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I=1/2?


Didymus

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I've always been drawn to impossibilities, which is why I so enjoy the concept of I as the square root of -1... But I just thought to question what sqrt(I) is.

 

Now, I've always loved math, but I never had the patience for the tedious homework necessary to make it into higher math classes... But when I searched for the answer, I was dissatisfied. According to the internets sqrt(I) is sqrt+/-(1/2).

 

My problem is the insinuation that I=+/-(1/2). No?

 

The thing about 1 is that if x>1, x^2>x. And if x<1, x^2<x. So "I" must be right on this magic Li'l circle where x^2=x, who h can only be 1.... If on some wonky imaginary number line. I could see if sqrt(I)= -1.... That makes perfect sense because -1 is clearly the cubic root of -1.

 

Sqrt 1/2 is not the cubic root of -1. This is blasphamy and madness.

 

So, am I making an elementary mistake, or is what I read on the internet an evil communist plot designed to torment me?

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I've always been drawn to impossibilities, which is why I so enjoy the concept of I as the square root of -1... But I just thought to question what sqrt(I) is.

 

Now, I've always loved math, but I never had the patience for the tedious homework necessary to make it into higher math classes... But when I searched for the answer, I was dissatisfied. According to the internets sqrt(I) is sqrt+/-(1/2).

 

My problem is the insinuation that I=+/-(1/2). No?

 

The thing about 1 is that if x>1, x^2>x. And if x<1, x^2<x. So "I" must be right on this magic Li'l circle where x^2=x, who h can only be 1.... If on some wonky imaginary number line. I could see if sqrt(I)= -1.... That makes perfect sense because -1 is clearly the cubic root of -1.

 

Sqrt 1/2 is not the cubic root of -1. This is blasphamy and madness.

 

So, am I making an elementary mistake, or is what I read on the internet an evil communist plot designed to torment me?

The sqrt(I) is equal to 0.707106781 + 0.707106781i. I don't see where the internet got the other answer from.

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The sqrt(I) is equal to 0.707106781 + 0.707106781i. I don't see where the internet got the other answer from.

I've seen the logic for that, but it seems like faulty reasoning like the proof that .999=1. The processes add up... But the logic doesn't hold in that it implies that the number you stated would be the cubic root of -1....

 

Do you disagree that -1^3=-1?

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I've seen the logic for that, but it seems like faulty reasoning like the proof that .999=1. The processes add up... But the logic doesn't hold in that it implies that the number you stated would be the cubic root of -1....

 

Do you disagree that -1^3=-1?

The way to understand i is geometrically. It's a counterclockwise rotation of the plane through an angle of pi/2 radians, or 90 degrees. As you can see, if you do it once and then do it a second time, you end up at the complex number -1, or the point on the plane (-1,0). So i^2 = -1 is a geometric triviality.

 

Now, what is sqrt(i)? Well, what rotation can you do two times in a row to get to i? Answer: A rotation of pi/4 or 45 degrees. And what complex number corresponds to a rotation of pi/4? It's:

 

sqrt(2) / 2 + i * sqrt(2) / 2

 

You can multiply it out to see that it works; and you can observe that if I rotate 45 degrees and then again 45 degrees, I have in effect rotated 90 degrees.

Edited by HalfWit
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More dyslexia.... Sqrt of sqrt would be 4th rather than 3rd root. Cubic root of -1 is definitely -1, but the 4th root of -1 is the square root of I... Not third. Knew I was off. Still if -1 is it's own 3rd, 5th, 7th, etc. Root, it would stand to reason that the even roots would repeat, thus sqrt I would be I.

 

Anywho... So if sqrt I is the formula listed by halfwit.... What would it's sqrt be? Since the odd roots of -1 repeat, wouldn't the evens? Or do the even roots diverge while the odd roots repeat.... And if so, how could that be possible?

 

I'm not doubting the geometry.... But this is a situation where the geometry seems counterintuitive (unless I'm just missing something obvious.)

Edited by Didymus
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More dyslexia.... Sqrt of sqrt would be 4th rather than 3rd root. Cubic root of -1 is definitely -1, but the 4th root of -1 is the square root of I... Not third. Knew I was off. Still if -1 is it's own 3rd, 5th, 7th, etc. Root, it would stand to reason that the even roots would repeat, thus sqrt I would be I.

 

Anywho... So if sqrt I is the formula listed by halfwit.... What would it's sqrt be? Since the odd roots of -1 repeat, wouldn't the evens? Or do the even roots diverge while the odd roots repeat.... And if so, how could that be possible?

 

I'm not doubting the geometry.... But this is a situation where the geometry seems counterintuitive (unless I'm just missing something obvious.)

It's very intuitive once you get it. It's hard to explain without diagrams. But draw your unit circle. i is an angle of pi/2 radians. Its square root is an angle of pi/4 radians. because

 

**** Multiplying complex numbers is just adding their angles!! ****

 

So when I have pi/4 and I square it, that's the same as rotating pi/4 and then rotating pi/4 again ... which would leave you at pi/2.

 

So if you want the square root of pi/4, it must be pi/8 for the same reason. You just keep bisecting its angle on your unit circle.

 

Now I must mention that I'm playing loose with terminology. What I really mean by pi/4 is "The complex number that lies on the unit circle and makes an angle of pi/4 between it, the origin, and the positive x-axis."

 

If you just draw this I believe you will be enlightened about this whole business. It's just **adding angles**.

Edited by HalfWit
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You can simply use De Moivre's , which is what HalfWit is more or less doing, to prove that [math]\sqrt{i}= \frac{1+i}{\sqrt{2}}[/math], which is what Unity+ has given, just not in surd form. Or of course you can just square the expression I have given.

Edited by ajb
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You can simply use De Moivre's , which is what HalfWit is more or less doing, to prove that [math]\sqrt{i}= \frac{1+i}{\sqrt{2}}[/math], which is what Unity+ has given, just not in surd form. Or of course you can just square the expression I have given.

Yes, which the following would occur.

 

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

 

[math]i= (\frac{1 + i + i - 1}{2})[/math]

 

[math]i= (\frac{i + i}{2})[/math]

 

[math]i= (\frac{2i}{2})[/math]

 

 

[math]i= (\frac{i}{1})[/math]

 

[math]i= i[/math]

Edited by Unity+
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Unless I overlooked something, all the posts seem to say there is one square root, that is, the one with the + sign.

Right, so it depends exactly what one is asking for. I gave the "positive root" (which is a little ambiguous here). The other root is just given by multiplying through by minus one.

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(1 + i)^2 = 1 + 2i - 1.


Noob question here... I get all the other steps, and on this one you're just hitting each piece with a square root..... How are you getting from 1+I to (1+i+i-1)?

My comment above was in reply to this post.

Edited by mathematic
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