1 = .999?

[ZetZet]

Explain why? If you let that recurring decimal be expressed as n and then have it so that you don't have the decimal places anymore (*9) then you divide both sides by 9 you get a fraction of 9/9.

you just get n again, which you know is 0.9 recurring. It doesn't make any sense.

[LeapFrogMasterRace]

One can never equal 0.9... no matter how many nines. It will always fall short by an incredibly small amount.

 

That's a false. .9999999999 isn't and will never be 1

Not according to every mathematician, it is  just the concept that is hard to believe.   

[TheRandomness]

you just get n again, which you know is n recurring. It doesn't make any sense.

It makes sense logically, but not mathematically. It seems like it should work like that, but it doesn't. Because math.

@TheRandomness

How about.... we let this be because it's a bit too confusing.

[Elehat]

Proof that 1=0.9999 recurring:

Let x=1, let y=0.999...=lim(n->inf) 1-1/10^n. Let 1-y=z. z=lim(n->inf)1/10^n. If z=/0, then 1/z is a finite number, so there exists some N such that N>1/z. Pick an N=10^k for some k.  Then z>10^-k. but lim(n->inf)1/10^n<1/10^k, as 1/10^n is a monotonic decreasing sequence. This is a contradiction.    

[ThinkWithPortals]

How about.... we let this be because it's a bit too confusing.

You would be right in any other case. That method is perfectly mathematically sound, but not in this case. It's weird.

[ThinkWithPortals]

Proof that 1=0.9999 recurring:

Let x=1, let y=0.999...=lim(n->inf) 1-1/10^n. Let 1-y=z. z=lim(n->inf)1/10^n. If z=/0, then 1/z is a finite number, so there exists some N such that N>1/z. Pick an N=10^k for some k.  Then z>10^-k. but lim(n->inf)1/10^n<1/10^k, as 1/10^n is a monotonic decreasing sequence. This is a contradiction.    

You lost me at (n->inf)

[Elehat]

limit as n tends to infinity:

edit: Consider  the sequence 0.9, 0.99, 0.999, 0.9999 etc. This sequence is bounded and monotonically increasing, so it has a limit. 

[babadoctor]

You lost me at (n->inf)

what is ->

[ZetZet]

You would be right in any other case. That method is perfectly mathematically sound, but not in this case. It's weird.

 

How about.... we let this be because it's a bit too confusing.

that is because 0.999 can't be written as a fraction, you can't divide it like you can with 1/3

 

0.3 recurring

 

n = 0.3333

10n = 3.3333

9n = 3

n = 3/9 or 1/3

with this you get n=1/3 so it's sound, with your last idea you just get n itself again, which you know was 0.9999, you know it can't be 1.

 

what is ->

limit it's getting really close to that point but never actually equals that point. 

 

Calculus man, it's fucked.

[babadoctor]

You lost me at (n->inf)

it is not a mathmatical symbol

[ThinkWithPortals]

what is ->

Frankly my dear, I don't give a damn.

[Speedstack79]

0.9 recurring can be written as a fraction.

 

n = 0.9999

10n = 9.9999

9n = 9

 

9/9 (= 1 )

n = .5

10n = 9.5

 

9n = 9

n = 1

 

1 = .5 ???

 

whyh is it wrong

[ZetZet]

n = .5

10n = 9.5

 

9n = 9

n = 1

 

1 = .5 ???

 

whyh is it wrong

same problem as with 0.99999999, you can't write 0.55555555555 as a fraction.

[Elehat]

what is ->

tends to. i can't use normal notation; lim(n->inf) 1/10^n is a representation for the limit  of 1/10^n as n tends to infinity. As noted, this is well defined. 

[Speedstack79]

same problem as with 0.99999999, you can't write 0.55555555555 as a fraction.

5/9 = .555555.... ??

[Nineshadow]

same problem as with 0.99999999, you can't write 0.55555555555 as a fraction.

You can?

5/9

 

Another example :

1.0245245245... = (10245 - 10)/9990 = 10235/9990

 

0.9999... can be written as a fraction.

It's 9/9 = 1.

[Emanuel_M]

Does 1 = .999?

 

 

Also if its called 0.999(repeating) then why isnt 1 called 0.999?

so because B is after A A=B?

[ZetZet]

5/9 = .555555.... ??

nvm, it actually works with 0.5555, you just did it wrong. 

 

 

 x = 0.55555555... 

10x = 5.555555555.... 

Subtract the first from the second giving, notice that the entire decimal part cancels, so we have 

10x - x = 5.0 

9x = 5 

so, solving for x we find: 

x = 5/9 

so 

0.5 recurring = 5/9

 

 

You can?

5/9

 

Another example :

1.0245245245... = (10245 - 10)/9990 = 10235/9990

 

0.9999... can be written as a fraction.

It's 9/9 = 1.

well no it's not. it's because 0.999 is THE exception. http://mathyquirks.blogspot.lt/2011/08/are-all-repeating-decimals-fractions-or.html

[manikyath]

i'm just gonna add this:

math makes a lot of sense, until the point infinity comes in.

 

and thats where @ZetZet is clearly losing it, because he doesnt realise at the point of infinity math rules change.

we cannot apply regular math rules to problems that include some form of infinity, because the math we're all used to is based on the fact we can press "equals" at the end, which would break the concept of infinity.

 

you can NEVER press equals at the end of a problem with infinity, so we use derivative means of solving the problem, to bring it back to a system where we have a form that allows us to press equal at the end.

 

its like using the "unknown factors" in math like the ever-popular X. you cannot press equals until you have all occurances of X on one side of the equasion.

the -1/12 video i linked earlier links trough to another video with some "knowledge prior to watching", and both of them contain a lot of this kind of problemsolving.

 

should also mentioned that both videos i mentioned are from numberphile, and the private channel of one of the people on numberphile, numberphile being a math oriented channel hosted by vareous math and physics university teachers.

--

if you're interested in this kind of complex math problems, the numberphile guys have a lot of these problems explained in a manner average joe can understand them.

[Speedstack79]

why is this even a thread lol, I thought infinity was not a number, why are we pretending an infinite number of 9's are added to a number (.99999...)

[ZetZet]

i'm just gonna add this:

math makes a lot of sense, until the point infinity comes in.

 

and thats where @ZetZet is clearly losing it, because he doesnt realise at the point of infinity math rules change.

we cannot apply regular math rules to problems that include some form of infinity, because the math we're all used to is based on the fact we can press "equals" at the end, which would break the concept of infinity.

exactly, in math 0.9 recurring is never 1. Only using calculus we can imply that it is in some cases. And we can only do that because we understand that 0.9 recurring is incredibly close to 1.

[ALwin]

n = .5

10n = 9.5

 

9n = 9

n = 1

 

1 = .5 ???

 

whyh is it wrong

 

There is a flaw in your math.

 

9n = 10n - 1

9.5 - 1 = 8.5

[Nineshadow]

exactly, in math 0.9 recurring is never 1. Only using calculus we can imply that it is in some cases. And we can only do that because we understand that 0.9 recurring is incredibly close to 1.

But 0.9 recurring is really just another way to write 1.

[Speedstack79]

There is a flaw in your math.

 

9n = 10n - 1

9.5 - 1 = 8.5

my flaw was that I set n = .5 in the first line and n = .95 on the second line, notice how 10 x n (.5) does not equal anywhere close to 9.5

[ZetZet]

But 0.9 recurring is really just another way to write 1.

it's not? we can only imply that it's infinitely close to 1. it's not actually equal to 1.