what do you mean by iota?

What is (i)^(1/2)?

 

 

 

What is the position of iota in counting dimension?

Where is the position of iota

 

 

What is (i)^(1/2)?

 

[math]\sqrt{i} = \frac{1}{\sqrt{2}} \left( 1 \pm i \right) [/math]

 

I have no idea what your other question is asking because it is using some very non-standard terminology & pictures.

Edited January 12, 201312 yr by Bignose

what do you mean by iota?

 

I think you're referring to the imaginary unit: [math]i^2=-1[/math]. If so, then it's rarely represented by the Greek iota. Rather, the modern equivalents of the lowercase Roman " i " are used.

 

What is (i)^(1/2)?

 

[math]i^{\frac{1}{2}}=\sqrt{i}[/math]

 

It's better to ask yourself, what squared equals [math]i[/math]? Or [math](a+bi)^2=i[/math], under the assumption the solution is neither purely real or imaginary.

 

This may help get you to derive the solution.

 

What is the position of iota in counting dimension?

 

What is a counting dimension? What's with the image? Is it supposed to resemble the real projective line?

Edited January 12, 201312 yr by Amaton

I guess it's the "number line" which is sometimes (rather confusingly) represented as a circle with infinite radius and where + and - infinity coincide (on the dubious basis that their reciprocals are the same).

 

If that's the case then the answer is fairly simple.

Imaginary numbers are not on that line, they are on the imaginary number line which crosses the real number line at zero.

 

http://en.wikipedia.org/wiki/Complex_plane

http://mathworld.wolfram.com/ArgandDiagram.html

I guess it's the "number line" which is sometimes (rather confusingly) represented as a circle with infinite radius and where + and - infinity coincide (on the dubious basis that their reciprocals are the same).

 

If that's the case then the answer is fairly simple.

Imaginary numbers are not on that line, they are on the imaginary number line which crosses the real number line at zero.

 

http://en.wikipedia.org/wiki/Complex_plane

http://mathworld.wolfram.com/ArgandDiagram.html

so it means that we can travel in time or 4d by driving car on iota speed.

The question is meaningless.

How do you travel at an imaginary speed?

I can say, without anyone being able to show that I'm wrong, that if you travel at an imaginary speed you turn into a rubber chicken.

so it means that we can travel in time or 4d by driving car on iota speed.

What would give you that idea?

 

Just like I wrote in another thread here yesterday, 'i' is part of the language of mathematics, and as part of that language, you have to use it according to correct grammar. It is meaningless to describe something real like a car driving at an imaginary speed, any more than describing a car driving at 'purple' speed or 'lightbulb' speed. You have to use the terms correctly to have something that makes sense.

 

'i' does have circumstances where it has a physical meaning. Describing the impedance in an electrical circuit, or describing a potential flow of a fluid. But, these phenomena have their own language terminology as well. You can't describe a car or a car's speed as 'invisid' like you can a potential flow. Words and terms only make sense when used correctly, and i is no exception to that.

_

Edited January 15, 201312 yr by NaxAlpha

Math is used to solve universal questions and the iota is a part of math.

Let the speed of object is 81 km/s.

Apply a method so that the speed of object becomes square root of its speed.

The speed of object equals to {9,-9}.

Apply again that method.

After this the speed of object becomes {3,-3,3i }

I want to find this method...

 

 

 

Extra Question:

Is it true that -infinity = +infinity?

Edited January 15, 201312 yr by Time And Space

time and speed - please stop writing in various colours it is visually unappealing and distracts from the discussion. and it is i - the english letter - that we use to represent the square root of minus one and not iota - the greek letter - which has no dot above it and is not generally used in mathematics.

Math is used to solve universal questions and the iota is a part of math.

Let the speed of object is 81 km/s.

Apply a method so that the speed of object becomes square root of its speed.

The speed of object equals to {9,-9}.

Apply again that method.

After this the speed of object becomes {3,-3,3i }

I want to find this method...

 

 

I think it is funny you forgot -3i in your answers.

 

Furthermore, when you take the square root of something like 81 km/s, you also have to take the square root of the units. So, "9" isn't your answer, but 9 (km/s)^(1/2). I sure hope that has some physical meaning in another step, but it sure isn't a velocity anymore. There are some physical units that have non-integer units in them, but not many, and really they only arise in special circumstances.

 

You can take the square root of any number you want... that doesn't mean that the answer will have any physical significance. Any really, just because you can take the square root of a negative velocity and get something imaginary in return... that still doesn't mean that is has any kind of meaning. You might as well ask, "why can't I take the square root of a banana and eat the imaginary part?" It just doesn't have any physical meaning.

 

 

Extra Question:

Is it true that -infinity = +infinity?

 

No.

 

Equivalence of infinities is not straightforward. But, [math]-\infty \ne \infty[/math]

..."why can't I take the square root of a banana and eat the imaginary part?"...

 

Haha!

 

No.

 

Equivalence of infinities is not straightforward. But, [math]-\infty \ne \infty[/math]

1/0 = +Infinity

1/0 = 1/(-0) = -1/0 = -(1/0) = -Infinity

so Infinity = -Infinity

1/0 = +Infinity

1/0 = 1/(-0) = -1/0 = -(1/0) = -Infinity

so Infinity = -Infinity

 

Not so - you play around with infinities and dividing by zero like that at your peril.

1/0 = +Infinity

1/0 = 1/(-0) = -1/0 = -(1/0) = -Infinity

so Infinity = -Infinity

 

It is true that

 

[latex]\displaystyle \lim_{x \to 0} \frac{1}{x} = \infty[/latex]

 

however

 

[latex]\frac{1}{0} \ne \infty[/latex]

 

If 1/0 = n, where n is anything but undefined, then there would have to exist a variable n which when multiplied by zero equals 1. There is no such number (of any type), thus the correct way to represent 1/0 is by simply stating that it is undefined.

Sorry guys, when I said that the basis for believing it was the same was "dubious" I hadn't realised that people might not understand.

Perhaps I should have said "on the dubious ridiculous basis that their reciprocals are the same".

[limit(x->+0)1/x=+Inf]

[limit(x->-0)1/x=-Inf]

That's why +Inf is unequal to -Inf

 

[latex]\displaystyle \lim_{x \to +0} \frac{1}{x} = +\infty[/latex]

 

[latex]\displaystyle \lim_{x \to -0} \frac{1}{x} = -\infty[/latex]

 

That's why +Inf is unequal to -Inf

"+0" and "-0" are exactly the same quantity - thus rendering your statement as incorrect.

Edited January 30, 201312 yr by x(x-y)

[limit(x->+0)1/x=+Inf]

[limit(x->-0)1/x=-Inf]

That's why +Inf is unequal to -Inf

 

I believe x(x-y) implicitly pointed this out, and here I am also. It may clear things up to denote your limits towards 0⁺ and 0^-.Now that I think of it, one-sided limits denoted incorrectly as coefficients can terribly miskew and confuse a problem.