Wormholes - physically possible?

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Jonatan
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#21
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#21
(Original post by polthegael)
Squishy's sig is a topic on many threads!

I thought that any root is taken as positive unless it is stated or implied otherwise.

Using Squishy's logic I could claim (cube root 1)(cube root 1)(cube root 1) is 1 or either of the complex conjugate roots (depends which roots I decided to multiply together). This is stupid. It could be argued that the function isn't one-one so the whole question is meaningless. I disagree. To me the answer is 1 by convention. Likewise the answer to Squishy's (root minus one) squared is minus 1.
If you take a look at his signature youd see that he doesnt follow that convention. Thats what makes the derivation possible.
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Jonatan
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#22
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(Original post by Squishy)
I didn't mean for it.

√x is defined as the positive square root only. When x < 0, then you take the "positive" complex root (e.g. +i instead of -i).

There are all sorts of proofs that have a mistake in them but not everyone spots it. I mentioned to someone else the other day:

Let x < 1
ln(x) < ln(1)
ln(x) < 0

Divide through by ln(x):

1 < 0
Its an inequality and not an equation. Thus you have to be careful about the direction of the < sign when you use an operator on both sides. if ln(x) is to be defined then x must be greater than 0. However, if x is between 1 and 0 then ln(x) must be negative. Since ln x is negative you are supposed to turn the < into a > when you divide with it on both sides. Thus the propper result is 1 > 0 which is perfectly true.
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Squishy
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#23
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#23
(Original post by polthegael)
i is imaginary, not complex :rolleyes:
Imaginary and real numbers are both subsets of the complex numbers...so technically it's both.
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Poc ar buile
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#24
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#24
(Original post by Squishy)
Imaginary and real numbers are both subsets of the complex numbers...so technically it's both.
i is not a real number!

To say that i is complex is strictly true - 0+1i and all that, but why not just accept it is imaginary..?!

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Squishy
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#25
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(Original post by polthegael)
i is not a real number!

To say that i is complex is strictly true - 0+1i and all that, but why not just accept it is imaginary..?!

No, I meant that i is imaginary and complex. Zero is the only number that is complex, real and imaginary.
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elpaw
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#26
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(Original post by polthegael)
To say that i is complex is strictly true - 0+1i and all that, but why not just accept it is imaginary..?!
why be so pedantic? "i is complex" is a perfectly acceptable statement
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username9816
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#27
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(Original post by elpaw)
why be so pedantic? "i is complex" is a perfectly acceptable statement
If it's good enough for Jelly, it's good enough for us. I agree though, "i is complex" is valid.

Infact, in the module I do for Further Maths, the chapter is called "Complex numbers" and not "Imaginary numbers".
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Poc ar buile
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#28
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#28
(Original post by Invisible)
Infact, in the module I do for Further Maths, the chapter is called "Complex numbers" and not "Imaginary numbers".
Possibly because it contains numbers which are complex and not just imaginary ones!

If you look at the earlier post, however, it appeared Squishy was saying that i is both real and complex or something.

I stated it's not real and is complex without a real component, i.e. imaginary.

Are you saying this is wrong? :confused:
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Nylex
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#29
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(Original post by polthegael)
Possibly because it contains numbers which are complex and not just imaginary ones!

If you look at the earlier post, however, it appeared Squishy was saying that i is both real and complex or something.

I stated it's not real and is complex without a real component, i.e. imaginary.

Are you saying this is wrong? :confused:
The set of complex numbers is all those in the form x + iy. An imaginary number is a complex number with x = 0.
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jpowell
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#30
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I think the point being made here is that reals and imaginarys are subsets of the complex numbers. Mathematicians usually consider any number with an imaginary component to be complex, even if it can also be classed as purely imaginary.
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Jonatan
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#31
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(Original post by polthegael)
Possibly because it contains numbers which are complex and not just imaginary ones!

If you look at the earlier post, however, it appeared Squishy was saying that i is both real and complex or something.

I stated it's not real and is complex without a real component, i.e. imaginary.

Are you saying this is wrong? :confused:
Do you mean real as in "It is real" or real as in "It belongs to the class real".

Some physical concepts are not explainable without imaginary numbers so I would say imaginary numbers are quite real , although they are not real numbers (well the are real numbers but they are not REAL numbers).
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jpowell
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#32
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I think the best way to put it is as follows. Complex numbers exist. They are not however real numbers .
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