isn't square root of -2 equal plus/minus [root 2]i ?
true, I forgot this .
I often forget the minus possiblity....
I often forget the minus possiblity....
I assume that you don't live in Victoria, Australia, as in VCE, "further maths" is the easiest type of maths (for people who can't do maths but still want to.Towelly wrote:
Oo, imaginary numbers (that's not sarcasm just what we call them), that stuff gets complicated, thank god I'm not doing further maths next year.-=raska=- wrote:
kmal : square root of -2 = 1,41i
now, argue.
We are talking about one number - the universal concept of one - but two different ways of representing it.Des.Kmal wrote:
because a number is a number, not more than 1 number.........Gulf_War_Syndrome wrote:
It's a good thing people like Einstein didn't post on forums otherwise they may have been discouraged from their work by kids like Des.Kmal calling them "dumbfucks".
1 = -i2 = lne = log10 = x0 = cos(0) = sin(pi/2) = 0.9~
They are all the same number, just different ways of coming by that number.
Like I said, some people will never understand some concepts. They are driven by their emotions and the fear of the unknown.
Personally, I found learning pure maths a mind expanding experience - yes, similar to taking magic mushrooms - and I'm glad I was open mined enough to try both.
Nope, in the UK in our A-levels you pick 3-5 subjects you want to do for AS for your first year, this includes maths (which I'm doing) and then further maths which is insanely hard.some_random_panda wrote:
I assume that you don't live in Victoria, Australia, as in VCE, "further maths" is the easiest type of maths (for people who can't do maths but still want to.Towelly wrote:
Oo, imaginary numbers (that's not sarcasm just what we call them), that stuff gets complicated, thank god I'm not doing further maths next year.-=raska=- wrote:
kmal : square root of -2 = 1,41i
now, argue.
After doing the AS exams you then go onto A2, I think you can do further maths for them as well but only those who are considering it for a degree or are just that clever do it.
Yup 10x - x is not 9... its 9x, go back to school OP!!!Bertster7 wrote:
You think wrong then.deadawakeing wrote:
I think 10x - x= 1Onionpaste wrote:
That means that x is equal to .99999~ and 1 at the same time. Therefore, .99999~ = 1!
That would equal 9x.
Here's an example, let x = 10 for arguments sake, then 10*10 - 10 = 9? No it doesn't, it equals 90. If x = 5? 10*5 - 5 = 45.
0.999999... can be proved to be =1 because it is so close that we as humans do not care for the difference.
0.666666666 can be proven to equal 2/3 even though it isn't. And we readily accept that it is.
Because as I said before we don't care for the difference.
That is the plain and simple case. There is nothing more to it.
0.666666666 can be proven to equal 2/3 even though it isn't. And we readily accept that it is.
Because as I said before we don't care for the difference.
That is the plain and simple case. There is nothing more to it.
And I say its 54!
you mean this thread isn't dead yet?!?!
its called splitting hairs... go out and bang a chick and dont talk about this... lol
Love is the answer
Oh my fucking god, I was going to post something like this today...
Anyhow...
As for 0.99999... = 1, that is sort of true:
For simplicities sake, let n equal the number of digits after the decimal place. As n-> infinity, 0.999... = 1.
That is similar to limits, ie. 1/x = 0 , as x -> infinity.
------------------------------
As for the proof for 2=1, I'll post the problem now. I didn't read the last two pages so I don't know if I'm the first.
THE EQUATION:
Let x = y
x2 = xy
x2 - y2 = xy - y2
(x+y)(x-y) = y(x-y)
(x+y) = y
As x=y , let x=1 and y=1
Therefore as, x+y=y,
1+1=1... 2=1
THE PROBLEM:
One of the steps involves dividing both sides by (x-y) to eliminate it from the equation. The problem is that if:
x=y, then (x-y) must equal ZERO! Dividing by zero gives the answer to be infinity... That's where the proof of 2=1 fails.
Mcminty.
Anyhow...
As for 0.99999... = 1, that is sort of true:
For simplicities sake, let n equal the number of digits after the decimal place. As n-> infinity, 0.999... = 1.
That is similar to limits, ie. 1/x = 0 , as x -> infinity.
------------------------------
As for the proof for 2=1, I'll post the problem now. I didn't read the last two pages so I don't know if I'm the first.
THE EQUATION:
Let x = y
x2 = xy
x2 - y2 = xy - y2
(x+y)(x-y) = y(x-y)
(x+y) = y
As x=y , let x=1 and y=1
Therefore as, x+y=y,
1+1=1... 2=1
THE PROBLEM:
One of the steps involves dividing both sides by (x-y) to eliminate it from the equation. The problem is that if:
x=y, then (x-y) must equal ZERO! Dividing by zero gives the answer to be infinity... That's where the proof of 2=1 fails.
Mcminty.
I FIGURED IT OUT!!!!!!
Ok think of it this way. .3333333..... plus .3333333333..... plus .333333333333...... equals .999999999999... right?
But .33333333333333..... is actualy one third
and one third plus one third plus one third equals............?
You guessed it! ONE!!!!!!!
So yes, .999999999999... does equal one.
Ok think of it this way. .3333333..... plus .3333333333..... plus .333333333333...... equals .999999999999... right?
But .33333333333333..... is actualy one third
and one third plus one third plus one third equals............?
You guessed it! ONE!!!!!!!
So yes, .999999999999... does equal one.
This falls down at step one. There's no such thing as a 9.9999 followed by an infinite number of 9`s. The number of 9's can only head towards infinity and therefore the rest of the maths doesn't work as you cannot ever complete the first step.Onionpaste wrote:
Dunno if this has already been covered, if so, sorry for the spam, not intended.
So:
Let's call .99999~ (that's a decimal followed by an infinite number of 9's) a variable x.
So that means that x = .9999~
So by that logic, 10x = 9.9999~
If you subtract x from 10x, you get 9.00000~, as in the equation 10x - x = 9.00000~
This means 9.0000~, or just 9 (I put the 0's there to show that all the 9's after decimal went away), equals 9x.
9x = 9
Divide each side by 9, and you get x = 1.
That means that x is equal to .99999~ and 1 at the same time. Therefore, .99999~ = 1!
If any math majors have found some sort of error in this reasoning, do not hesitate to reply so I can debate it with you
-Onion
Last edited by .:XDR:.PureFodder (2006-11-30 02:49:03)
That's not hard. You try doing proper maths, once you get to uni and they've got you doing Laplace, Fourier and Z-Transforms on AC signals represented as complex numbers (or polar coordinates) - then it gets hard. Most maths is easy, it's just a case of precticing it loads, which I hate doing, so I forget it all. But then it's very frustrating when you get a problem and you know you once knew how to solve it, but you can't remember how you used to do it......Towelly wrote:
Nope, in the UK in our A-levels you pick 3-5 subjects you want to do for AS for your first year, this includes maths (which I'm doing) and then further maths which is insanely hard.some_random_panda wrote:
I assume that you don't live in Victoria, Australia, as in VCE, "further maths" is the easiest type of maths (for people who can't do maths but still want to.Towelly wrote:
Oo, imaginary numbers (that's not sarcasm just what we call them), that stuff gets complicated, thank god I'm not doing further maths next year.
After doing the AS exams you then go onto A2, I think you can do further maths for them as well but only those who are considering it for a degree or are just that clever do it.
..... or maybe that's just me.
*edit* You lot have got it easy anyway, back in my day we did proper A-levels.... (and that wasn't that long ago, I think my year were the last to do them)
Last edited by Bertster7 (2006-11-30 03:02:10)
No, you can work with infinite numbers ok. It's all about limits, but I can't remember any of it......:XDR:.PureFodder wrote:
This falls down at step one. There's no such thing as a 9.9999 followed by an infinite number of 9`s. The number of 9's can only head towards infinity and therefore the rest of the maths doesn't work as you cannot ever complete the first step.Onionpaste wrote:
Dunno if this has already been covered, if so, sorry for the spam, not intended.
So:
Let's call .99999~ (that's a decimal followed by an infinite number of 9's) a variable x.
So that means that x = .9999~
So by that logic, 10x = 9.9999~
If you subtract x from 10x, you get 9.00000~, as in the equation 10x - x = 9.00000~
This means 9.0000~, or just 9 (I put the 0's there to show that all the 9's after decimal went away), equals 9x.
9x = 9
Divide each side by 9, and you get x = 1.
That means that x is equal to .99999~ and 1 at the same time. Therefore, .99999~ = 1!
If any math majors have found some sort of error in this reasoning, do not hesitate to reply so I can debate it with you
-Onion
Will the upper limit of 0.9999999 recuring be 1?
Why are you guys debating a fact? If your highest level of math education is 8th grade algebra you should not be posting here. This thread shouldn't even be in this forum seeing that it is merely listing a fact which can't be debated. There is no gray area. It is not "sort of" true. It is as true as saying 1 = 1. There is plenty of proof supporting this fact. There is no proof that suggests otherwise, only conjecture. Saying "every number is its own number" or any other opinion you might have on the concept of infinity is not proof.
Read this: http://en.wikipedia.org/wiki/0.999...
THREAD OVER
Read this: http://en.wikipedia.org/wiki/0.999...
THREAD OVER
Last edited by Fancy_Pollux (2006-11-30 03:25:21)
It depends.Fancy_Pollux wrote:
Why are you guys debating a fact? If your highest level of math education is 8th grade algebra you should not be posting here. This thread shouldn't even be in this forum seeing that it is merely listing a fact which can't be debated. There is no gray area. It is not "sort of" true. It is as true as saying 1 = 1. There is plenty of proof supporting this fact. There is no proof that suggests otherwise, only conjecture. Saying "every number is its own number" or any other opinion you might have on the concept of infinity is not proof.
Read this: http://en.wikipedia.org/wiki/0.999...
THREAD OVER
I always studied engineering maths, where problems like this are not considered 'real', because they aren't. It is just an anomaly of the mathematical system we use, not a truely identical number.
You will find a lot of mathematicians, Physicists and Engineers totally disagreeing on this one - no matter what the absolute proofs say. It's not a problem you would ever encounter in a practical problem because no matter how much accuracy your calculations work to you are never going to actually be using an infinitely recurring number. All real numbers in the real world are of finite length so this problem will never arise, so it could be said not to exist.
No. It is a fact. This is not disputable. Again, no more conjecture. Don't say "well we live in a world of finite length etc etc". Provide mathematical proof that .9 recurring is not equal to 1 or this debate is over.Bertster7 wrote:
I always studied engineering maths, where problems like this are not considered 'real', because they aren't. It is just an anomaly of the mathematical system we use, not a truely identical number.
You will find a lot of mathematicians, Physicists and Engineers totally disagreeing on this one - no matter what the absolute proofs say. It's not a problem you would ever encounter in a practical problem because no matter how much accuracy your calculations work to you are never going to actually be using an infinitely recurring number. All real numbers in the real world are of finite length so this problem will never arise, so it could be said not to exist.
Last edited by Fancy_Pollux (2006-11-30 03:49:16)
But why is it a fact?Fancy_Pollux wrote:
No. It is a fact. This is not disputable. Again, no more conjecture. Don't say "well we live in a world of finite length etc etc". Provide mathematical proof that .9 recurring is not equal to 1 or this debate is over.Bertster7 wrote:
I always studied engineering maths, where problems like this are not considered 'real', because they aren't. It is just an anomaly of the mathematical system we use, not a truely identical number.
You will find a lot of mathematicians, Physicists and Engineers totally disagreeing on this one - no matter what the absolute proofs say. It's not a problem you would ever encounter in a practical problem because no matter how much accuracy your calculations work to you are never going to actually be using an infinitely recurring number. All real numbers in the real world are of finite length so this problem will never arise, so it could be said not to exist.
Because the numerical system we use says so.
The numerical system we use is a tool, not what is real. You find a computer system that says 0.99 recuring = 1 that doesn't use specific coding to express this anomaly (as most computers/compilers do). You won't be able to, because computers represent finite numbers in a finite way. Each 2's complement additive operation loses half a bit of accuracy, not to mention a finite range of numbers that can be represented by a finite number of bits. So you find me a non specifically coded computer operation that will demonstrate that 1=0.999 recuring. If you can't then this debate is very much alive and well.
Not that that is real anyway. But it is a closer approximation than theoretical maths.
Are you suggesting that 1 divided by 3 does not exist because a computer cannot list the infinite number of recurring digits in its decimal representation? That is a very weak argument. Questioning "why is it a fact" when there is already infallible mathematical proof is not something that can sustain a debate. Again, provide mathematical proof, or whatever conjecture you post means nothing.Bertster7 wrote:
But why is it a fact?
Because the numerical system we use says so.
The numerical system we use is a tool, not what is real. You find a computer system that says 0.99 recuring = 1 that doesn't use specific coding to express this anomaly (as most computers/compilers do). You won't be able to, because computers represent finite numbers in a finite way. Each 2's complement additive operation loses half a bit of accuracy, not to mention a finite range of numbers that can be represented by a finite number of bits. So you find me a non specifically coded computer operation that will demonstrate that 1=0.999 recuring. If you can't then this debate is very much alive and well.
Not that that is real anyway. But it is a closer approximation than theoretical maths.
Last edited by Fancy_Pollux (2006-11-30 04:11:23)
No, I'm saying that an infinitely recurring number isn't a real thing. Any proofs that show a number that doesn't exist to be the same as one are meaningless because it isn't real.Fancy_Pollux wrote:
Are you suggesting that 1 divided by 3 does not exist because a computer cannot list the infinite number of recurring digits in its decimal representation? That is a very weak argument. Questioning "why is it a fact" when there is already infallible mathematical proof is not something that can sustain a debate. Again, provide mathematical proof, or whatever conjecture you post means nothing.Bertster7 wrote:
But why is it a fact?
Because the numerical system we use says so.
The numerical system we use is a tool, not what is real. You find a computer system that says 0.99 recuring = 1 that doesn't use specific coding to express this anomaly (as most computers/compilers do). You won't be able to, because computers represent finite numbers in a finite way. Each 2's complement additive operation loses half a bit of accuracy, not to mention a finite range of numbers that can be represented by a finite number of bits. So you find me a non specifically coded computer operation that will demonstrate that 1=0.999 recuring. If you can't then this debate is very much alive and well.
Not that that is real anyway. But it is a closer approximation than theoretical maths.
It is a flaw in the matematical system. Nothing more.
It is interesting to understand such anomalies but they are of no practical use (except from the perspective of avoiding them in calculations).
Placing blind faith in any system as you seem to do is not wise. All the systems used are human approximations for expressing real values, they are not infailable.
oh good gosh, finally, thank you for proving my point. ffs.Gulf_War_Syndrome wrote:
We are talking about one number - the universal concept of one - but two different ways of representing it.Des.Kmal wrote:
because a number is a number, not more than 1 number.........Gulf_War_Syndrome wrote:
It's a good thing people like Einstein didn't post on forums otherwise they may have been discouraged from their work by kids like Des.Kmal calling them "dumbfucks".
the equation/concept is correct. i cant argue with math, but one number is only one number.
thanks gulf war syndrome, even though u meant to prove me wrong, u just further perpetuated my point.
Add me on Origin for Battlefield 4 fun: DesKmal
Whats about 1/3 is 0.333333333333333333333333333333~
and 2/3 is 0.666666666666666666666666666666~
then logically 3/3 is 0.99999999999999999999999999999999999999999999~
isn't it?!
and 2/3 is 0.666666666666666666666666666666~
then logically 3/3 is 0.99999999999999999999999999999999999999999999~
isn't it?!
Actually, 1/3 doesn't exactly equal 0.3333333333~ But like 0.99999999999~ it is so close we say it is to simplify things.
We'd live in an entirely different world if we didn't assume that.
We'd live in an entirely different world if we didn't assume that.
Actually it is exactly equal. The '~' means recurring. There is no estimation. Both values are exactly the same.Stags wrote:
Actually, 1/3 doesn't exactly equal 0.3333333333~ But like 0.99999999999~ it is so close we say it is to simplify things.
We'd live in an entirely different world if we didn't assume that.
Last edited by Fancy_Pollux (2006-11-30 07:58:01)