most people WITH a math background will tell you that you can represent rational number as a infinitely repeating decimal, but it's not right.

ATo note, 0.1 is NOT a rational number, 1/10 IS. Rational numbers aren't decimals PERIOD if you look at it as a higher math problem, the set "Q" isn't built that way at all, it's built using a diagonal line through 2 quadrants of a graph, it's only a series of fractions. The reason 0.1 "is" a rational number is because you can represent it "as 1/10." Decimals are, in the strictest (and every) sense, REPRESENTATIONS, not rational numbers. Most teachers and people in general don't argue cause they don't care. Sure, 0.1, 1.7, 317317.317 can all be equated to rational numbers, but they are NOT rational numbers.

</argue>

I wonder what RossGr.'s math background is, and whether he's trying to pull one over our eyes or just making fun of all the laymen. You see, his proof is RIGHT mathematicially, but he's making ONE point and then pretending he proved something ELSE. He shouldn't be trying to pass off a calculus "=" as the boolean "=". I should hope anyone with a math background should catch on.

The problem with the proof isn't the equals sign, so don't try to argue that. He's right, his series DOES "=" 1. Don't argue the fraction either, sorta pointless at this point as most people will try to argue points which can be easily pushed aside.

The problem is that it is NOT "=" 1. "equals" and "equal to" (or as Dr. Pizza said "equals exactly one") are two ENTIRELY different things by the time you reach convergent series (the "meat" of his proof). "=" is no different than any other calculus/physics symbol, it's overloaded up the arse. The terms "equal" and "stricly equal" are different. THAT'S where he gets you.

You have to argue the proof, and find the implication which is wrong. Trying to prove the sum of an infinite series "stricly equals" or "equals exactly" a finite number cannot be done (there may be some VERY special cases that I might not have encountered, but this aint one). It doesn't happen here. You CANNOT use infinite series in this way, BIG BAD GLARING NO NO, because finding the sums of an infinite series is based (at it's core) off of improper intigrals. Improper intigrals use LIMITS of t as t=>inf. Once you use a limit sign the boolean "=" goes out the window. And ANYONE who argues that you can use a boolean operator "=" on a limit should be forced to read a copy of Anton and Anton every night for the rest of their life. Good behavior gets it replaced by a copy of Anton Bevins and Davis, although that's not that much better.

His proof IS right, but his IMPLICATION FROM THE PROOF is flawed. Take a discrete math class to learn all about that.

You can always confuse people with math, and there are LOTS of tricks (used for tricking freshman buisness majors into thinking 1=2 with a division by 0 is always fun). A favorite is proving that all the cats in the world are black using induction. Doesn't make it right.

Heaven forfend a mathematician ever run for office.

</rant>

 

[Lynx516]

ignore

 

[RossGr]

Nitannty Lion
Your arguments seem to only address the sum of a series argument. This is of course completely valid, as any student of calculus should be able to tell you.

None of your arguments seem to even touch the second proof. This proof is not from your first time though Calculus course. It is based on something called Real Analysis, which is one of the courses required of all math majors.

The question of equality on the real line is a very valid issue. In my second proof I claim equality based on the statement

1-10^(-N) < .999... < 1+ 10^(-N) being true for all N where N is a positive integer.

This of course requires proof. I did not present this as part of my proof, have been meaning to, just haven't got around to it. The fact is equality on the real line is different from boolean equality, but thats ok, we permit the logic people to use our symbols, as long as they define what they mean in their special cases. The root of all such definitions lies in Real Analysis. Your arguments are basically this, my special case is different from the general case, therefore the general case is wrong.

Bad logic on your part. Seems to me that is a well known logical fallacy.
By the way, while qualifications mean little or nothing on the web, it is content that you are judged on, here are mine.

I have the following degrees
BS Physics 1977
BS Math 1984

Both from Oregon State U.

I have also completed Masters Level course work in Math. (Including Grad Level Real Analysis courses)
At Lehigh and Oregon State

Edit: Spell check

 

[ZeroNine8]

In the second method of proving this, you claim that for all N>0
0.99...-0.1^N < 1
and
0.99...+.1^N > 1

The first is trivial, however the second one seems to hinge on the assumption that 0.99...=1 before it can even work. It seems to me that if you do not assume 0.99...=1 initially, then you cannot say 0.99...+.1^N>1 for all N. I would be curious to see the proof of this statement on its own merit, rather than initially assuming it is then deriving the proof from that. If I'm off-base here, please explain how you can do this without assuming 0.99...=1.

Edit: I shouldn't say it hinges on the assumption that 0.99...=1, rather it hinges on the assumption that 0.99...<1 is false. The argument is the same, though the syntax may not be exactly as it should.

 

[RossGr]

Originally posted by: ZeroNine8
In the second method of proving this, you claim that for all N>0
0.99...-0.1^N < 1
and
0.99...+.1^N > 1

The first is trivial, however the second one seems to hinge on the assumption that 0.99...=1 before it can even work. It seems to me that if you do not assume 0.99...=1 initially, then you cannot say 0.99...+.1^N>1 for all N. I would be curious to see the proof of this statement on its own merit, rather than initially assuming it is then deriving the proof from that. If I'm off-base here, please explain how you can do this without assuming 0.99...=1.

Edit: I shouldn't say it hinges on the assumption that 0.99...=1, rather it hinges on the assumption that 0.99...<1 is false. The argument is the same, though the syntax may not be exactly as it should.

Perhaps you should actually read the proof! I did not assume anything. I very carefully developed that relationship. Now go back to the proof and actually make an effort to understand it.

 

[ZeroNine8]

nm, i started reading the 'second' proof too far down, i just scrolled past all the sigma stuff thinking it was part of the first, not realizing it was developed up there. Besides, reading the proof has nothing to do with it, I've read it several times and still managed to skim right over that section the last time.

 

[Lynx516]

Sorry but what you said is that you can skim a proof (which makes you miss out vital details) then pick at a point without fully understanding the proof. That stinks of bad mathmatical practice and ilogical thought. Can you please clarify your mathmatical background?

edit: You should refer to the academic paper " R.V. Churchill and J.W. Brown. Complex Variables and Applications. 0.9999... = 1 ed., McGraw-Hill, 1990."
That should solve this issue for ever.