pi
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Vikram A Punathambekar wrote:
By definition, i * i = -1
Thankyou! Someone who gets the definition correct! :-D I see most people say that i = sqrt(-1), which is NOT correct - it implies that i2 = 1
Ryan
"Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
Hang on. If the definition if i is i^2 = -1 then i = +/- sqrt(-1) Nothing wrong with i := sqrt(-1) cheers, Chris Maunder
CodeProject.com : C++ MVP
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Somebody has probably said this already, and from the perspective of almost all numbers people think about in real life, it's absolutly true, but really the vast majority of real numbers are transcendental. So there are algebraic numbers, which are solutions to polynomial equations with rational coefficients. So 2 is algebraic(solution to x-2=0), Sqrt(2), which solves x^2-2=0, is, and so on. Then there are transcendental numbers, which are numbers that aren't solutions to ANY polynomial. Pi and e are two examples; both are pretty deep proofs, and pretty much have to be taken at face value unless you're a pro and have a spotter. Since there are as many polynomials as there are finite length sequences of rational numbers, there are as many algebraic equations as there are fractions and, in turn, as many algebraic equations as natural numbers. (by "as many as", I mean that we can pair each algebraic number with a natural number-1,2,3,4,...- so that each one from each group has exactly one partner from the other group.) Since the set of algebraic equations has the same size as the natural numbers, we say that it's a countable set. So for both algebraic equations and fractions, there is a way to count them up, 1,2,3, so that you hit every one of them. (I was blown away the first time I heard there was a way to count the fractions.) And since each algebraic eqation has a finite number of solutions, the set of all algebraic numbers is countable. You can also prove that the set of all real numbers is not countable. You can prove that any way you pair up natural numbers and real numbers, there will always be real numbers left over. A casual way to think about these sizes is that the size of the real numbers is as much bigger as infinity is from a some finite number. So in exactly the same way that (infinity - 10)=infinity, the uncountable real numbers-the countable algebraic numbers leaves an uncountable amount left over. So therefore "almost all" real numbers are not solutions to algebraic equations. And the "almost all" here means exactly the same thing as if I said that I'm thinking of a number between 1 and infinity, and you were to guess, that you have "almost no" chance of guessing correctly. So, "almost all" real numbers are transcendental. Was that kid sister safe?
Nathan A. wrote:
as if I said that I'm thinking of a number between 1 and infinity, and you were to guess, that you have "almost no" chance of guessing correctly
On the opposite I would argue you probably thought of 42 or some other integer number in the vicinity :-D However I have some friend having a Math PhD and they tend to think to number such exp(log(pi+1))... ;P No, seriously, you're right!
Nathan A. wrote:
So, "almost all" real numbers are transcendental.
Well, yeah... But truth to tell, real human being don't use them ;P
Nathan A. wrote:
Was that kid sister safe?
:laugh: Well, sort of :-D
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I'm trying to find a good way to explain why pi is infinite (not what it is). And I'm drawing up blanks. Any math gurus care to shed me some light please? Jeremy Falcon
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Same as to David stone. Although it's funny your:
Vikram A Punathambekar wrote:
Like I said, if a number is not real, it's square is negative.
what about 1+i ? it square is 2i which is not negative, hence it is a real?
Super Lloyd wrote:
what about 1+i ?
It's a complex number, which means it has a real part and an imaginary part. Cheers, Vikram.
I don't know and you don't either. Militant Agnostic
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Super Lloyd wrote:
what about 1+i ?
It's a complex number, which means it has a real part and an imaginary part. Cheers, Vikram.
I don't know and you don't either. Militant Agnostic
if complex number don't coun't, what's the point of your definition then? :omg: :laugh:
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I'm trying to find a good way to explain why pi is infinite (not what it is). And I'm drawing up blanks. Any math gurus care to shed me some light please? Jeremy Falcon
Pi is irrational, like my ex-wife. Fortunately, neither is infinite. Both go on forever, without end, for no good reason, never repeating any sensible pattern. Thank God that Pi can't hold a credit card. "...a photo album is like Life, but flat and stuck to pages." - Shog9
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I'm trying to find a good way to explain why pi is infinite (not what it is). And I'm drawing up blanks. Any math gurus care to shed me some light please? Jeremy Falcon
Let me approach this another way. You are just wondering - why the heck must a simple thing such as the ratio of the circumference to the diameter be so darn complex. For gawds sake it is a ratio ! Assuming you are able to measure both the circumference and diameter to a great precision with an electron microscope thing. Then PI would simply be c/d - right ? So why on earth is Pi so enormously complicated. Why is Pi irrational/ trancedental/ math-mumbojumbo word - when both circumference and diameter are simple, measurable real numbers. The answer to that lies in the fact that there is no such thing as a measurement. We humans cannot measure, we can only count. We can count inches, millimeters, microns, nanometers and call it a measurement, but we are fooling no one. Still, you wonder how did this monstrosity creep up on us. Why cant PI be something simpler like a square root (3). I think the best answer I have heard so far is "yeah, it might be a really simple number in another universe parallel to ours, we might have just got unlucky in our universe". The moment I heard this from a scifi geek buddy - I understood I had hit a culdesac. There was no point trying to get to the root of this without some tools such as a wormhole-stargate.
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Ryan Binns wrote:
i * i = sqrt(-1) * sqrt(-1) i * i = sqrt(-1 * -1)
The property that sqrt(a) * sqrt(b) = sqrt(a * b) only applies to real x >= 0. So you really can't do that.
They dress you up in white satin, And give you your very own pair of wings In August and Everything After
I'm after everything
Aah. Ok. Perhaps you could explain that to every maths lecturer I ever had. 5 years of maths, and they all said the same thing :~
Ryan
"Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
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Hang on. If the definition if i is i^2 = -1 then i = +/- sqrt(-1) Nothing wrong with i := sqrt(-1) cheers, Chris Maunder
CodeProject.com : C++ MVP
Chris Maunder wrote:
If the definition if i is i^2 = -1 then i = +/- sqrt(-1)
I thought that was implicit. SQRT(36) is -6 as much as it is 6. Cheers, Vikram.
I don't know and you don't either. Militant Agnostic
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Hang on. If the definition if i is i^2 = -1 then i = +/- sqrt(-1) Nothing wrong with i := sqrt(-1) cheers, Chris Maunder
CodeProject.com : C++ MVP
Chris Maunder wrote:
If the definition if i is i^2 = -1 then i = +/- sqrt(-1)
I thought that was implicit. SQRT(36) is -6 as much as it is 6. Cheers, Vikram.
I don't know and you don't either. Militant Agnostic
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Chris Maunder wrote:
If the definition if i is i^2 = -1 then i = +/- sqrt(-1)
I thought that was implicit. SQRT(36) is -6 as much as it is 6. Cheers, Vikram.
I don't know and you don't either. Militant Agnostic
Right - every square root has two solutions so you can't define a single quantity as the solution to a square root, otherwise you'd be implying sqrt(-1) = -sqrt(-1). And that would cause a few problems ;) cheers, Chris Maunder
CodeProject.com : C++ MVP
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Right - every square root has two solutions so you can't define a single quantity as the solution to a square root, otherwise you'd be implying sqrt(-1) = -sqrt(-1). And that would cause a few problems ;) cheers, Chris Maunder
CodeProject.com : C++ MVP
Chris Maunder wrote:
every square root has two solutions so you can't define a single quantity as the solution to a square root
Except, of course, SQRT(0). ;P Cheers, Vikram.
I don't know and you don't either. Militant Agnostic
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if complex number don't coun't, what's the point of your definition then? :omg: :laugh:
The set of complex numbers is a superset of the set of real numbers. So real numbers are included in the set of complex numbers. Unreal numbers are just the set of all numbers that are not in the reals.
They dress you up in white satin, And give you your very own pair of wings In August and Everything After
I'm after everything
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Aah. Ok. Perhaps you could explain that to every maths lecturer I ever had. 5 years of maths, and they all said the same thing :~
Ryan
"Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
That really sucks. Those profs should be dragged out back and shot. (In the mathematical sense, of course. ;P)
They dress you up in white satin, And give you your very own pair of wings In August and Everything After
I'm after everything
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The set of complex numbers is a superset of the set of real numbers. So real numbers are included in the set of complex numbers. Unreal numbers are just the set of all numbers that are not in the reals.
They dress you up in white satin, And give you your very own pair of wings In August and Everything After
I'm after everything
hu.... ??? for the remark my remark was related the Vikram definition of real number. Vikram definition: number whose square value is not negative.
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Chris Maunder wrote:
every square root has two solutions so you can't define a single quantity as the solution to a square root
Except, of course, SQRT(0). ;P Cheers, Vikram.
I don't know and you don't either. Militant Agnostic
There's always one... :rolleyes: cheers, Chris Maunder
CodeProject.com : C++ MVP
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I thought you could still have an infinite irrational number though. Or, is that not the case? Jeremy Falcon
That's one of these things one can't be certain of. I think. What's to say that pi doesn't plane out on zero decimals halfway to infinity? :)
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I'm trying to find a good way to explain why pi is infinite (not what it is). And I'm drawing up blanks. Any math gurus care to shed me some light please? Jeremy Falcon
Isn't Pi a coëfficiënt for the circle? I'm thinking that it keeps on going because a real circle doesn't exist, it's always a rounding of some polygone. The more decimals you have, the closer you get to the real circle and because a real circle doesn't exist, you can keep on going. But that's just a theory :-). There are other rational numbers you know: e, root of 2, ... :-D No hurries, no worries. -- modified at 2:25 Friday 17th March, 2006
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That really sucks. Those profs should be dragged out back and shot. (In the mathematical sense, of course. ;P)
They dress you up in white satin, And give you your very own pair of wings In August and Everything After
I'm after everything
David Stone wrote:
Those profs should be dragged out back and shot. (In the mathematical sense, of course. ;P)
Of course. We should mathematically prove that the shot will kill them before firing it ;)
Ryan
"Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
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Chris Maunder wrote:
every square root has two solutions so you can't define a single quantity as the solution to a square root
Except, of course, SQRT(0). ;P Cheers, Vikram.
I don't know and you don't either. Militant Agnostic
Vikram A Punathambekar wrote:
Except, of course, SQRT(0).
You mean you can't have -0 ? :rolleyes: The Intel 387 maths coprocessor could. It differentiated between +0 and -0 :~
Ryan
"Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"