Both Phi and the Fibonacci numbers can be found numerous places in nature.Jenspm wrote:
Does infinity exist in Nature?-=raska=- wrote:
Even in nature, 0.999~ = 1. There is no difference between nature and maths when nature incorporates math phenomenons. Why should the answer differ ? We have several algebric proofs that 0.999~ = 1. You probably think that 0.99~ =/= 1 in nature because its hard to find a phenomenon where 0.999~ exists. But when it is there, its 1 !Jenspm wrote:
We discussed this in that other thread, and in math 0.999~ = 1.
In Nature, 0.999~ =/= 1
Math is human made, not made by nature.
I believe that is the conclusion we came to last time.
He/she kept the buck and bought a coke from the vending machine._1_MAN-ARMY.17 wrote:
receptionist acts smart as he put 1$ on his pocket
Each guy pays $9.
The total in the till is now $27.
The receptionist takes $2.
That leaves $25 in the till, 1 with each man, (making 3) and 2 with the receptionist. That's 25 + 1 + 1 + 1 + 2 = 30.
---
*wonders if he just made a very obvious mistake*
The total in the till is now $27.
The receptionist takes $2.
That leaves $25 in the till, 1 with each man, (making 3) and 2 with the receptionist. That's 25 + 1 + 1 + 1 + 2 = 30.
The receptionist has the last two dollars.Andoura wrote:
Wheres the last dollar?
---
*wonders if he just made a very obvious mistake*
http://upload.wikimedia.org/wikipedia/c … occoli.jpgblisteringsilence wrote:
Both Phi and the Fibonacci numbers can be found numerous places in nature.Jenspm wrote:
Does infinity exist in Nature?-=raska=- wrote:
Even in nature, 0.999~ = 1. There is no difference between nature and maths when nature incorporates math phenomenons. Why should the answer differ ? We have several algebric proofs that 0.999~ = 1. You probably think that 0.99~ =/= 1 in nature because its hard to find a phenomenon where 0.999~ exists. But when it is there, its 1 !
Romanesco Brocoli, which is a fractal, that is a theoric representation of infinity and square root of -1.
Of course, in practice its not a real fractal due to biological reasons, cells, etc..
Last edited by -=raska=- (2007-06-27 12:36:47)
Am I right or what?liquidat0r wrote:
Each guy pays $9.
The total in the till is now $27.
The receptionist takes $2.
That leaves $25 in the till, 1 with each man, (making 3) and 2 with the receptionist. That's 25 + 1 + 1 + 1 + 2 = 30.The receptionist has the last two dollars.Andoura wrote:
Wheres the last dollar?
---
*wonders if he just made a very obvious mistake*
people say stuff like
.33333~ = 1/3 so...
.33333~ X 3 = .99999~
and that 1/3 X 3 = 1
but... if .9999~ = 1 then....
.333333~ = .34 and .34 x 3 doesnt equal .99999~ so .999999~ cant = 1???
~ = recurring
[edit] Algebra
[edit] Fractions
One reason that infinite decimals are a necessary extension of finite decimals is to represent fractions. Using long division, a simple division of integers like 1⁄3 becomes a recurring decimal, 0.333…, in which the digits repeat without end. This decimal yields a quick proof for 0.999… = 1. Multiplication of 3 times 3 produces 9 in each digit, so 3 × 0.333… equals 0.999…. And 3 × 1⁄3 equals 1, so .[9]
Another form of this proof multiplies 1/9 = 0.111… by 9.
- from wiki
.333~ = .34 so 1/3 doesnt = .333 , assuming .999~ = 1
.33333~ = 1/3 so...
.33333~ X 3 = .99999~
and that 1/3 X 3 = 1
but... if .9999~ = 1 then....
.333333~ = .34 and .34 x 3 doesnt equal .99999~ so .999999~ cant = 1???
~ = recurring
[edit] Algebra
[edit] Fractions
One reason that infinite decimals are a necessary extension of finite decimals is to represent fractions. Using long division, a simple division of integers like 1⁄3 becomes a recurring decimal, 0.333…, in which the digits repeat without end. This decimal yields a quick proof for 0.999… = 1. Multiplication of 3 times 3 produces 9 in each digit, so 3 × 0.333… equals 0.999…. And 3 × 1⁄3 equals 1, so .[9]
Another form of this proof multiplies 1/9 = 0.111… by 9.
- from wiki
.333~ = .34 so 1/3 doesnt = .333 , assuming .999~ = 1
Last edited by I0973 (2007-06-27 12:50:29)
Wise up your comparing apples with oranges while changing horses in mid-stream.
I LOL how people break their heads to a simple answer, and how the OP skips fractional numers.
This guy solved it.twiistaaa wrote:
how is this even a problem?
$25 plus $3 returned = $28, $28 + $2 = $30
they never pay $27 or $9. they always pay $25 or $8.3 each. $8.3 x 3 = 25
the missing dollar is the .3 x 3 and we know that they still have $1 in their pocket each and the other guy has $2.
to those who are going between decimals and fractions and say well 1/3 x 3 = 1 but .333~ x3 = .999~, assuming 1/3 = .333~ the decimals just are an imperfect way of expressing the fraction. The more exact answer will always be a fraction and therefore a decimal must be adjusted accordingly to meet the product of a fractional equation. That i think is the simplest way to explain this.
there is no missing $1 - they paid $9 each total including the $2 tip. You don't add 2 + 27 = 29, you subtract 2 from 27 = 25
can't believe some of the answers lol..
can't believe some of the answers lol..
Last edited by IG-Calibre (2007-06-27 18:13:23)
How is this still going?
well it wasnt for an hour..... untill you bumped it and im bumping it again by saying thisBubbalo wrote:
How is this still going?
It sounds like you're the only one here who doesn't get it.Andoura wrote:
look they all payd 10$ in 1st when the manager give them back 1$ each it means they payd 9$ ... wich is 9 * 3 = 27 then the manager took 2$ for him wich makes 27 + 2 = 29 !!!!twiistaaa wrote:
how is this even a problem?
$25 plus $3 returned = $28, $28 + $2 = $30
they never pay $27 or $9. they always pay $25 or $8.3 each. $8.3 x 3 = 25
the missing dollar is the .3 x 3 and we know that they still have $1 in their pocket each and the other guy has $2.
Dont try to change things lool
The text was bolded, I count it as going.weasel_thingo wrote:
well it wasnt for an hour..... untill you bumped it and im bumping it again by saying thisBubbalo wrote:
How is this still going?
Hold up. How is .333~ rounded to .34? That makes no sense.I0973 wrote:
people say stuff like
.33333~ = 1/3 so...
.33333~ X 3 = .99999~
and that 1/3 X 3 = 1
but... if .9999~ = 1 then....
.333333~ = .34 and .34 x 3 doesnt equal .99999~ so .999999~ cant = 1???
~ = recurring
[edit] Algebra
[edit] Fractions
One reason that infinite decimals are a necessary extension of finite decimals is to represent fractions. Using long division, a simple division of integers like 1⁄3 becomes a recurring decimal, 0.333…, in which the digits repeat without end. This decimal yields a quick proof for 0.999… = 1. Multiplication of 3 times 3 produces 9 in each digit, so 3 × 0.333… equals 0.999…. And 3 × 1⁄3 equals 1, so .[9]
Another form of this proof multiplies 1/9 = 0.111… by 9.
- from wiki
.333~ = .34 so 1/3 doesnt = .333 , assuming .999~ = 1
No one cares anymore, the "problem" was solved long ago, go to sleep.Smithereener wrote:
Hold up. How is .333~ rounded to .34? That makes no sense.I0973 wrote:
people say stuff like
.33333~ = 1/3 so...
.33333~ X 3 = .99999~
and that 1/3 X 3 = 1
but... if .9999~ = 1 then....
.333333~ = .34 and .34 x 3 doesnt equal .99999~ so .999999~ cant = 1???
~ = recurring
[edit] Algebra
[edit] Fractions
One reason that infinite decimals are a necessary extension of finite decimals is to represent fractions. Using long division, a simple division of integers like 1⁄3 becomes a recurring decimal, 0.333…, in which the digits repeat without end. This decimal yields a quick proof for 0.999… = 1. Multiplication of 3 times 3 produces 9 in each digit, so 3 × 0.333… equals 0.999…. And 3 × 1⁄3 equals 1, so .[9]
Another form of this proof multiplies 1/9 = 0.111… by 9.
- from wiki
.333~ = .34 so 1/3 doesnt = .333 , assuming .999~ = 1
Someone does care - me. And I know, there was a thread about the problem a few months ago. Went on for like 12+ pages till it got closed. But my post wasn't about the .99999~=1 problem. I just wanted to know this guy's reasoning for .3333~ being rounded to .34.Yaocelotl wrote:
No one cares anymore, the "problem" was solved long ago, go to sleep.Smithereener wrote:
Hold up. How is .333~ rounded to .34? That makes no sense.I0973 wrote:
people say stuff like
.33333~ = 1/3 so...
.33333~ X 3 = .99999~
and that 1/3 X 3 = 1
but... if .9999~ = 1 then....
.333333~ = .34 and .34 x 3 doesnt equal .99999~ so .999999~ cant = 1???
~ = recurring
[edit] Algebra
[edit] Fractions
One reason that infinite decimals are a necessary extension of finite decimals is to represent fractions. Using long division, a simple division of integers like 1⁄3 becomes a recurring decimal, 0.333…, in which the digits repeat without end. This decimal yields a quick proof for 0.999… = 1. Multiplication of 3 times 3 produces 9 in each digit, so 3 × 0.333… equals 0.999…. And 3 × 1⁄3 equals 1, so .[9]
Another form of this proof multiplies 1/9 = 0.111… by 9.
- from wiki
.333~ = .34 so 1/3 doesnt = .333 , assuming .999~ = 1
Worse math than average.Smithereener wrote:
Someone does care - me. And I know, there was a thread about the problem a few months ago. Went on for like 12+ pages till it got closed. But my post wasn't about the .99999~=1 problem. I just wanted to know this guy's reasoning for .3333~ being rounded to .34.Yaocelotl wrote:
No one cares anymore, the "problem" was solved long ago, go to sleep.Smithereener wrote:
Hold up. How is .333~ rounded to .34? That makes no sense.
$30 for a hotel room suggests they're either in a third world hotel, or paying by the hour. Three guys paying for one hour in a hotel room, makes you wonder...
x, y, z pay $10 each.
$30 paid so far.
Manager gives x, y, and z $1 back.
$30 - $3 = $27
Manager keeps $2 to himself from initial $30.
$27 - $2 = $25.
$25 = proper cost of Hotel. Problem solved.
It just comes down to semantics and how you frame the problem, bordering on relativity. It depends on how you look at it; the riddle involves throwing you for a loop by adding in the $9 each deal, when that clearly isn't the case. It's what it could *seem* to have been paid, but the Manager doesn't have $27. He has $30 to start with prior to the change.
$30 paid so far.
Manager gives x, y, and z $1 back.
$30 - $3 = $27
Manager keeps $2 to himself from initial $30.
$27 - $2 = $25.
$25 = proper cost of Hotel. Problem solved.
It just comes down to semantics and how you frame the problem, bordering on relativity. It depends on how you look at it; the riddle involves throwing you for a loop by adding in the $9 each deal, when that clearly isn't the case. It's what it could *seem* to have been paid, but the Manager doesn't have $27. He has $30 to start with prior to the change.
what happens if $4.50 suddenly appears after being vomited from a dimensional gate. How does that change things. I donno, but it would definitely be wierd
discuss
discuss
Compression of time relative to c?Blehm98 wrote:
what happens if $4.50 suddenly appears after being vomited from a dimensional gate. How does that change things. I donno, but it would definitely be wierd
discuss
suppose x = .999....
then 10x = 9.999....
10x - x = 9x = 9.999... - .999... = 9
9x = 9
x = 1 = .999...
therefore, .999... = 1
then 10x = 9.999....
10x - x = 9x = 9.999... - .999... = 9
9x = 9
x = 1 = .999...
therefore, .999... = 1
Last edited by i3igpete (2007-06-28 00:19:46)
It works, but only if .99999.... suggests taht there is an infinite amount of 9s, which, according to many different things that i forgot, is so close to 1 that it can be considered there. It is, in fact, infinitely close to 1, without being there...
the moment however, that it is no longer an infinite number of numbers, it is no longer infinitely close, and therefore it is not close enough to be exact. And as such it deviates from the above formula, although it still can get pretty close
the moment however, that it is no longer an infinite number of numbers, it is no longer infinitely close, and therefore it is not close enough to be exact. And as such it deviates from the above formula, although it still can get pretty close