|  ajinx999 | Sep 30, 2007 6:58am | What is 1/i ? {i is the complex number = (-1)^(1/2)}
First, let me show you the value of i^2
i = (-1)^(1/2)
i^2 = -1
This value of i^2 is unique and true.
But, I get a different value by this algorithm;
1/i = 1/[(-1)^(1/2)]
= [(1)^(1/2)]/[(-1)^(1/2)]
= [1/(-1)]^(1/2)
= (-1)^(1/2)
= i
Therefore, 1/i = i
i^2 = 1
This is fallacious.
Certainly, there is an anomaly in the latter algorithm (or else i = 1 or -1 = 1). But, I don't know where is it. |
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Sponsor |  Morosoph | Sep 30, 2007 7:17am | You second approach contains a fallacy: (-1)^(1/2) is not unique. It has two values: i and -i; just because i^2 is unique doesn't impose the same restriction on the inverse function. The last step should read:
... (-1)^(1/2)
= ħi |
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| | ajinx999 | Sep 30, 2007 10:12am | @ Morosoph:
I'm sorry to say that your reasoning is wrong.
(-1)^(1/2) = i only
Conform according to this,
if atleast one of a or/and b is positive,
[(a)^(1/2)][(b)^(1/2)] = (ab)^(1/2)
if a and b both are negative
[(a)^(1/2)][(b)^(1/2)] = -(ab)^(1/2)
For example,
(4)^(1/2) = 2
It is not equal to -2 (or else 2 = -2 which is undoubtedly false)
You probably got confused. Square function is a many one function but square root function is a one-one function. You said the reverse.
Since we are dealing with complexes. According to me, square of i comes to be unique. |
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|  Fephisto | Sep 30, 2007 10:34am | No, I'm pretty sure Morosoph is in the right here. (4)^(1/2) = 2 or -2, since y = x^(1/2) is defined as y such that y^2 = x. So, by definition -2 satisfies this since (-2)^2 = 4.
You could instead look at i in terms of looking at the cyclic group:
{1,i,-1,-i} = {i^0,i^1, i^2, i^3}
And so 1/i = i^(-1) = i^3 = -i. |
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Sponsor | | Morosoph | Sep 30, 2007 11:27am | Thank-you Fephisto.
Indeed 4^½ = ħ2, since (-2)^2 = 4.
(x^2 = y^2) <=> (x = ħy).
The squaring function x => x^2 is not one-to-one, which implies that the inverse function is not unique. Squaring, however, does give a unique result, even though it is many-to-one.
I expect that a little visualisation will help here. Consider a circle representing a set on the left of a sheet of paper. On the right draw another. Every point in the right hand set has two values from the left hand set that map to it, so pick a single point on the right, and draw two lines from the left hand set to it.
What you will (I hope) observe is that whereas the arrows from left to right are many to one (like the square function), following the arrows from right to left is one to many (like the square-root function). Another way of putting this is that the inverse function is not unique.
When we define a function such as sqrt, we often define which of the values we wish to choose; when we fail to do so, it is often implicit in order to make sense of the problem or the solution. Sometimes we use ħsqrt(x) in order to demonstrate that we mean both values, as with the quadratic formula. |
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| | ajinx999 | Oct 2, 2007 2:34am | Come on guys! I didn't expect such an opposition. I think I need to elucidate some basic concepts. Try this.
Firstly, let me start with modulus function or absolute value
function.
Modulus or absolute value of a number (non-mathematical definition) is actually the distance of the number from the origin. Modulus function is applicable to all complex numbers. I hope you know about the Complex plane. (or Argand plane or Gaussian plane, whatever you say) In the complex plane, for any complex number z = x+iy (where x is Re(z) and y is Im(z). Re(z) and Im(z) imply real and imaginary components of the complex number)
abs(z) = [(x^2)+(y^2)]^(1/2)
This can be proposed using many algorithms. (Simplest one is the Pythagorean theorem)
Now, consider a real number 'a',(Real numbers Set is a subset of Complex numbers Set), its imaginary component is zero. Therefore, the modulus function becomes
abs(a) = [(x^2)]^(1/2)
This is the square root function for real numbers.
Generalising,
sqrt(x) = (x)^(1/2) where x>0 or x=0
Square function for real numbers
f(x) = x^2
has a domain which contains real numbers.
But, square root function for real numbers has a range which contains all non-negative real numbers. So, the above two functions are not invertible.
I hope you know the graph of square root function (active in I quadrant only)
So, although (-2)(-2) = 4, (4)^(1/2) = 2 and not -2
One more reasoning is that if you consider (4)^(1/2) = (+-)2, then square root function doesn't remain a function. It becomes a one-to-many relation which contradicts the definition of a function.
Do I need to explain anything else?
One of my friend gave me a good simple reason for sqrt(x^2) = abs(x)
He said sqrt(x^2) is a 1 degree term, so it has only one root and x^2 is a 2 degree term, so it has 2 roots.
@Fephisto: My question actually is to find the anomaly in the second algorithm which gives 'i' as the answer. The answer of 1/i is definitely -i. There are multitudinous proofs for that.
According to me the anomaly probably lies in taking the radical sign common for the numerator and denominator because the numbers have different signs. But I wasn't able to justify it properly.
Also, refer to the formula about square roots of 'a' and 'b' which I wrote in my previous thread. It is applicable to complex numbers (which include real numbers).
@Morosoph: Whatever you said in your last paragraph is absolutely correct. Every positive number has 2 square roots. For instance,
2 = (+-)(4)^(1/2)
And hah, about that circle, I know you are trying to expain my how a relation is mapped. I agree with you. I also wrote correctly that square function is a many-to-one and square root function is one-to-one, previously |
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Sponsor | | Morosoph | Oct 2, 2007 6:12am | 6: You're actually hitting upon a deeper problem, which is why Riemann Surfaces were developed.
Sqrt() as you define it is not (and cannot be) the inverse of z^2. Rather, the positive solution is preferred in order to make the function "single-valued". This means that one of the values cannot be returned to.
Consider: y = sqrt[(x)^2] where x is -2. By choosing the positive root, we preclude y from being equal to x. This means that sqrt only works as a full inverse if you restrict your domain for x. This is why abs(x)=sqrt(x^2), and why sqrt isn't an inverse function for all x.
Your demonstration over the value of 1/i relies upon sqrt being an inverse function, yet the solutions to (z^2 = -1) are that z = ħi.
In the case of Pythagorus's Theorem, the sides of the triangle are known to be positive, giving the required restricted domain. Negative lengths make no sense.
Reading your post, you do grasp a lot of the concepts, but this problem that you raise calls for a return to common sense. You've performed a trick with complex numbers above, and this trick is that you've used the way in which you resolve which solution to pick to give different answers.
By analogy, if given a choice of two people, I always pick the one on the left, you can make me pick the other one by placing me behind them. If you're going to place me behind them anyway, it matters whether I pick the one on the left afterwards or beforehand. |
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| | ajinx999 | Oct 2, 2007 9:03pm | @Morosoph: You are considering square-root function as a multi-valued function. But, I think, in case of function inversion, one should not take into consideration multi-valued nature of a function. Multi-valued functions has always freaked me in my study of complex logarithm and de Moivre's formula. So, the only thing I comprehend about multi-valued functions is that their reimann surfaces are artistic.
The negative lengths are considered as a convention which solely indicates their positions with respect to axes.
What I meant is that the complex numbers ħaħib (a>0, b>0) are at equal distances from the origin and there moduli are same.
z = ħi is the correct solution for z^2 = 1. Here, you have 2 solutions because you are considering square function.
If f and g are 2 functions, for f and g to be invertible,
Dom(f) = Range(g) and Range(f) = Dom(g)
This isn't observed in case of square and square root functions. They'll be invertible only when the domain of the square function is restricted to non-negative real numbers.
Have you thought about the "taking the radical sign common for numerator and denominator" step? I mean, what is [a^(1/2)/b^(1/2)] when a>0 and b<0 or vice versa. Is there a general rule like the rule for multiplication of roots? (I have stated that in my second post.)
One more thing, How did you paste that picture in your post? |
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Sponsor | | Morosoph | Oct 3, 2007 6:56am | Answer to your last question:
<a href="http://en.wikipedia.org/wiki/Riemann_surface"><img src="http://upload.wikimedia.org/wikipedia/en/b/b5/Riemann_sqrt.jpg"></a>
As to the rest. It is because making sqrt single-valued involves a decision that the problem arises. That decision appears to be made consistently, but in practice is somewhat like an instruction to "pick the person on the left". Standing behind the pair, I pick the other individual. |
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| | Fephisto | Oct 3, 2007 8:36pm | | Sorry ajinx, looks like I misunderstood your question. |
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Mathematics → Discussion