Is 1 = 0.9999......

[Beau]

OMG! NO! AIIIIIIIIIIIEIEIEIEIEIEIE!E!EEEE!!!!!

Not this question again!?! :|

 

[StageLeft]

if .99999999999 = 1 then it would be 1 not, .99999999

 

[Kyteland]

Originally posted by: Skoorb
if .99999999999 = 1 then it would be 1 not, .99999999

Yes, but aren't they different representations for the same thing? By your logic i=sqrt(-1) are not the same thing because if i=sqrt(-1) then it would be sqrt(-1) not i.

And I know that this is a silly argument, but it kinda illustrates the point that your perception of them being equal or not before hand influences if you think they are equal when you are asked.

If that made any sense.....

 

[TuxDave]

Originally posted by: Skoorb
if .99999999999 = 1 then it would be 1 not, .99999999

Just like 4/2 can't equal 2 because then it would be written as 2, not 4/2

 

[josphII]

1 does equal 0.999..., it is mathematical FACT

heres the proof:

Proof: 0.9999... = Sum 9/10^n
(n=1 -> Infinity)

= lim sum 9/10^n
(m -> Infinity) (n=1 -> m)

= lim .9(1-10^-(m+1))/(1-1/10)
(m -> Infinity)

= lim .9(1-10^-(m+1))/(9/10)
(m -> Infinity)

= .9/(9/10)

= 1

 

[Mday]

this question has been asked before. check highly techical or something.

the sequence of numbers .9 .99 .999 .99.... etc does converge to 1.

while it's not proper to say that .9999... = 1, it's a matter of rhetoric.

yes, i've taken real analysis.

 

[silverpig]

What some ignorant people may think:

0.999... = the "last" number in 0.9, 0.99, 0.999, 0.9999... etc

WRONG

0.999... is not a progression. You don't add nines, or "take" nines, or approach anything. How can a number approach another number? It doesn't. Sums approach numbers, numbers are just numbers.

0.999... is 1.


For all of you who say it isn't:

1. In order to prove that two numbers a and b are different, you must be able to explicitly define another number, c, that is larger than a and smaller than b. For example:

Let a = 4
Let b = 5
4 < 5 because if we let c = 4.5 we have a < c < b ===> 4 < 4.5 < 5

Therefore 4 < 5.

Now try that with 0.999... and 1.

There is no way to write that difference because it does not exist. But, a common way to write differences is by doing it this way:

0.9r = w + x
1 = y + z

w,x,y, and z must be defined numbers though.

w < a < y
x < b < z
w + x < a + b < y + z

Therefore 0.9r < 1.

All you have to do now, is come up with numbers for w,x,y,z,a, and b.

 

[Dark4ng3l]

This is easey, 9.9999... is not exact(like pi,it's an estimated value) it essentially represents 3/3 witch is 1

 

[Zugzwang152]

Originally posted by: Dark4ng3l
This is easey, 9.9999... is not exact(like pi,it's an estimated value) it essentially represents 3/3 witch is 1

exactly.

 

[bleeb]

Originally posted by: silverpig
What some ignorant people may think:

0.999... = the "last" number in 0.9, 0.99, 0.999, 0.9999... etc

WRONG

0.999... is not a progression. You don't add nines, or "take" nines, or approach anything. How can a number approach another number? It doesn't. Sums approach numbers, numbers are just numbers.

0.999... is 1.


For all of you who say it isn't:

1. In order to prove that two numbers a and b are different, you must be able to explicitly define another number, c, that is larger than a and smaller than b. For example:

Let a = 4
Let b = 5
4 < 5 because if we let c = 4.5 we have a < c < b ===> 4 < 4.5 < 5

Therefore 4 < 5.

Now try that with 0.999... and 1.

There is no way to write that difference because it does not exist. But, a common way to write differences is by doing it this way:

0.9r = w + x
1 = y + z

w,x,y, and z must be defined numbers though.

w < a < y
x < b < z
w + x < a + b < y + z

Therefore 0.9r < 1.

All you have to do now, is come up with numbers for w,x,y,z,a, and b.

i like this answer but it still doesn't prove that 0.9999.... is equal to 1. Because if two numbers are different, then you can find a number that is between them. But with this particular case, the difference will be 1-0.99999.....

 

[deerslayer]

You guys are giving me a headache

 

[Confused]

I bent my wookie




Confused

 

[Haircut]

Originally posted by: Dark4ng3l
This is easey, 9.9999... is not exact(like pi,it's an estimated value) it essentially represents 3/3 witch is 1

So, why exactly is 9.9999... or indeed pi not exact?

 

[bleeb]

forget all this math crap and become a retarded porn star

 

[McPhreak]

Originally posted by: TuxDave
That proof looks good to me. I believe 0.9999... = 1. For those who don't believe still, show me 1-0.999... doesn't equal 0.

And if you say it equals 0.000000.....001... there's an infinite number of zeros, so that '1' never really exists, so technically 1-0.999... = 0

Does that mean 0.99999.....98 = 0.999999999.....99?

 

[Skyclad1uhm1]

Let's be nice annoying as usual, and confuse people

10*(1-0.9)=1
1 decimal, one zero.

100*(1-0.99)=1
2 decimals, 2 zeroes

1000*(1-0.999)=1
3 decimals, 3 zeroes.

1000...*(1-0.999...)=1?
infinite decimals and the same infinite number of zeroes?




(Edit: According to 0.999...=1 that would give 0 of course)

 

[bleeb]

Originally posted by: Skyclad1uhm1
Let's be nice annoying as usual, and confuse people

10*(1-0.9)=1
1 decimal, one zero.

100*(1-0.99)=1
2 decimals, 2 zeroes

1000*(1-0.999)=1
3 decimals, 3 zeroes.

1000...*(1-0.999...)=1?
infinite decimals and the same infinite number of zeroes?




(Edit: According to 0.999...=1 that would give 0 of course)

heheh confusing everyone was my whole point. I bet even the people who thought they knew for sure, have taken a second, more deeper look into the problem.

 

[josphII]

Originally posted by: bleeb

Originally posted by: silverpig
What some ignorant people may think:

0.999... = the "last" number in 0.9, 0.99, 0.999, 0.9999... etc

WRONG

0.999... is not a progression. You don't add nines, or "take" nines, or approach anything. How can a number approach another number? It doesn't. Sums approach numbers, numbers are just numbers.

0.999... is 1.


For all of you who say it isn't:

1. In order to prove that two numbers a and b are different, you must be able to explicitly define another number, c, that is larger than a and smaller than b. For example:

Let a = 4
Let b = 5
4 < 5 because if we let c = 4.5 we have a < c < b ===> 4 < 4.5 < 5

Therefore 4 < 5.

Now try that with 0.999... and 1.

There is no way to write that difference because it does not exist. But, a common way to write differences is by doing it this way:

0.9r = w + x
1 = y + z

w,x,y, and z must be defined numbers though.

w < a < y
x < b < z
w + x < a + b < y + z

Therefore 0.9r < 1.

All you have to do now, is come up with numbers for w,x,y,z,a, and b.

i like this answer but it still doesn't prove that 0.9999.... is equal to 1. Because if two numbers are different, then you can find a number that is between them. But with this particular case, the difference will be 1-0.99999.....

firstly i cant believe there are 42 idiots that think 1 is not equal to 0.9999..... but anyways

the difference between 1 and 0.999.... is 0.0000..... !!

 

[josphII]

Originally posted by: Haircut

Originally posted by: Dark4ng3l
This is easey, 9.9999... is not exact(like pi,it's an estimated value) it essentially represents 3/3 witch is 1

So, why exactly is 9.9999... or indeed pi not exact?

they are not exact because pi and 0.999... have an infinite number of digits

 

[bleeb]

Originally posted by: josphII

Originally posted by: bleeb

Originally posted by: silverpig
What some ignorant people may think:

0.999... = the "last" number in 0.9, 0.99, 0.999, 0.9999... etc

WRONG

0.999... is not a progression. You don't add nines, or "take" nines, or approach anything. How can a number approach another number? It doesn't. Sums approach numbers, numbers are just numbers.

0.999... is 1.


For all of you who say it isn't:

1. In order to prove that two numbers a and b are different, you must be able to explicitly define another number, c, that is larger than a and smaller than b. For example:

Let a = 4
Let b = 5
4 < 5 because if we let c = 4.5 we have a < c < b ===> 4 < 4.5 < 5

Therefore 4 < 5.

Now try that with 0.999... and 1.

There is no way to write that difference because it does not exist. But, a common way to write differences is by doing it this way:

0.9r = w + x
1 = y + z

w,x,y, and z must be defined numbers though.

w < a < y
x < b < z
w + x < a + b < y + z

Therefore 0.9r < 1.

All you have to do now, is come up with numbers for w,x,y,z,a, and b.

i like this answer but it still doesn't prove that 0.9999.... is equal to 1. Because if two numbers are different, then you can find a number that is between them. But with this particular case, the difference will be 1-0.99999.....

firstly i cant believe there are 42 idiots that think 1 is not equal to 0.9999..... but anyways

the difference between 1 and 0.999.... is 0.0000..... !!

PROVE IT =)

 

[aux]

Someone may have posted this but I'm too tired to read the whole thread.
Here it goes: 0.9999... = 9*0.1111... = 9 * (10^(-1) + 10^(-2) + 10^(-3) + ....)
now, 10^(-1), 10^(-2), 10^(-3), ... is a geometric progression that sums up to (10^(-1) )/ ( 1- 10^(-1)), which is exactly 1/9
therefore 0.9999... = 9* (10^(-1) + 10^(-2) + 10^(-3) + ....) = 9 * 1/9 = 1

 

[911paramedic]

This is nuts.

I see how you could do the 10x-x=10 (x being 0.99999) so reduced it is x=1, but that just sucks my butt.

Try to prove it without manipulating the number at all. Does 1=0.99999... NO! Does it even look the same, no? Is it the same? Close, but no cigar.

 

[Czar]

Originally posted by: josphII

Originally posted by: bleeb

Originally posted by: silverpig
What some ignorant people may think:

0.999... = the "last" number in 0.9, 0.99, 0.999, 0.9999... etc

WRONG

0.999... is not a progression. You don't add nines, or "take" nines, or approach anything. How can a number approach another number? It doesn't. Sums approach numbers, numbers are just numbers.

0.999... is 1.


For all of you who say it isn't:

1. In order to prove that two numbers a and b are different, you must be able to explicitly define another number, c, that is larger than a and smaller than b. For example:

Let a = 4
Let b = 5
4 < 5 because if we let c = 4.5 we have a < c < b ===> 4 < 4.5 < 5

Therefore 4 < 5.

Now try that with 0.999... and 1.

There is no way to write that difference because it does not exist. But, a common way to write differences is by doing it this way:

0.9r = w + x
1 = y + z

w,x,y, and z must be defined numbers though.

w < a < y
x < b < z
w + x < a + b < y + z

Therefore 0.9r < 1.

All you have to do now, is come up with numbers for w,x,y,z,a, and b.

i like this answer but it still doesn't prove that 0.9999.... is equal to 1. Because if two numbers are different, then you can find a number that is between them. But with this particular case, the difference will be 1-0.99999.....

firstly i cant believe there are 42 idiots that think 1 is not equal to 0.9999..... but anyways

the difference between 1 and 0.999.... is 0.0000..... !!

its actually 0.0000...1

 

[bleeb]

Here is a prove that is somewhat feasible:

Given:

Two numbers are not equal if the difference doesn't equal zero...

1 - 0.9 = 0.1

1 - 0.99 = 0.01

1 - 0.999 = 0.001

1 - 0.9999 = 0.0001

1 - 0.99999 = 0.00001

so on and so forth. You can clearly see there will always be a difference. Now the main point that everyone is saying that as this difference a super infinitely small number, it will converge to zero. Thus proving that the numbers are equal. But that only occurs when we reach the end of infinity. Which you can obviously tell will never happen. >=)

 

[bigredguy]

If 1 does equal .99... then 1-.99... would equal 0, but it wouldn't 1-.99... = some miniscule decimal, something greater than 0.
.99...+x=1 if x has a value then they can't be the same.