Does 0.999999999........ = 1?

Originally posted by iamatiger
1/2+1/4+1/8..... = a

multiply both sides by 2: 1 + 1/2 + 1/4 + 1/8.... = 2a

substitute in a from the first equation: 1 + a = 2a

take a from both sides: 1 = a

-----------------

6/7 + 6/49 + 6/343.... = b

multiply both sides by 7: 6 + 6/7 + 6/49 + 6/343... = 7b

substitute in the value of b: 6 + b = 7b

take b from both sides: 6 = 6b
...[text shortened]... above:

(x-1) + c = xc

take c from both sides

(x-1) = (x-1)c

divide by (x-1)

1 = c

That acually seems quite solid.
What sets these series apart?
I mean 1/2+1/3+1/4 does not equal 1

Are infinite small numbers always 0?

Originally posted by chasparos
I would use your exact argument to prove 0.99999 is NOT 1 :-)
It all depends on which side of the fence you start.

Go ahead. 🙂

-Ray.

Originally posted by TheMaster37
I can make a sum that comes out on 0, mess around with it so it ends up in 1, then mess it up again to is becomes less then 0, and again to make it more then 1...

You mean the series S = 1 - 1 + 1 - 1 + 1 - 1 + ...
Add parentheses in different ways and obtain either -1, 0, 1.

-Ray.

Originally posted by chasparos
Look at 1/9=0.1111...
I could just as easily say that equality is false since
9/9 = 1 not 0.99...... or is this reversal faulty?

But, it is equal. Just do the long division. You could easily say that the equality is false, but you'd be wrong. 🙂

-Ray.

Originally posted by TheMaster37
0.9... is 1 per definition 🙂

Let me ask you this;

The number 0 is also written as 0.0...
What is the difference between 1 and 0.9...?

Thats the part that gets me...
No matter what base you choose
the same way of writing, that is an infinite stretch of the highest digit allowed after the decimal point describes a series with this characteristic. A very beautiful thing :-)

Can the same type of behavior be seen in other
infinite repetitive decimalseries, like 1/7=0.14285714....?

a=1/7
1000000a-a/2=142857.07142857142857
multiply with 100 and subtract a
100 000 000a-51a=14285707
99 999 949a=14285707
divide with 14285707
7a=1 ... hmm that surprised me a little :-)
thus.. all repetitive decimals can be truncated this way?
Cool :-)

Originally posted by rgoudie
But, it [b]is equal. Just do the long division. You could easily say that the equality is false, but you'd be wrong. 🙂

-Ray.
[/b]

Thats exactly my point in that case..
If you start with the division adding 1:s as you go along or
you could start the multiplication adding 9:s as you go along..
You will be doing that for quite some time to before you get that 1 to carry :-)

depending on what youre trying to prove you'll succeed...

And for the record: I know I am the one trying to prove the negative here... An impossible task :-)

To be sure, I am disputing your assertion that 1/9 = 0.111... is false.
How do you justify this assertion?

-Ray.

Originally posted by rgoudie
You mean the series S = 1 - 1 + 1 - 1 + 1 - 1 + ...
Add parentheses in different ways and obtain either -1, 0, 1.

-Ray.

That one works too, i had this one in mind:

1, -1/2, 1/2, -1/3, 1/3, -1/4, 1/4, ...

You can see the sum is one because every term except the first cancels.

If you now place brackets around every two numbers in the sum and calculate a bit:

1-1/2 = 1/1*2
1/4-1/5 = 1/4*5

You get S = 1/1*2 + 1/2*3 + 1/3*4 + ... = 1

Also 1/3*4 = -2/3 + 3/4. Writing that out for all fractions gives:

1/2 - 1/2 + 2/3 - 2/3 + 3/4 - 3/4... =0 (and all terms in S are positive!)

Changing the order of the sequence can result in showing that S > 1 or S < 0 as well 🙂

Originally posted by chasparos
Thats exactly my point in that case..
If you start with the division adding 1:s as you go along [b]or
you could start the multiplication adding 9:s as you go along..
You will be doing that for quite some time to before you get that 1 to carry :-)

depending on what youre trying to prove you'll succeed...

And for the record: I know I am the one trying to prove the negative here... An impossible task :-)[/b]

There is where it goes wrong. The 1 NEVER carries.

Originally posted by TheMaster37
There is where it goes wrong. The 1 NEVER carries.

Yes. That was my point.

9*0.1111 is 0.999999..

now you can make one of two observations:
a) 1/9 is not 0.11111...
or
b) 1=0.999999...

And my original quesion was why it was obvious
that b not a is true, based on the reasoning in the 9/9=1
proof. This is a statement that is not supported by any logic,
is is just proclaimed obvious. An unsupported premise.

Originally posted by rgoudie
To be sure, I am disputing your assertion that 1/9 = 0.111... is false.
How do you justify this assertion?

-Ray.

well i just turn your proof of 1=0.999.. on its head.

assume:
a) 9/9=1
b) 1/9 * 9 = 9/9

examine 1/9=0.111111....

0.1111...*9 = 0.99999... not equal to 1 -> 1/9 not equal to 0.1111...

So.. to be able to refute this you have to prove 0.9999..=1
Which was the not obvious part. Otherwise its a circular argument, proving a theory based on the assumption that its true. Makes it a "Its so because i said it is" kind of proof.

Originally posted by TheMaster37
0.9... is 1 per definition 🙂

Let me ask you this;

The number 0 is also written as 0.0...
What is the difference between 1 and 0.9...?

I have to adress this too :-)

I must agree that I cant name that quantity.
I cant however tell you in whatway pi differs from 4
or 0.9999... differs from say.. 0.888...
in all these instances the exact answer is undefined.


[Edit] I have accepted the basic statement that 0.9999..=1 I just dont agree with some of the arguments made to prove it.
BTW lets put another truth on its head all proofs here has taken 0.999... which we only can represent imperfectly and rooven it equal to 1. however, can the reverse be proven? Can it be proven that not both 1=0.9999 and 1>0.999... are true? Does such dualities exist? What is 1 - [smallest possible number]?
1-0.9999... is proven here to be equal to 0 what is then the smallest possible number larger than zero?

divide 1 by 9 and your answer will be 0.111... Calculation is a strong proof 🙂 The division process will always go on, and will keep adding ones to the row. Therefore 1/9 = 0.111... (in wich the ... means that there are infinite ones behind the.) and multiplying both sides with 9 calculates the 1 = 0.999... Both are calculations, both with authorised operations 🙂. You said you accepted it, i hope i can show you a reason to believe it as well 😉

To answer your other question; There is no smallest number greater then 0. For every number 0 < E we can come up with, there exists a number B with 0 < B < E, for example B = E/2.

With this you've pointed out a difficulty with Q and R (rationals and reals). You can't point out a smallest number in every set of numbers anymore. But the rewards are great, in Q we can divide, and in R we can do almost everything! The only thing we can't do yet is take roots of every number.

The price to pay for that is to lose the ordening structure (you can compare all reals and say wich one is greater, if they aren't equal). The complex numbers C don't have that, but you can take all the roots you want...

Originally posted by TheMaster37
divide 1 by 9 and your answer will be 0.111... Calculation is a strong proof 🙂 The division process will always go on, and will keep adding ones to the row. Therefore 1/9 = 0.111... (in wich the ... means that there are infinite ones behind the.) and multiplying both sides with 9 calculates the 1 = 0.999... Both are calculations, both with authorised op ...[text shortened]... en't equal). The complex numbers C don't have that, but you can take all the roots you want...

Thank you for taking the time to teach :-)
I will only make one last observation on this matter:
My thoughts on using calculation as definitive proof in this
case can cut both ways.
I dont belive 1/9 is any thing other than 0.111...
I just dont think you can prove it by testing it in calculus.
because for every extra decimal i calculate to prove it I can make a
multiplication to one extra decimal to refute it.
9/9 =1
1/9 =0.1 -> 9*0.1=0.9 (not 1)
1/9 =0.11 -> 9*0.11=0.99 (not 1)
1/9 =0.111 -> 9*0.111=0.999 (not 1)
1/9 =0.1111 -> 9*0.1111=0.9999 (not 1)

Unless you first prove 0.999... is 1 (that has been done here and I'm impressed and intrigued) you cannot prove that 1/9 is 0.111...

You can probably safetly assume the pattern will keep, but does calculating one more decimal acually prove it?

Again thank you all for the lesson. :-)

Originally posted by chasparos
That acually seems quite solid.
What sets these series apart?
I mean 1/2+1/3+1/4 does not equal 1

Are infinite small numbers always 0?

The series 1/2 + 1/4 + 1/8... and 1/7 + 1/49 + 1/343 are examples of series that converge on 1. It is quite easy to prove that they do in the way I gave, but there are other series that converge on 1, e.g
(pi/2) - (pi/2)^3/3! + (pi/2)^5/5! - (pi/2)^7/7!....
You would need another way to prove the convergence of that series. Of course there are other series that converge to other numbers than 1 and there are series that never converge - they keep on going up to infinity 1/2 + 1/3 + 1/4 + 1/5... is one of those.

Sadly (or happily) there is no one way to tell what a series converges to, each one is a different mathematical problem.