Is (i*i*i*i)^0.25=(i*i)?

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  • #1
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Main Question or Discussion Point

http://www.google.com/search?hl=com&q=(i*i*i*i)^0.3928".
(i*i*i*i)^x=1
Where x is any number
So:
(i*i*i*i)=1
(i*i*i*i)^0.25=1^0.25
(i*i)=1
But in default (i*i)=-1
So:
1=-1

Therefor (i*i*i*i)^0.25≠(i*i)
Which I find quite funny, weird, and interesting, since you can usually just put ^x numbers down when using root, like:
(3^3)=27
(3^3)^0.5=(3^2)=9

Any comments?
 
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Answers and Replies

  • #2
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http://www.google.com/search?hl=com&q=(i*i*i*i)^0.3928".
(i*i*i*i)^x=1
Where x is any number
So:
(i*i*i*i)=1
(i*i*i*i)^0.25=1^0.25
(i*i)=1
But in default (i*i)=-1
So:
1=-1

Therefor (i*i*i*i)^0.25≠(i*i)
Which I find quite funny, weird, and interesting, since you can usually just put ^x numbers down when using root, like:
(3^3)=27
(3^3)^0.5=(3^2)=9

Any comments?

First be careful, 1^0.25 has 4 solutions!

1, -1, i, and -i. if you raise either one of them to the 4th power you end up with 1.

Secondly, I have no idea how you get from the second to the third line(in the quote below), it just does not follow.

"So:
(i*i*i*i)=1
(i*i*i*i)^0.25=1^0.25
(i*i)=1"
 
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  • #3
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(i*i*i*i)=1 //Default
(i*i*i*i)^0.25=1^0.25 //Multiplied by itself by 0.25 on both side of equation
(i*i*i*i)^0.25=(i*i) //By default, this would be the answer with real numbers
i^0.25=1 //Most relevant
(i*i)=1 //Done

Whole point was with real numbers this works:
(3^3)^0.5=9
(3^3)^0.5=(3^2) //It's multiplied with itself one time less: 3*3*3 to 3*3
(3)^2=3*3
3*3=9

So this function works with real numbers:
(3*3*3)^0.5=(3*3)
But it doesn't work with complex/imaginary numbers:
(i*i*i*i)^0.5≠(i*i*i)

I'm just a teenager that stopped paying attention to school due to great problems at home and similar after 3rd-4th grade, and now that I got alot of spare time, I'm just philosophizing.
This isn't some revolutionary piece of mathematical code, just a short philosophy of the difference between real numbers and complex/imaginary numbers.

Edit: Just figured I'm wrong, we were taught that (numbers) are always calculated first, but I guess ^ is an exception, because (3*3)^0.5=(sqrt 3 * sqrt 3), not (3*3=sqrt 9).
 
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  • #4
HallsofIvy
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In general, the "law" axbx= (ab)x is true for real values of a and b, not complex values.
 
  • #5
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In general, the "law" axbx= (ab)x is true for real values of a and b, not complex values.
Yes, that was my philosophy.
Thank you for confirming.
 
  • #6
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(i*i*i*i)^0.25=(i*i) //By default, this would be the answer with real numbers
I'm not sure that I understand where that is coming from...

Whole point was with real numbers this works:
(3^3)^0.5=9
(3^3)^0.5=(3^2) //It's multiplied with itself one time less: 3*3*3 to 3*3
(3)^2=3*3
3*3=9
(3*3*3)^0.5 = 3^(3/2) = (3^0.5)*(3^0.5)*(3^0.5) = 3*(3^0.5) = 5.19615242, does it not?
 
  • #7
Redbelly98
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  • #8
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(i*i*i*i)=1 //Default
(i*i*i*i)^0.25=1^0.25 //Multiplied by itself by 0.25 on both side of equation
(i*i*i*i)^0.25=(i*i) //By default, this would be the answer with real numbers
i^0.25=1 //Most relevant
(i*i)=1 //Done
That wouldn't even work if i was a real number. What you most likely mean is,
(i*i*i*i)^0.5=(i*i)

Whole point was with real numbers this works:
(3^3)^0.5=9
(3^3)^0.5=(3^2) //It's multiplied with itself one time less: 3*3*3 to 3*3
(3)^2=3*3
3*3=9
No, it doesn't. Try doing 27^0.5 on your calculator. I get something like 5.2. In general:
(a^b)^c = a^(bc)
For real numbers, so your example should be.
(3^4)^0.5=9

You somehow seem to think that ^0.5 will remove one factor, but that isn't the case. You should review the basic rules for real exponentiation before you get started on complex.
 
  • #9
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http://www.google.com/search?hl=com&q=(i*i*i*i)^0.3928".
(i*i*i*i)=1
(i*i*i*i)^0.25=1^0.25
(i*i)=1
If they were REAL numbers, you'd have used the one of the exponetial laws wrong, simple math. So, exteneded the sum should look something like this:

(i^4) = 1
(i^4)^0.25=1^0.25
4*0.25=1
i=1^0.25

and as Diffy said, 1^0.25 has a few possibilities. So it would be one of those, most likely being i.

Which I find quite funny, weird, and interesting, since you can usually just put ^x numbers down when using root, like:
(3^3)=27
(3^3)^0.5=(3^2)=9
Once again, when there's a power outside the brackets, you multiply it by the one in the brackets. So what you're really doing is this:

(3^3)=27
(3^3)^0.5=27^0.5
3^1.5=5.19615.. etc.

The proper way of doing what you wanted to do with the threes would be this:

(3^3)=27
(3^3)^2/3=27^2/3
3^2=9
 
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  • #10
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You somehow seem to think that ^0.5 will remove one factor, but that isn't the case. You should review the basic rules for real exponentiation before you get started on complex.
I'm a philosopher, not a mathematician.
Quite sad that is, since I presumably should technically have some potential, due to the fact that I got an IQ of above 130 (excluding social intelligence), and been considering an education within physics, but I've figured mathematics is just pissing me off, and everyone seem to be of greater understanding relative to time, than me, so it's not worth it.

PS: yes, I thought it would remove one factor, since I temporary relayed on the fact that ^0.5 removes one factor when the exponent is equal to 2 (2 -> 1).
Kaimyn is right, I'm just.. not very clear in my thinking now a'days..
Too much on the computer, prioritizing only the easiest solutions/shortcuts, and haven't been doing mathematics for years, also only knowing the very very basics of it, due to lack of ability to concentrate @ school.

Thanks for replies anyways.
 
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