0.99 (Repeating) = 1?

[Lukeje]

Well I have a question on this topic as well, is .999 (repeating) a rational number? There was a back and forth about that on the original .999=1 topic.

As far as I can tell, no, which is why I posted above. I cannot think of any ratio whose decimal approximation is .999repeating.

3*(1/3) = 0.(9) == 1.

 

[Novajam]

Yup. 0.9 recurring does equal to one.

The only problem is that this trick has no practical application, outside of impressing friends and starting debates.

 

[WeedWorm]

This is bullshit topic, math is boring and no, they were not very intelligent.

Youre boring

 

[Halceon]

The difference is 0.1^inf.

Also 0.(9) seems to equal 1 only in decimal numbering.

That and

god no not this thread again!!!!!!

 

[Maze1125]

The difference is 0.1^inf.

Yep, and 0.1^infinity = 0.
Hence 0.999... = 1.

Also 0.(9) seems to equal 1 only in decimal numbering.

0.999... only exists because of decimal numbering.

 

[Arawak]

It is easy to figure out with using 3/3 = 1 , so that 1/3 = .9999 and if you and (1/3)+(1/3)+(1/3)= (3/3) = also equals .9999 and 1 so thats it its called convergence

no 1/3 is .33333333

also the .99999999 = 1 is a form of simplification and since math is always constant it would never be equal to 1 unless you were a lazy prick or never understood simplification in the first place.

edit; and on the teacher with -2 +3 = -5 probably read it wrong or was tired and wanted to go home after grading a ton of other papers.

The difference is 0.1^inf.

Yep, and 0.1^infinity = 0.
Hence 0.999... = 1.

Also 0.(9) seems to equal 1 only in decimal numbering.

0.999... only exists because of decimal numbering.

flatout wrong, infinity is infinity and even if you take out the infinite part and put it into say the 3rd place as a set maximum, you get .111 which is if you simplify it, 1/10 of a whole number which is a large difference.

the logic is wrong, and unless there is an outside force acting upon this it will still be wrong, as its simply a form of perverting numbers for a stupid cause.

edit;again, lotta edits. you put 0.1^inf and all that would do is move the decimal one point to the left which would multiply it by ten, so 0.1 becomes .01 to infinity, but if this happens as a whole equation .9999 loses some value thus becoming .81 .729 to infinity so if by the logic presented in this thread the value of the number would become null.

hence .99 repeating = 0.

 

[Halceon]

The difference is 0.1^inf.

Yep, and 0.1^infinity = 0.
Hence 0.999... = 1.

Also 0.(9) seems to equal 1 only in decimal numbering.

0.999... only exists because of decimal numbering.

And 0 =/= 0.

 

[Maze1125]

edit;again, lotta edits. you put 0.1^inf and all that would do is move the decimal one point to the left which would multiply it by ten, so 0.1 becomes .01 to infinity, but if this happens as a whole equation .9999 loses some value thus becoming .81 .729 to infinity so if by the logic presented in this thread the value of the number would become null.

hence .99 repeating = 0.

Dude, you really need to check up on your decimal multiplication.
And your general comprehension skills...

 

[magic8BALL]

The idea is this.

x = 0.99 Therefore...

10x = 9.99 Which means!

10x - x = 9.99 - x Which means!

9x = 9 Which means!

x = 1.

This is a valid mathematical proof. Anyone who disagrees with this is incorrect. I'm not even sure why there is a debate over ths.

 

[AdamAK]

Another way of putting it is expressing 0.9999... as a geometric series. In this case it would be : 0.5 + 0.5^2 + 0.5^3 + ... repeated til infinity.
In the end the sum of this geometric series would be a/1-k = 0.5/0.5 = 1

 

[dirk45]

Let's try it a bit different (and mathematically correct):
1.111111.. =
(1/10)^0+(1/10)^1+(1/10)^2+(1/10)^3+...
A mathematical sentence says:
SUM(i=0..INF;x^i) = 1 / (1 - x) if 0 < x < 1.
So 1.1111 = SUM(i=0..INF;(1/10)^i) = 1 / ( 1-(1/10)) = 10/9.
So .111111.. = 1.11111.. - 1 = 10/9 - 1 = 1/9.
Now .99999.. = .11111.. * 9 = 1/9 * 9 = 1.
q.e.d.

 

[USSR]

Just a quick question: does .999repeating actually exist? Nothing actually equals it. You can't divide any integer by any other integer and get it (unlike other repeating decimals, like .111repeating, which is 1/9) Please, mathy people only for answering. I want a proof (or disproof), not just logic.

You would first have to "prove" that repeating numbers exist. If you assert that they do (and you do, with your claim of 1/9 = 0.111r), then you're saying that operations can be performed on them. 9 * 0.111r = 0.999r using conventional maths, although we've already defined that 0.111r = 1/9. 9 * 1/9 = 9/9 = 1. I would consider that a fairly trivial proof of 0.999r = 1, unless you claim that multiplication doesn't work like that on recurring numbers. I can assure you it does, and that no matter how infinitesimally small you go down 0.111r, the end result will always be 0.999r.

I don't know if you were trying to prove the theory wrong with your little 0.1 repeating problem, but you didn't really disprove anything. Yes 9 * 0.1 repeating = 0.9 repeating, so I don't know what you are trying to say. Are you trying to state that you couldn't normally take a whole number and times it by a decimal?

I'm not downing your idea..I just completely misunderstood what you tried to disprove =/

 

[Maze1125]

Another way of putting it is expressing 0.9999... as a geometric series. In this case it would be : 0.5 + 0.5^2 + 0.5^3 + ... repeated til infinity.
In the end the sum of this geometric series would be a/1-k = 0.5/0.5 = 1

That's completely the wrong series to use.
The correct one is 9x0.1 + 9x(0.1^2) + 9x(0.1^3)... so a = 0.9 and r = 0.1, so a/1-r = 0.9/0.9 = 1

 

[Maze1125]

Just a quick question: does .999repeating actually exist? Nothing actually equals it. You can't divide any integer by any other integer and get it (unlike other repeating decimals, like .111repeating, which is 1/9) Please, mathy people only for answering. I want a proof (or disproof), not just logic.

You would first have to "prove" that repeating numbers exist. If you assert that they do (and you do, with your claim of 1/9 = 0.111r), then you're saying that operations can be performed on them. 9 * 0.111r = 0.999r using conventional maths, although we've already defined that 0.111r = 1/9. 9 * 1/9 = 9/9 = 1. I would consider that a fairly trivial proof of 0.999r = 1, unless you claim that multiplication doesn't work like that on recurring numbers. I can assure you it does, and that no matter how infinitesimally small you go down 0.111r, the end result will always be 0.999r.

I don't know if you were trying to prove the theory wrong with your little 0.1 repeating problem, but you didn't really disprove anything. Yes 9 * 0.1 repeating = 0.9 repeating, so I don't know what you are trying to say. Are you trying to state that you couldn't normally take a whole number and times it by a decimal?

I'm not downing your idea..I just completely misunderstood what you tried to disprove =/

The basic concept was that 0.111... = 1/9
So 0.111... x 9 = 1/9 x 9
So 0.999... = 9/9 = 1

And yes, he was arguing for the equality, not against it.

 

[magic8BALL]

RTP 0.9999(recurring)= 1
Let x = 0.9999
It the follows that:
10x = 10 * 0.9999
10x = 9.999
subtract x from each side:
10x - x = 9.999 - 0.9999
Simplify:
9x = 9
solve:
x = 9/9
x = 1
but x = 0.9999
therefore
0.9999(recurring)= 1
QED

Questions from those without a Maths degree? Can we move on with our lives?

 

[3rd rung]

It is easy to figure out with using 3/3 = 1 , so that 1/3 = .9999 and if you and (1/3)+(1/3)+(1/3)= (3/3) = also equals .9999 and 1 so thats it its called convergence

no 1/3 is .33333333

also the .99999999 = 1 is a form of simplification and since math is always constant it would never be equal to 1 unless you were a lazy prick or never understood simplification in the first place.

edit; and on the teacher with -2 +3 = -5 probably read it wrong or was tired and wanted to go home after grading a ton of other papers.

The difference is 0.1^inf.

Yep, and 0.1^infinity = 0.
Hence 0.999... = 1.

Also 0.(9) seems to equal 1 only in decimal numbering.

0.999... only exists because of decimal numbering.

flatout wrong, infinity is infinity and even if you take out the infinite part and put it into say the 3rd place as a set maximum, you get .111 which is if you simplify it, 1/10 of a whole number which is a large difference.

the logic is wrong, and unless there is an outside force acting upon this it will still be wrong, as its simply a form of perverting numbers for a stupid cause.

edit;again, lotta edits. you put 0.1^inf and all that would do is move the decimal one point to the left which would multiply it by ten, so 0.1 becomes .01 to infinity, but if this happens as a whole equation .9999 loses some value thus becoming .81 .729 to infinity so if by the logic presented in this thread the value of the number would become null.

hence .99 repeating = 0.

So for my mistype on (1/3)=.3333 I have changed that and flushed out my argue ment a little better but the overal math does hold true it is not just simplification when you look at it it is true being that (1/3)+(1/3)+(1/3) will give you an answer of .999999 and (3/3). Since (3/3) is also equal 1 the arguement holds true

and as I said before I have just writen .3333 and .999 to several decimal places for convenance but I do mean them to be reapeating to infintiy as they actually would

 

[Littlee300]

.999(bar notation) does equal to 1. Your teacher is ignorant.

What happens to the .111111111111....?

 

[thylasos]

According to logs, yeah. Apparently.

 

[Corkydog]

.999(bar notation) does equal to 1. Your teacher is ignorant.

What happens to the .111111111111....?

It's not .111111....

It's .0000000000....with a hypothetical 1 at the end. Of infinity. Which is why .9 repeating does in fact equal one.

http://en.wikipedia.org/wiki/0.999...