Poll: 0.999... = 1

[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.

 

[Piflik]

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.

I did in the first post you quoted, but here it is again:

1-1 = 0
(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1) = 0
(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+..... = 0
1+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+..... = 0
1 = 0

 

[robotam]

The shirt divides both sides by (a - b). Because a = b, (a - b) = 0. Thus, the proof is invalid.

I thought that was the point of the shirt. If you are drunk while doing your algebra you will doing something strange and end up concluding that 2 is equal to 1.

 

[Syntax Error]

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...

I have my doubts on the bolded part. Shouldn't you have subtracted x on the right side as well?

Also, 11 PAGES! I expect to be ninja'ed.

 

[grammarye]

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 ^^

You would appear to be missing that the statement is not 0.999 = 1, but 0.999... = 1; the latter being a mathematical symbol for recurring.

They are equal, because we have the concept in mathematics of the same real number having multiple equivalent representations.

Shut up.

Oh dear. Please go and read the rules.

Insults are not welcome.

Please quote his post properly, instead of pulling two words out of context...

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.

Actually it is you who has failed to parse the concept of infinity correctly. In your example, multiplication by ten is more accurately represented as:

0.99000000000000... * 10 = 9.90000000000000...

Just as

0.99999999999999... * 10 = 9.99999999999999...

Multiplication by 10 in a base-10 numerical system is a decimal place shift. Nothing more, nothing less. There is no insertion of zero happening.

 

[Redingold]

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.

I did in the first post you quoted, but here it is again:

1-1 = 0
(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1) = 0
(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+..... = 0
1+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+..... = 0
1 = 0

Mathematics fail.

Since each bracket is equal to zero (-1 + 1 = 0), you've just written 1+0+0+0+0+0+0+... = 0

Which is wrong.

 

[robotam]

I did in the first post you quoted, but here it is again:

1-1 = 0
(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1) = 0
(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+..... = 0
1+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+..... = 0
1 = 0

Did you just add 1 to the left side of your equation but not to your right?

 

[Soraryuu]

I stand by the opinion that there's no such thing as infinity. Maybe for time, maybe for the multiverse, but not for matter/energy. Therefore, any mathematic equation that uses infinity is not valid in my eyes.

 

[Coldie]

I did in the first post you quoted, but here it is again:

1-1 = 0
(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1) = 0
(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+..... = 0
1+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+..... = 0
1 = 0

So it's either an infinite sum of [1 - 1], which equals 0; or 1 + inf.sum[-1 + 1] = 1. If you actually have a infinite sum of (alternating) [1] + [-1], then the result is undetermined.

Your proof is invalid.

 

[Piflik]

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.

Actually it is you who has failed to parse the concept of infinity correctly. In your example, multiplication by ten is more accurately represented as:

0.99000000000000... * 10 = 9.90000000000000...

Just as

0.99999999999999... * 10 = 9.99999999999999...

Yes: 0.99000000000... * 10 = 9.9000000000...
No: 0.999999999999--- *10 = 9.99999999999...90

You add a 0 as last decimal in both examples.

 

[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.

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.

There is an n + 1th decimal, because there are an infinite number of decimal places, all of which are 9. Take a number n. The number n + 1 exists, because there is an infinite amount of numbers. Consider then nth decimal place. The n + 1th decimal place exists, because there are an infinite number of decimal places.

http://en.wikipedia.org/wiki/Hilbert's_hotel

Read that and come back when you understand.

Shut up.

Oh dear. Please go and read the rules.

Insults are not welcome.

You're not going to actually provide a reasonable argument, are you? Say, one that involves disproving my completely correct mathematical proof that 0.999... = 1.

Could you try?

 

[C95J]

ummm... no?

yeah, that's the best I can do.

 

[grammarye]

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.

Actually it is you who has failed to parse the concept of infinity correctly. In your example, multiplication by ten is more accurately represented as:

0.99000000000000... * 10 = 9.90000000000000...

Just as

0.99999999999999... * 10 = 9.99999999999999...

Yes: 0.99000000000... * 10 = 9.9000000000...
No: 0.999999999999--- *10 = 9.99999999999...90

You add a 0 as last decimal in both examples.

You're still missing it. There is no insertion of zero. There is no need to insert a zero. There is no point in that numerical sequence at which you could insert a zero. It's infinite. In the first example, you already have all the zeros you could ever possibly need, and inserting one would break the universe, because it is an infinite sequence of zeros.

Edit: Multiplication, as stated in my previous edit (clearly this thread is moving way too fast for that), is not 'shift by one and add a zero on the end'. That for a lot of cases that happens to be the outcome does not make it so for all cases. There is a distinction between happy coincidence and correct application of theory.

 

[Piflik]

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.

I did in the first post you quoted, but here it is again:

1-1 = 0
(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1) = 0
(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+..... = 0
1+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+..... = 0
1 = 0

Mathematics fail.

Since each bracket is equal to zero (-1 + 1 = 0), you've just written 1+0+0+0+0+0+0+... = 0

Which is wrong.

That is the whole point of the proof...it is wrong, but if you allow for infinity, it is also true

Did you just add 1 to the left side of your equation but not to your right?

No such thin...I merely shifted the brackets to the right, which is perfectly fine and doesn't change anything.

 

[Redingold]

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.

I did in the first post you quoted, but here it is again:

1-1 = 0
(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1) = 0
(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+..... = 0
1+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1)+..... = 0
1 = 0

Mathematics fail.

Since each bracket is equal to zero (-1 + 1 = 0), you've just written 1+0+0+0+0+0+0+... = 0

Which is wrong.

That is the whole point of the proof...it is wrong, but if you allow for infinity, it is also true

Did you just add 1 to the left side of your equation but not to your right?

No such thin...I merely shifted the brackets to the right, which is perfectly fine and doesn't change anything.

Two things. How, if you "allow for infinity" is this so-called proof correct?

Secondly, shifting the brackets makes an enormous difference.

(3+5)*7 = 56

3+(5*7) = 38

You see?

SNIP

I'd have been happy to discuss things with you.

However, I shan't now. As I prefer to not talk to those who tell me to shut up.

Enjoy your day/evening.

Your refusal to debate indicates you are uncertain as to the validity of your own position.

Do you actually have a good argument? Perhaps you could explain to someone else what is, if you won't talk to me (which is very rude, by the way).