Poll: 0.999... = 1

[Lma0nade]

If x = 0.999999...
Then 10x = 9.9999...
Therefore, 10x - x = 9
Which implies 9x = 9
Thus, x = 1
x also = 0.99999...

In conclusion, I have just proven 1 = 0.9999...

Uh, wait, the second part will have one less 9 after the decimal point than the first part.

x = 0.9999999999
10x = 9.999999999

So the 10x - x would actually equal 8.9999999991, not 9.

Back on topic, .999999999999999999999999999999999999999999999999999999999999999999999999999(and so on) does not equal 1 because 1 is 1.0000000000000000 not (above .999...). They are two different numbers which can be rounded to each other, but are not equal to each other.

 

[Redingold]

No.

0.999 is 0.999.

1 is 1.

Yes, very good, but can you demonstrate why 0.999... is not equal to 1? Here, I'll prove that it is, and you can try and find a problem with my proof.

M'kay. The number 0.999... is equal to an infinite series 0.9 + 0.09 + 0.009 + 0.0009 and so on. If you know anything about slightly advanced maths, you'll know that the sum of an infinite geometric series is equal to a/(1-r) when |r| < 1 (explained below for those who aren't so good at maths)

In our example here, a, the first term, is 0.9, and r, the common ratio, is 0.1 (because each term is the previous term multiplied by 0.1).

So we have 0.9/(1-0.1) which equals 0.9/0.9 which equals 1.

Explanation of maths involved:

A geometric sequence is one where each term is the previous term multiplied by some number r. The first term is a, the second term is ar, the third term is ar[sup]2[/sup] and so on. The nth term is ar[sup]n-1[/sup].

The sum of a geometric series to n terms, which we shall call S[sub]n[/sub], is therefore equal to a + ar + ar[sup]2[/sup]...+ ar[sup]n-2[/sup] + ar[sup]n-1[/sup]

Multiplying by r, we get rS[sub]n[/sub] = ar + ar[sup]2[/sup] + ar[sup]3[/sup]...+ ar[sup]n-1[/sup] + ar[sup]n[/sup]

Subracting rS[sub]n[/sub] from S[sub]n[/sub] leads to S[sub]n[/sub] - rS[sub]n[/sub] = a - ar[sup]n[/sup]

This means S[sub]n[/sub](1-r) = a(1 - r[sup]n[/sup])

And S[sub]n[/sub] = a(1 - r[sup]n[/sup])/(1-r)

Now, to find the sum to infinity, n must be equal to infinity. If |r| > 1, r[sup]infinity[/sup] is infinite. If |r| < 1, r[sup]infinity[/sup] is equal to zero. (If |r| = 1, we end up with 0/0, and I don't wanna go there (it's not 1)).

Thus, S[sub]infinity[/sub] = a(1 - r[sup]infinity[/sup])/(1-r) = a(1-0)/(1-r) = a/(1-r) when |r| < 1

Satisfied now?

yes, 0.99999 is 1, but only in the weak human interpretation of the concept of Mathematics.

Really? I suppose you have some strong, alien mathematics where 0.999... is not equal to 1, then. Could I see it?

 

[Redingold]

No.

0.999 is 0.999.

1 is 1.

Yes, very good, but can you demonstrate why 0.999... is not equal to 1? Simply saying that they are each equal to themselves does not show that they are not equal to each other. Here, I'll prove that 0.999... is equal to 1, and you can try and find a problem with my proof.

M'kay. The number 0.999... is equal to an infinite series 0.9 + 0.09 + 0.009 + 0.0009 and so on. If you know anything about slightly advanced maths, you'll know that the sum of an infinite geometric series is equal to a/(1-r) when |r| < 1 (explained below for those who aren't so good at maths)

In our example here, a, the first term, is 0.9, and r, the common ratio, is 0.1 (because each term is the previous term multiplied by 0.1).

So we have 0.9/(1-0.1) which equals 0.9/0.9 which equals 1.

Explanation of maths involved:

A geometric sequence is one where each term is the previous term multiplied by some number r. The first term is a, the second term is ar, the third term is ar[sup]2[/sup] and so on. The nth term is ar[sup]n-1[/sup].

The sum of a geometric series to n terms, which we shall call S[sub]n[/sub], is therefore equal to a + ar + ar[sup]2[/sup]...+ ar[sup]n-2[/sup] + ar[sup]n-1[/sup]

Multiplying by r, we get rS[sub]n[/sub] = ar + ar[sup]2[/sup] + ar[sup]3[/sup]...+ ar[sup]n-1[/sup] + ar[sup]n[/sup]

Subracting rS[sub]n[/sub] from S[sub]n[/sub] leads to S[sub]n[/sub] - rS[sub]n[/sub] = a - ar[sup]n[/sup]

This means S[sub]n[/sub](1-r) = a(1 - r[sup]n[/sup])

And S[sub]n[/sub] = a(1 - r[sup]n[/sup])/(1-r)

Now, to find the sum to infinity, n must be equal to infinity. If |r| > 1, r[sup]infinity[/sup] is infinite. If |r| < 1, r[sup]infinity[/sup] is equal to zero. (If |r| = 1, we end up with 0/0, and I don't wanna go there (it's not 1)).

Thus, S[sub]infinity[/sub] = a(1 - r[sup]infinity[/sup])/(1-r) = a(1-0)/(1-r) = a/(1-r) when |r| < 1

Satisfied now?

yes, 0.99999 is 1, but only in the weak human interpretation of the concept of Mathematics.

Really? I suppose you have some strong, alien mathematics where 0.999... is not equal to 1, then. Could I see it?

 

[Coldie]

It isn't equivalent to zero. That was my point, infinity is the concept of "as close as you can get without being it"... Kinda. 1x10^(-infinity) would equal 0.(an infinite number of zero's)1.
Basically, the difference between 0.9 recurring and 1 is infinitesimally, immeasurably small. But they are not equal.

Infinity is not "as close as you can get without being it", the term you're thinking of is asymptote - approached infinitely close, but never met. Infinity is anything it damn wants to be, it could approach, it could meet, it could intersect. There's no such thing as a 1 in the end of an infinite string of 0s. 10[sup]-infinity[/sup] equals 0. It is equivalent to zero. It's not asymptotically close to zero (as you propose), it is zero.

I need some popcorn.

 

[DTWolfwood]

but the more u add the more the difference between the 2 becomes apparent no? i mean no 1 will consider 999,999.99 = 1,000,000.00 your always short. but i guess if you never expand .999+ sure its equal to 1. I believe the entire movie of Office Space depends on this lol

 

[Piflik]

I know there is no end to infinity...that's why I called it theoretical 'end'

There's no theoretical end to infinity, either. Infinity is infinite, it has no end. At all. It might have a beginning, but never an end. It might be countable or uncountable, but it never, ever ends. Ever.
There is never a zero, there are only 9s. When you shift the digits left, it's still just 9s going on forever. If there's ever a shortage of digits, you could just shift it left and mine the integer part for a nine, then repeat (forever!).

Math is so simple, yet so easily misunderstood.

No...sorry, but you're wrong. You cannot shift the digits in any number and just call it a day. Infinite or not. If you take 0.9999... and multiply it with 10, there is a 0 at the end, not a 9. Regardless of how many 9s you write in that number. If you want to prove me wrong, you would have to actually write that number with infinite 9s, and even if you can pull that one of, I still assume that there would be a 0 at the end when you multiply it with 10...

Actually, there really is no such thing as infinity. It is a theoretical concept, but it doesn't exist. Every number has an end. Always.

 

[Lyx]

It isn't equivalent to zero. That was my point, infinity is the concept of "as close as you can get without being it"... Kinda. 1x10^(-infinity) would equal 0.(an infinite number of zero's)1.
Basically, the difference between 0.9 recurring and 1 is infinitesimally, immeasurably small. But they are not equal.

Infinity is not "as close as you can get without being it", the term you're thinking of is asymptote - approached infinitely close, but never met. Infinity is anything it damn wants to be, it could approach, it could meet, it could intersect. There's no such thing as a 1 in the end of an infinite string of 0s. 10[sup]-infinity[/sup] equals 0. It is equivalent to zero. It's not asymptotically close to zero (as you propose), it is zero.

I need some popcorn.

That counterargument is so....... i dont even know why i pressed the reply button. The only thing 0.000... has in common with 0.999... is the instruction to repeat digits - except of here: THEY SERVE NO PURPOSE - THE TARGET WAS ALREADY REACHED BEFOREHAND.

 

[brunothepig]

Infinity is not "as close as you can get without being it", the term you're thinking of is asymptote - approached infinitely close, but never met. Infinity is anything it damn wants to be, it could approach, it could meet, it could intersect. There's no such thing as a 1 in the end of an infinite string of 0s. 10[sup]-infinity[/sup] equals 0. It is equivalent to zero. It's not asymptotically close to zero (as you propose), it is zero.

I need some popcorn.

Sorry you're right, that was a clumsy way of saying it. I'm tired...
Anyway, I remember talking about my math teacher with this, and he told me how to debunk it, but I've forgotten... Like so many other things from my Specialist Mathematics class... I'm just gonna have to say, they aren't equal, but I forget how to prove it. I believe it's a proof by contradiction... Google it if you want.
Enjoy your popcorn.

 

[Redingold]

I know there is no end to infinity...that's why I called it theoretical 'end'

There's no theoretical end to infinity, either. Infinity is infinite, it has no end. At all. It might have a beginning, but never an end. It might be countable or uncountable, but it never, ever ends. Ever.
There is never a zero, there are only 9s. When you shift the digits left, it's still just 9s going on forever. If there's ever a shortage of digits, you could just shift it left and mine the integer part for a nine, then repeat (forever!).

Math is so simple, yet so easily misunderstood.

No...sorry, but you're wrong. You cannot shift the digits in any number and just call it a day. Infinite or not. If you take 0.9999... and multiply it with 10, there is a 0 at the end, not a 9. Regardless of how many 9s you write in that number. If you want to prove me wrong, you would have to actually write that number with infinite 9s, and even if you can pull that one of, I still assume that there would be a 0 at the end when you multiply it with 10...

Actually, there really is no such thing as infinity. It is a theoretical concept, but it doesn't exist. Every number has an end. Always.

Wrong. Consider multiplying 0.999... by 10. The second decimal place becomes the first decimal place, the third decimal place becomes the second decimal place and so on. The n+1th decimal place becomes the nth decimal. However, the n+1th decimal place is a 9, because every decimal place is a 9. Thus, when we've multiplied out number by 10, every nth decimal place will be a 9.

Also, pi goes on forever.

 

[Coldie]

Actually, there really is no such thing as infinity. It is a theoretical concept, but it doesn't exist. Every number has an end. Always.

Infinity is very prominent a mathematical concept. It's quite real, I assure you. And, as redundant as it is redundant, infinitely long numbers are, in fact, infinitely long. They never have an end. Here, let me show you, an infinitely long number that has an infinite number of digits after the decimal point and no end ever:

Quite a famous transcendent number, especially prominent in trigonometry and geometry. Also it's infinitely long and has a finite value.

Some infinitely long numbers have a repeating pattern, for instance 1/3 is just infinitely repeating threes. The notation for such numbers is 0.(periodic pattern), so 1/3 = 0.(3) or 1/7 = 0.(142857).

So if you take 0.(9): 1 - 0.(9) = 1 x 10[sup]-infinity[/sup] = 0.
You can find the detailed calculations in the first couple pages of the thread.

Don't try to understand it, just accept it as the universal and absolute truth. Because that's what it is. Elementary Arithmetics.

 

[Redingold]

SNIP

Nope. Still wrong.

One is one.

Example: I am holding one cup. I am holding 0.999 cup.

Which one of these is wrong?

Pretty easy stuff, guys ^^

Please address my point. I have used mathematics, you have repeatedly stated absolutely nothing noteworthy.

Find a flaw in my maths or shut up.

 

[fix-the-spade]

I will post a proof after a few replies, but other people are free to post proofs before that.

We had this thread a while ago, it all hinges on whether or not you consider the numbers absolute.

.999=/=1, if it were 1 it would be represented as 1, representing as .999 implies a difference to 1, but a difference that is beyond your means or comprehension to measure. Being unable to measure a difference/quantity does not mean it does not exist, so it's different.

Also that proof is flawed, 9x would be 8.999 and 9x1 would be 9. 10x would be 9 and 10x1 would be 10, proves they are not equal as the difference grows with multiplication. Were they equal it would not.

 

[Redingold]

I will post a proof after a few replies, but other people are free to post proofs before that.

We had this thread a while ago, it all hinges on whether or not you consider the numbers absolute.

.999=/=1, if it were 1 it would be represented as 1, representing as .999 implies a difference to 1, but a difference that is beyond your means or comprehension to measure. Being unable to measure a difference/quantity does not mean it does not exist, so it's different.

Also that proof is flawed, 9x would be 8.999 and 9x1 would be 9. 10x would be 9 and 10x1 would be 10, proves they are not equal as the difference grows with multiplication. Were they equal it would not.

For the third time:

M'kay. The number 0.999... is equal to an infinite series 0.9 + 0.09 + 0.009 + 0.0009 and so on. If you know anything about slightly advanced maths, you'll know that the sum of an infinite geometric series is equal to a/(1-r) when |r| < 1 (explained below for those who aren't so good at maths)

In our example here, a, the first term, is 0.9, and r, the common ratio, is 0.1 (because each term is the previous term multiplied by 0.1).

So we have 0.9/(1-0.1) which equals 0.9/0.9 which equals 1.

Explanation of maths involved:

A geometric sequence is one where each term is the previous term multiplied by some number r. The first term is a, the second term is ar, the third term is ar[sup]2[/sup] and so on. The nth term is ar[sup]n-1[/sup].

The sum of a geometric series to n terms, which we shall call S[sub]n[/sub], is therefore equal to a + ar + ar[sup]2[/sup]...+ ar[sup]n-2[/sup] + ar[sup]n-1[/sup]

Multiplying by r, we get rS[sub]n[/sub] = ar + ar[sup]2[/sup] + ar[sup]3[/sup]...+ ar[sup]n-1[/sup] + ar[sup]n[/sup]

Subracting rS[sub]n[/sub] from S[sub]n[/sub] leads to S[sub]n[/sub] - rS[sub]n[/sub] = a - ar[sup]n[/sup]

This means S[sub]n[/sub](1-r) = a(1 - r[sup]n[/sup])

And S[sub]n[/sub] = a(1 - r[sup]n[/sup])/(1-r)

Now, to find the sum to infinity, n must be equal to infinity. If |r| > 1, r[sup]infinity[/sup] is infinite. If |r| < 1, r[sup]infinity[/sup] is equal to zero. (If |r| = 1, we end up with 0/0, and I don't wanna go there (it's not 1)).

Thus, S[sub]infinity[/sub] = a(1 - r[sup]infinity[/sup])/(1-r) = a(1-0)/(1-r) = a/(1-r) when |r| < 1

Satisfied now?

There is a mathematical proof that 0.999... = 1. Disprove it or shut up.

 

[Sebenko]

Depends.

What degree of accuracy am I recording to?

If it's an integer, and I'm using the round function, then it's one. If it's not using round, then 0.9 = 0. Yay truncation!

Short? Long?
Single? Double?

 

[Piflik]

I know there is no end to infinity...that's why I called it theoretical 'end'

There's no theoretical end to infinity, either. Infinity is infinite, it has no end. At all. It might have a beginning, but never an end. It might be countable or uncountable, but it never, ever ends. Ever.
There is never a zero, there are only 9s. When you shift the digits left, it's still just 9s going on forever. If there's ever a shortage of digits, you could just shift it left and mine the integer part for a nine, then repeat (forever!).

Math is so simple, yet so easily misunderstood.

No...sorry, but you're wrong. You cannot shift the digits in any number and just call it a day. Infinite or not. If you take 0.9999... and multiply it with 10, there is a 0 at the end, not a 9. Regardless of how many 9s you write in that number. If you want to prove me wrong, you would have to actually write that number with infinite 9s, and even if you can pull that one of, I still assume that there would be a 0 at the end when you multiply it with 10...

Actually, there really is no such thing as infinity. It is a theoretical concept, but it doesn't exist. Every number has an end. Always.

Wrong. Consider multiplying 0.999... by 10. The second decimal place becomes the first decimal place, the third decimal place becomes the second decimal place and so on. The n+1th decimal place becomes the nth decimal. However, the n+1th decimal place is a 9, because every decimal place is a 9. Thus, when we've multiplied out number by 10, every nth decimal place will be a 9.

No...you're understanding infinity incorrectly. There is no such thing. You can write as many 9s as you want (let's call that number n), the n+1th decimal will not be a 9, since there is no n+1th decimal.

Little example: Let n be 2

0.99 * 10 = 9.90 not 9.99

It is the same thing if you assume 'infinite' 9s...although, as I said, there is no such thing. If you want to do maths with infinity, you cannot use traditional maths....they are not meant to be used with that concept. If you do, you can prove errors like 1 = 0 or 0.9999... = 1, since they have no means to correctly symbolize infinity.

 

[Piflik]

Actually, there really is no such thing as infinity. It is a theoretical concept, but it doesn't exist. Every number has an end. Always.

Infinity is very prominent a mathematical concept. It's quite real, I assure you. And, as redundant as it is redundant, infinitely long numbers are, in fact, infinitely long. They never have an end. Here, let me show you, an infinitely long number that has an infinite number of digits after the decimal point and no end ever:

Quite a famous transcendent number, especially prominent in trigonometry and geometry. Also it's infinitely long and has a finite value.

Some infinitely long numbers have a repeating pattern, for instance 1/3 is just infinitely repeating threes. The notation for such numbers is 0.(periodic pattern), so 1/3 = 0.(3) or 1/7 = 0.(142857).

So if you take 0.(9): 1 - 0.(9) = 1 x 10[sup]-infinity[/sup] = 0.
You can find the detailed calculations in the first couple pages of the thread.

Don't try to understand it, just accept it as the universal and absolute truth. Because that's what it is. Elementary Arithmetics.

So you agree with my prof that 0 = 1? Because if you want to do traditional maths (or Elementary Arithmetics as you call it...) with infinity, you would have to...

 

[Coldie]

So you agree with my prof that 0 = 1? Because if you want to do traditional maths with infinity, you would have to...

Infinity is an integral part of the so-called "traditional" math. Infinity and infinite numbers are also very prominent in the Set Theory, look it up.

You seem to have no understanding of how math actually works, so would you kindly post your alleged "proof" of 0 = 1?

I find you lack of math disturbing.