Is 1 = 0.9999......

[MadRat]

Funny how the same logic to take .999... to its limit of one and define .333... as 1/3 can be abused to make 1 into a ...0001.000... value.

They are consistent with their madness.

 

[silverpig]

Originally posted by: MadRat
Funny how the same logic to take .999... to its limit of one and define .333... as 1/3 can be abused to make 1 into a ...0001.000... value.

They are consistent with their madness.

0.999... doesn't have a limit d00d. What's the limit of 2?

If you have a problem with seeing 1 as ...0001.000... then I suggest you re-read your grade 5 math book and re-learn place value. I don't know if that's what your problem as stated is as not much of that makes sense, but if that is indeed it...

 

[CTho9305]

LMAO@bleeb and madrat. I am embarrased for all those who voted no. You must be english majors or something .

 

[silverpig]

Originally posted by: CTho9305
LMAO@bleeb and madrat. I am embarrased for all those who voted no. You must be english majors or something .

[non-personal-bad-joke] 0.999999 of a McDonalds french fry might as well be 1 full fry, but it's still less than a full one[/non-personal-bad-joke]

 

[stonecold3169]

Very dependant upon who you talk to, and your own personal beliefs.

The first proof is just wrong. You assume the conclusion in the begining and then use those "facts" to go where you want to. Thus, your logic is wrong.

Now where you actually go with this varies a whole lot. I'm a dual major (CS and Math) and when I was shown the fraction proof it was used to show that our representation of numbers is incredibly flawn, which is quite true. Also, this is one of the main problems that the sect of mathmatics that has a problem with infinity lies. Any mathmetician on this board has seen proofs proving different levels and sizes of infinity. Our definition of infinity is kinda sketchy to begin with, so by .3 repeating, for "infinity", what does that mean? It's a hard question to answer accurately until you've taken number theories and the such (I'm sure there are many, many people on this board much better with math then I am, so I hope I don't offend anyone).

To give an easy example, try and prove that there are an infinite number of twin primes (a twin prime is a set of prime numbers 2 numbers away from each other, such as 5 and 7, 17 and 19, etc.). We can prove to a very, very large digit that they exist, but we don't know enough properties about infinity to know if patterns break at certain levels. I realize the above is a whole lot different from proving infinity, but I want a quick example I knew was unsolved.

To all the people proving this with calculus... calc is based upon simplifying very very complicated (sometimes so hard we find in unfathomable to solve it otherwise) by taking certain shortcuts. When done correctly, your percent error will typically be so small that it is negligable for our purposes (When building a bridge it is pretty irrelevent in the tension force on a support is off is incorrect at the 100,000,000th digit behind the decimal. However, to accomidate we DO have coefficents of safety, etc which partly tie to this, although that is different altogether). Calc is not an exact science, it isn't meant to be. It is a powerful tool for very close approximations, not exactness.

 

[silverpig]

Originally posted by: stonecold3169
Very dependant upon who you talk to, and your own personal beliefs.

The first proof is just wrong. You assume the conclusion in the begining and then use those "facts" to go where you want to. Thus, your logic is wrong.

Now where you actually go with this varies a whole lot. I'm a dual major (CS and Math) and when I was shown the fraction proof it was used to show that our representation of numbers is incredibly flawn, which is quite true. Also, this is one of the main problems that the sect of mathmatics that has a problem with infinity lies. Any mathmetician on this board has seen proofs proving different levels and sizes of infinity. Our definition of infinity is kinda sketchy to begin with, so by .3 repeating, for "infinity", what does that mean? It's a hard question to answer accurately until you've taken number theories and the such (I'm sure there are many, many people on this board much better with math then I am, so I hope I don't offend anyone).

To give an easy example, try and prove that there are an infinite number of twin primes (a twin prime is a set of prime numbers 2 numbers away from each other, such as 5 and 7, 17 and 19, etc.). We can prove to a very, very large digit that they exist, but we don't know enough properties about infinity to know if patterns break at certain levels. I realize the above is a whole lot different from proving infinity, but I want a quick example I knew was unsolved.

To all the people proving this with calculus... calc is based upon simplifying very very complicated (sometimes so hard we find in unfathomable to solve it otherwise) by taking certain shortcuts. When done correctly, your percent error will typically be so small that it is negligable for our purposes (When building a bridge it is pretty irrelevent in the tension force on a support is off is incorrect at the 100,000,000th digit behind the decimal. However, to accomidate we DO have coefficents of safety, etc which partly tie to this, although that is different altogether). Calc is not an exact science, it isn't meant to be. It is a powerful tool for very close approximations, not exactness.

Those two types of infinity are completely different.

The fraction proof is perfectly fine, ask any Ph.D mathematician.

The twin prime problem doesn't have to do with infinity, but with our inability to come up with a formula describing the prime numbers. We just don't know what the next one will be given a certain prime.

Calculus does give you exact answers. I don't know why people don't get that. Maybe it's because you were taught riemann summation first or something, but calculus finds the exact limit of a riemann sum, that limit being the area/volume/whatever.

If you don't believe me, try using calculus to find the area of a circle x^2 + y^2 = r^2. I guarantee you will get EXACTLY pi*r^2.

 

[stonecold3169]

Originally posted by: silverpig

Originally posted by: stonecold3169
Very dependant upon who you talk to, and your own personal beliefs.

The first proof is just wrong. You assume the conclusion in the begining and then use those "facts" to go where you want to. Thus, your logic is wrong.

Now where you actually go with this varies a whole lot. I'm a dual major (CS and Math) and when I was shown the fraction proof it was used to show that our representation of numbers is incredibly flawn, which is quite true. Also, this is one of the main problems that the sect of mathmatics that has a problem with infinity lies. Any mathmetician on this board has seen proofs proving different levels and sizes of infinity. Our definition of infinity is kinda sketchy to begin with, so by .3 repeating, for "infinity", what does that mean? It's a hard question to answer accurately until you've taken number theories and the such (I'm sure there are many, many people on this board much better with math then I am, so I hope I don't offend anyone).

To give an easy example, try and prove that there are an infinite number of twin primes (a twin prime is a set of prime numbers 2 numbers away from each other, such as 5 and 7, 17 and 19, etc.). We can prove to a very, very large digit that they exist, but we don't know enough properties about infinity to know if patterns break at certain levels. I realize the above is a whole lot different from proving infinity, but I want a quick example I knew was unsolved.

To all the people proving this with calculus... calc is based upon simplifying very very complicated (sometimes so hard we find in unfathomable to solve it otherwise) by taking certain shortcuts. When done correctly, your percent error will typically be so small that it is negligable for our purposes (When building a bridge it is pretty irrelevent in the tension force on a support is off is incorrect at the 100,000,000th digit behind the decimal. However, to accomidate we DO have coefficents of safety, etc which partly tie to this, although that is different altogether). Calc is not an exact science, it isn't meant to be. It is a powerful tool for very close approximations, not exactness.

Those two types of infinity are completely different.

The fraction proof is perfectly fine, ask any Ph.D mathematician.

The twin prime problem doesn't have to do with infinity, but with our inability to come up with a formula describing the prime numbers. We just don't know what the next one will be given a certain prime.

Calculus does give you exact answers. I don't know why people don't get that. Maybe it's because you were taught riemann summation first or something, but calculus finds the exact limit of a riemann sum, that limit being the area/volume/whatever.

If you don't believe me, try using calculus to find the area of a circle x^2 + y^2 = r^2. I guarantee you will get EXACTLY pi*r^2.

I never said that the fraction proof was not technically correct... it works, given our definitlion of numbers, I said that the first proof assumed the conclusion which is a no-no. I CAN however point you towards my calculus prof, who not only has his Phd in math, but is also critically acknowledged for an amazing discovery in a path of orbitals spaceshuttles can take to reach the moon at the cost of 1/4 the fuel it takes now.. at the cost of an obcene amount of time.), and he will tell you flat out that he buys the proof, but only because we have no basis for the notation we use (well, a weak foundation). To fight and say that any proof involving infinity, which is unproven, is absolutely true is ridiculous. Math is constantly changing, even though most people agree with the mere thought of infinity doesn't mean it exists. there is as I mentioned a brilliant sect which doesn't buy it at all.

To say that calc is an exact science is also considered an error by me. I can say that 2+2=4, and 2^2 =4, does this make x^x= x + x? no, because you would need to prove that this works for all values, and it's easy enough to find a value which disproves this statement. Perhaps my bridge example was flawn, I just wanted to show that I DO believe calc to be very highly accurate, esspecially for our purposes. However, with no way of testing many, many problems with anything but calc, it's hard to say it always gives perfect answers, right?

 

[MikeMike]

i know .99999999=1 but
y then doesn't .9999999999+.99999999=2=1+1?

sorry if already stated i am to lazy to read all posts

 

[silverpig]

Originally posted by: nourdmrolNMT1
i know .99999999=1 but
y then doesn't .9999999999+.99999999=2=1+1?

sorry if already stated i am to lazy to read all posts

It actually does... try it

 

[CTho9305]

Originally posted by: nourdmrolNMT1
i know .99999999=1 but
y then doesn't .9999999999+.99999999=2=1+1?

sorry if already stated i am to lazy to read all posts

It does.

 

[silverpig]

And actually, that right there is another proof in and of itself

 

All right, I should probably chime in here. I think the problem with Silverpig's side is that they understand the mechanics of mathematics but not the philosophy of it. Yes, mathematics have axioms that require consistency in the story line (I use that phrase loosely). The point is, one side is arguing based on the mechanics of mathematics, which boils that specific axioms and theorems derived from the axioms.

At the end, the question of technicality has a truth to it on whether 0.9999.... is equivalent to one from a mathematician's point of view. But there is a metaphysical aspect of it which is what is being argued. Those like Silverpig (seeming to be a realist) who seem to be strong believers in mathematics as abstract truths insist on the absoluteness of that statement. But the truth is, from an idealist's point of view and an empiricist, mathematics only matters in solving the problem of the world and physical sciences. So if the concept of infinity does nothing, then it doesn't matter what a pure mathematician believes. Zero was created not because it exists or we couldn't do without zero, but because it made mathematics much easier to work with and solve the problems of the physical sciences. Many aren't taught this at school, but for standard systems in mathematics, particularly set theory, there are competing theories (for instance axiom of choice vs other set theories).

In essence, this question has gone past technicality and is simply a discussion for philosophers of mathematics. It seems to me that Silverpig and those who share his view don't understand this. But to be fair, on the other end there are people arguing fruitlessly for something that is based on an accepted premise, yet denying what follows from the premise.

All right, I'm out. Gosh, so many nerds on this forum that would kill for their opinions on technical discussion. It kind of surprises me. I guess we all have our passions that we would die for. Enjoy your discussions without [personal] attacks, if I may add.

 

[NovaTone]

another proof...

.9999(repeating) DOES equal one by convergence. the key here is repeating. Interpreting the number in decimal form does NOT accurately represent repeating numbers. Here's an example to why it does equal one.

0.999(repeating) can be more accurately defined by using infinite sums...namely,
" 9 * [(the summation from k=1 to infinity) (1/10)^x] " would describe 0.99999(repeating).
i.e. 9 * (0.1 + 0.01 + 0.001 + etc...) describes 0.9999(repeating).

Now, from our knowledge of geometric series, we can calculate what this infinite geometric sum equals (or converges to) by using...

" Sum= a1 / (1-r) " where a1 = (1/10) and r = (1/10) for this example.

thus Sum= (1/10) / [1 - (1/10)] ==> (1/10) / (9/10) ==> (1/9)

Now, if we multiply this (1/9) with the constant 9 in front of the summation, we get 1.

so, 0.999(repeating) = "9 * [(the summation from k=1 to infinity) (1/10)^x]" = 1.

There appears to be a contradiction whereas in actuality there is not
0.999(repeating) = 1

 

[Skyclad1uhm1]

Originally posted by: silverpig

Originally posted by: 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.

OMG you are NOT adding or subtracting infinity. You are adding and subtracting an infinite number of decimals, but you are doing nothing with infinity itself.

That was my point basically, that's the mistake people are making: thinking 10*0.999... != 9+0.999...

 

[RossGr]

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

this algebraic manipulation, is not a proof, it is a demonstration. Just because this trick does not constitute a formal proof does not mean the end result is not correct. It is, .999.... =1

A mathmatician formally proves a statement like that by demonstrating that there is no real number between .999... and 1. This is easy to see. Suppose we were to add an arbitrally small number to .999.... say .1^N where you are free to pick N as large as you like as long it is a idenifiable number, if we do the addition we will get 1 + .000...99999 here the ellipsis (three decimals) represents a string of N-3 zeros. this is the path to show that, since there is no number between 1 and .999.... the difference is zero therefore they are equal.

I have also developed this proof starting from the Cantor set, it popped out as a side effect as I developed the binary number system. To restate what others have said, this is not an approximation or round off error this is a mathematical idendity, it is as true as 1+1 =2.

I do not buy the philosophical BS. Luvly, your are simply blowing hot air. A Philosopher can talk all they want about their concept of unity it simply does not apply, here we are talking about the real number system, if you want the correct result, you MUST talk to a mathematician, after all, who do you suppose built the real number system?

it is also true that

1= - e^(i pi)

It simply is not a fact that there is only (-e^(i pi)) ways of writing the concept of unity.

 

[MadRat]

...0001000100010001000100010001.0...

 

[josphII]

Originally posted by: BruceLee
Yes sorry for the notation fault. However, my point still remains: no one knows the exact value of either pi or e, yet with these 2 values in an expression an integer is output. That alone has always bugged me about calculus and is my main reason for not buying into the .999 repeating = 1. I understand that all integers have an infinite amount of zeros behind the decimal point, but the value is in no way approximated.

we dont know the exact value for e or pi becasue we cant, they are not exact numbers! there are an infinite number of decimal places with the numbers e and pi, and we use taylor series to find the values of these decimal places

e^x = SUM x^k/k! , for k= 0 to inf

pi is a little more difficult to calculate, it involves using the taylor series for arctan.

 

[ed21x]

if 1= .99999999....

then 1+1= .9999999... + .99999999.. = 1.99999999999...998 <2

1+1+1 = 2.999999..997 even more less than 3

and 1+1+1+1= .. differentiation increases...

so clearly, .999999 !=1

 

[Haircut]

Originally posted by: ed21x
if 1= .99999999....

then 1+1= .9999999... + .99999999.. = 1.99999999999...998 <2

1+1+1 = 2.999999..997 even more less than 3

and 1+1+1+1= .. differentiation increases...

so clearly, .999999 !=1

1.9999....998 cannot exist. It is impossible to have an infitinte string that has a last value.

 

[silverpig]

Originally posted by: josphII

Originally posted by: BruceLee
Yes sorry for the notation fault. However, my point still remains: no one knows the exact value of either pi or e, yet with these 2 values in an expression an integer is output. That alone has always bugged me about calculus and is my main reason for not buying into the .999 repeating = 1. I understand that all integers have an infinite amount of zeros behind the decimal point, but the value is in no way approximated.

we dont know the exact value for e or pi becasue we cant, they are not exact numbers! there are an infinite number of decimal places with the numbers e and pi, and we use taylor series to find the values of these decimal places

e^x = SUM x^k/k! , for k= 0 to inf

pi is a little more difficult to calculate, it involves using the taylor series for arctan.

e and pi are exact numbers. They are singular points on the number line. It's not like they're little fuzzy areas or anything.

Just because we don't have an exact decimal representation of them doesn't mean they're not points.

 

[spyordie007]

Originally posted by: spidey07
bleep,

you can believe what you want.

But there really isn't any debate to this.

They are the same number. Just as 1 + 1 = 2. (or 1.999...)

What Spidey said, I cant believe that so many people voted No, it just proves the lack of real technical or mathematical skills of ATOTers. Maybe if I rephrase it (again) it will help some of you understand...

.9999999........ is infinitely close to 1, it is not an approximation at all it is infinitely close. For you to be able to say that it is = ~1 would assume that it terminates eventually and if it terminates eventually it is ?? 1 however if it does not terminate ever .99999?c.. = 1

There?fs nothing to debate here for scientists, mathematicians and engineers alike, .999?c.. is equal to 1

-Spy

 

[Kyteland]

Wow, I had no idea that this would spark such a huge flame, I mean debate.

Just as an update the three mathematicians here at work (two with their phd) managed to convince all of the computer programmer types that 0.999... was indeed the same as 1. Whether you want to believe it or not, we converted all of the skeptics here. My mission is complete.

 

[Viper GTS]

Originally posted by: spyordie007

Originally posted by: spidey07
bleep,

you can believe what you want.

But there really isn't any debate to this.

They are the same number. Just as 1 + 1 = 2. (or 1.999...)

What Spidey said, I cant believe that so many people voted No, it just proves the lack of real technical or mathematical skills of ATOTers. Maybe if I rephrase it (again) it will help some of you understand...

.9999999........ is infinitely close to 1, it is not an approximation at all it is infinitely close. For you to be able to say that it is = ~1 would assume that it terminates eventually and if it terminates eventually it is ?? 1 however if it does not terminate ever .99999?c.. = 1

There?fs nothing to debate here for scientists, mathematicians and engineers alike, .999?c.. is equal to 1

-Spy

He speaks the truth. Many of you seem to have trouble wrapping your minds around the concept of a repeating, non-ending number like 0.999...

I too had trouble with the concept, & argued for quite some time with my dad when he presented the idea to me.

I believe I was 10 at the time.

Viper GTS

 

[josphII]

Originally posted by: silverpig

Originally posted by: josphII

Originally posted by: BruceLee
Yes sorry for the notation fault. However, my point still remains: no one knows the exact value of either pi or e, yet with these 2 values in an expression an integer is output. That alone has always bugged me about calculus and is my main reason for not buying into the .999 repeating = 1. I understand that all integers have an infinite amount of zeros behind the decimal point, but the value is in no way approximated.

we dont know the exact value for e or pi becasue we cant, they are not exact numbers! there are an infinite number of decimal places with the numbers e and pi, and we use taylor series to find the values of these decimal places

e^x = SUM x^k/k! , for k= 0 to inf

pi is a little more difficult to calculate, it involves using the taylor series for arctan.

e and pi are exact numbers. They are singular points on the number line. It's not like they're little fuzzy areas or anything.

Just because we don't have an exact decimal representation of them doesn't mean they're not points.

yeah your right, i meant to say they are exact, but irrational. the point is we know exactly what they are. what i mean by that is if we need to calculate pi to the nth decimal place we can do so w/ 100% certainty.

 

[bleeb]

SO HAS THE ISSUE BEEN RESOLVED WITHOUT A DOUBT??? =)