what is wrong with this?

[kep]

what do u think about this
in complex numbers {(-1)^0.5 = i}
-1=-1

(-1)^0.5 = (-1)^0.5

(1/-1)^0.5 = (-1)^0.5

{(1)^0.5}/{(-1)^0.5} = {(-1)^0.5}/{(1)^0.5}

(1)^0.5 * (1)^0.5 = (-1)^0.5 *(-1)^0.5

1= -1

this is one of fallacy in complex numbers
so what is the wrong step?

 

[Pommie]

1/(-1^½) != -1^½

Mike.

 

[Papabravo]

What is wrong is that in the second line you assumed that i was equal to 1/i and that is not true. The algebra of complex numbers is different than the algebra of real numbers. That is why we use a different symbol so that we won't be tempted to make inconsistent assertions like that one. The problem with your original assertion is that when combining real numbers and complex numbers the order of operations is important AND many of those operations are NOT commutative.

 

[kep]

thanx
got da idea 1/i != i
i miss da basics !!!!

 

[3iMaJ]

1/i = -i

Close though.

 

[3iMaJ]

What is wrong is that in the second line you assumed that i was equal to 1/i and that is not true. The algebra of complex numbers is different than the algebra of real numbers. That is why we use a different symbol so that we won't be tempted to make inconsistent assertions like that one. The problem with your original assertion is that when combining real numbers and complex numbers the order of operations is important AND many of those operations are NOT commutative.

The algebra of complex numbers is the same as the algebra for real numbers, one just has to be careful. The same rules apply. However, the calculus of complex numbers does get a little tricky. Order of operations is always important, there aren't times that you can just "forget" about them whether you're working with real, rational, complex, whole, or whatever type of number system you dream up. Although if you make one up, you're going to have to prove associtivity, transitive, communitivity etc.

 

[Papabravo]

In order to substantiate the claim that the two algebras are the same you need to start with elements of the set which are different. Now it is true that the algebra of real numbers is contained within the algebra of complex numbers. It is the set of all complex numbers with the inaginary part set equal to zero. There is a much closer relationsip between vector algebra and complex algerbra.

 

[3iMaJ]

In order to substantiate the claim that the two algebras are the same you need to start with elements of the set which are different. Now it is true that the algebra of real numbers is contained within the algebra of complex numbers. It is the set of all complex numbers with the inaginary part set equal to zero. There is a much closer relationsip between vector algebra and complex algerbra.

I made no such claim. I only said there certain rules that numbering systems follow. Yes the real numbers are a subset of the complex numbers, but that in no way implies that there are two different kinds of algebra?

And a side note, real numbers can be just as easily be manipulated using vector algebra as complex numbers, although they typically aren't because there is no reason to complicate it to such a degree.

In the end its all algebra, all following the same rules.

 

[43617373]

How is a number being obtained from a negative in the square root, or am I forgetting a large part of my algebra.

 

[dknguyen]

How is a number being obtained from a negative in the square root, or am I forgetting a large part of my algebra.

You break up the number underneath the square root:

sqr(-x) = sqr(-1)*xqr(x)

and then replace the sqr(-1) with i, making the result an imaginary number (or more specifically, a complex number with a real component of zero, at least for this example). It's the basis for math with complex numbers. Maybe you just never learned it?

 

[Papabravo]

I made no such claim. I only said there certain rules that numbering systems follow. Yes the real numbers are a subset of the complex numbers, but that in no way implies that there are two different kinds of algebra?

And a side note, real numbers can be just as easily be manipulated using vector algebra as complex numbers, although they typically aren't because there is no reason to complicate it to such a degree.

In the end its all algebra, all following the same rules.

I invite the other members to reread your original post to see if you made precisely that claim. It sure sounds like you did to me, but what do I know, I'm just a graduate engineer.

 

[Nigel Goodwin]

I invite the other members to reread your original post to see if you made precisely that claim. It sure sounds like you did to me, but what do I know, I'm just a graduate engineer.

My daughters doing A level Further Maths, and apparently you need to be 'weird' to be any good at it - she fits right in!

According to the teachers, A level Further Maths is more difficult than a normal maths degree, and where top universities require applicants to have grade A's in their subjects, even a grade E is acceptable if it's Further Maths.

Needless to say, I haven't got the faintest clue what she's doing with it! - means she can't ask me to help

I did offer to help if she took Physics, but she took Chemistry, Maths, Further Maths and Music Technology - so I'm safe!

 

[quixotron]

My daughters doing A level Further Maths, and apparently you need to be 'weird' to be any good at it - she fits right in!

According to the teachers, A level Further Maths is more difficult than a normal maths degree, and where top universities require applicants to have grade A's in their subjects, even a grade E is acceptable if it's Further Maths.

Needless to say, I haven't got the faintest clue what she's doing with it! - means she can't ask me to help

I did offer to help if she took Physics, but she took Chemistry, Maths, Further Maths and Music Technology - so I'm safe!

Further maths??

The 500 and 600 level math courses at Purdue require advisor consent or a minimum grade of B or better. A is the highest you can get on a 4.0 scale, never heard of an E or and "S". 700 level math courses require PHD status or consent of an advisor with superb background mathematical skills.

The highest I have taken is random variables and electromagnetics field theory, both 600 level courses. I got a B in both, they were rough courses, designed to select phd and research assistants, it dropped my gpa a bit though. If I ever take something like that again, it'd be two classes only.

 

[Nigel Goodwin]

Further maths??

An A Level course in the UK (notice my location is filled in, so you know where I am), it's considerably above normal A Level maths courses.

Interesting TV phone in the other day, apparently Maths degree courses have the highest drop out rate of any courses? - and a university maths lecturer rang in to say that it's mostly only the Further Maths students who can cope with the degree course?.

 

[DMW]

How about:

' = reoccurring

X = 0.9999' [* 10]
10X = 9.9999' [-X]
9X = 9.0 [/9]
x = 1

 

[Ratchit]

To the Ineffable All,
There seems to be some confusion about "i", which is sometimes wrongly called the square root of -1. The square root of -1 is a numeric constant, specifically 1/_90deg in polar form. "i" is correctly defined as a operator in correct math texts. 5i means take 5 and rotate it 90 degrees CCW. 2+5i defines a duplex number that has a real part and a orthogonal part. There is confusion because of bad notation. i^2 is not an exponential operation. It really means an operation that rotates 1 twice by 90 degrees. A better notation would be something like i(2). Now we get correct answers by treating "i" like a constant, but that is due to conformal similarity, not because the concept is correct. So again, Sqrt(-1) is not "i". It is 1/_90 in polar form or i(1) in rectangular form. Ratch

 

[3iMaJ]

Lets define i in a rigorous manner so we can be done with all of this nonsense.

Given an ordered pair (a,b), lets define two operations:
(a,b) + (c,d) = (a + b,c +d)
(a,b)*(c,d) = (ac - bd,bc + ad)

We will cleverly call our operations addition and multiplication. For our newly defined operations we would like them to satisfy a few properties, namely:

Closure under both + and *.
Both + and * are associative.
Both + and * are communitive.
The operation * is distributive over +.
Existance of additive identity.
Existance of multiplicitive identity.
Existance of additive inverses.
Existance of multiplicitve inverses.

If our operations have all of these properties then we've defined a field. A field is a communitive ring. A ring is simply a algebraic structure that contains two operations, in this case multiplication and addition. A communitive ring (or a field) is simply a ring where multiplication is communitive. You might note that the matrix operations do not form a field, as their multiplication is not communitive (generally speaking). As well as the problem with the existance of an inverse in all cases. An example of a field is the set of real numbers.

(a,b)*(1,0) = (a,b) = (1,0)(a,b) = (a,b): Our multiplicitive identity
(a,b) + (0,0) = (a,b) = (0,0) + (a,b): Our additive identity
(a,b)*(c,d) = (1,0)
(ac - bd, bc + ad) = (1,0)
ac - bd = 1
ad + bc = 0
Solving for c and d we get (c,d) = (a/(a^2 + b^2), -b/(a^2 + b^2))
which is our multiplicitive inverse.
(a,b) + (c,d) = (0,0)
a + c = 0
b + d = 0
c = -a
d = -b
(a,b) + (-a,-b) = (0,0): Our additive inverse

These properties are obviously hold and I will not prove them.

Closure under both + and *.
Both + and * are associative.
Both + and * are communitive.
The operation * is distributive over +.

So, where does i fit in?

This is the definition of i.

(a,b) = a(1,0) + b(0,1)
let (0,1) = i
(0,1)*(0,1) = (-1,0)
i^2 = -1

Since (a,b) uniquely specifies a + ib, the complex numbers are a one-to-one correspondence to the complex plane.

A question often arises what are complex numbers useful for? Since real numbers are a subset of complex numbers there aren't solutions to all equations with out invoking the more general complex number, namely

x^2 + 1 = 0, has no real number solution, but as we've just seen in the definition of i, the solution to that equation i.

To recap:

The "imaginary" number i is a consequence of how we define the operations + and * in our new field we're calling the complex numbers. A complex number is simply a number that has operations + and * that satisfy several properties that were listed earlier.

The unit i arises when we write (a,b) = a(1,0) + b(0,1). We can write it in this form because of the properties of our field. The interesting piece of this equation is (0,1), what exactly is (0,1)? It is (0,1)*(0,1) = i^2 = -1. So we can see where i = sqrt(-1) comes from.

Its VERY important to note, and this was the point made in the previous post is that the sqrt(.) function is only defined for real x >= 0. Using sqrt(-1) to manipulate numbers can result in false results namely:

-1 = i*i = sqrt(-1)*sqrt(-1) = sqrt(1) = 1. This is incorrect because the square root function IS NOT defined for x < 0, which is the reason for the bad result. To avoid this mistake simply do the following:

i = i*sqrt(1), so -1 = i*i = i*i*sqrt(1)*sqrt(1) = -1. Crisis averted.

 

[Ratchit]

3iMaJ,

I don't know why you are proving binary properties of associativity, communitivity, and distributivity on the unary operator i.

A question often arises what are complex numbers useful for?

I don't believe so. Anyone who has worked in engineering knows how useful duplex numbers are.

The "imaginary" number i is a consequence of how we define the operations + and * in our new field we're calling the complex numbers. A complex number is simply a number that has operations + and * that satisfy several properties that were listed earlier.

Actually, there is no "imaginary" number. That is a misnomer. The number most folks call imaginary is actually a number that is orthogonal to the real number taken as the reference. For instance, in electrical engineering, the energy in AC circuits within the magnetic and electrostatic fields represented by "imaginary" numbers is every bit as real as the energy dissipated by the resistive component.

Its VERY important to note, and this was the point made in the previous post is that the sqrt(.) function is only defined for real x >= 0.

Well, every negative number has two square roots. I already showed that -1 has a square root of plus or minus 1/_90.

This is incorrect because the square root function IS NOT defined for x < 0

Well, I say that that isn't so for the above reason.

Crisis averted.

Crisis that never was. Ratch

 

[3iMaJ]

3iMaJ,

I don't know why you are proving binary properties of associativity, communitivity, and distributivity on the unary operator i.



I don't believe so. Anyone who has worked in engineering knows how useful duplex numbers are.



Actually, there is no "imaginary" number. That is a misnomer. The number most folks call imaginary is actually a number that is orthogonal to the real number taken as the reference. For instance, in electrical engineering, the energy in AC circuits within the magnetic and electrostatic fields represented by "imaginary" numbers is every bit as real as the energy disipated by the resistive component.



Well, every negative number has two square roots. I already showed that -1 has a square root of plus or minus 1/_90.



Well, I say that that isn't so for the above reason.



Crisis that never was. Ratch

I was deriving complex numbers based on two operators. And I'd like to point out that i is not an operator. The operators of the field are + and *. i is simply (0,1) its a number within the complex set, not an operator.

You'll also note that I put "imaginary" in quotations just as I've done here. That is commonly what they're called, even though it is technically incorrect.

Let me be very specific here:

The square root operator is ONLY defined for real numbers >= 0. Now let me show you why you can use them on complex numbers.

Reminding you (0,1)(0,1) = i^2 = -1.
sqrt(-16) = sqrt(-1)*sqrt(16) = i*4. HOWEVER sqrt(-16) w/o separating it into sqrt(-1)*sqrt(16) is not defined for the square root function.

The mathematical definition of the square root is that its the inverse function of f(x) = x^2 for all real x >= 0. In other words the square root function is a mapping of non-negative real numbers to R+ union 0. The key there is non-negative.


Quit with all the orthogonal nonsense, of course a purely real number is orthogonal to a "imaginary" by construction.

(a,b) is our ordered pair construction, so (a,0)*(0,b) = (0,0). Which is by definition orthogonal. This true because of how we constructed our field, and not for any other reason.

(0,1) is 1<90 because of the mapping of the complex numbers to the complex plane. Your orthogonality isn't a consequence of the mapping like you seem to claim, but rather a consequence of the construction of the field.

 

[Ratchit]

3iMaJ,

And I'd like to point out that i is not an operator. The operators of the field are + and *. i is simply (0,1) its a number within the complex set, not an operator.

I disagree, as do many math textbooks that define i to be a operator. Yes, + and * are binary operators too, but that does not keep i from being a unary operator.

You'll also note that I put "imaginary" in quotations just as I've done here. That is commonly what they're called, even though it is technically incorrect.

We both agree on that. I think they should be called orthogonal numbers.

Reminding you (0,1)(0,1) = i^2 = -1.

You are treating i as a constant, which I don't agree with.

HOWEVER sqrt(-16) w/o separating it into sqrt(-1)*sqrt(16) is not defined for the square root function.

The Sqrt(-16) is calculated, not defined. I would not say that 2+2 is defined to be 4. It is calculated to be 4. I would, however, say that 0! can be defined to be 1 because its calculation is indefinite.

The mathematical definition of the square root is that its the inverse function of f(x) = x^2 for all real x >= 0. In other words the square root function is a mapping of non-negative real numbers to R+ union 0. The key there is non-negative.

You can map it anywhich way you want, but why restrict yourself. I know that the Sqrt(-4) is either -2/_90 or 2/_90. How? Because when I multiply each of those number by themselves, I get -4. That is the basic definition of a square root.

Quit with all the orthogonal nonsense, of course a purely real number is orthogonal to a "imaginary" by construction.

A orthogonal number is what it is. And it is not the mapping that makes it orthogonal. The orthogonal number is the dog that wags the tail (map).

(a,b) is our ordered pair construction, so (a,0)*(0,b) = (0,0). Which is by definition orthogonal. This true because of how we constructed our field, and not for any other reason.

The tail wagging the dog again.

(0,1) is 1<90 because of the mapping of the complex numbers to the complex plane. Your orthogonality isn't a consequence of the mapping like you seem to claim, but rather a consequence of the construction of the field.

The orthogonal number itself determines its properties, not the mapping. Ratch