Verifying I2=1: Right or Wrong?

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The discussion centers on the verification of the equation I²=1, where I is defined as the square root of -1. Participants clarify that the manipulation of exponents within radicals is incorrect, leading to the erroneous conclusion that -1 equals 1. They emphasize the importance of proper definitions in complex numbers, noting that complex exponents are multivalued and can lead to paradoxes. The conversation also highlights the necessity of defining "i" correctly and using ordered pairs to represent complex numbers. Overall, the conclusion is that the original assertion is incorrect due to fundamental misunderstandings in complex number theory.
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please tell me the solution i did is right or wrong and why?

we know,
I=√-1
I2=√-1*√-1 =√(-1)2 = √1 = 1
 
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Wrong, you cannot move the exponent inside the radical. That you arrived at -1=1 is a sign of a mistake.
 
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lurflurf said:
Wron, you cannot move the exponent inside the radical. That you arrived at -1=1 is a sign of a mistake.

A mistake? For sure. But a very interesting mistake. "Paradoxes" of these kind indicate that something very interesting is going on with complex exponents. In particular, they indicate that complex exponents (such as roots) are multivalued. Once you take this approach, all paradoxes vanish :-p
 
ranjitnepal said:
please tell me the solution i did is right or wrong and why?

we know,
I=√-1
I2=√-1*√-1 =√(-1)2 = √1 = 1

Would it be enough just to accept the definition so that we use both of these:

i*i=-1 and i=sqrt(-1)This is what happens with (-i).
(-i)*(-i)=(-1)(-1)*i*i=(1)*(-1)=-1
What went wrong there?
Apparantly i*i=-1 and (-i)(-i)=-1.

Still seem good. i*i still -1 and (-1)(-1)=1
 
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micromass said:
"Paradoxes" of these kind indicate that something very interesting is going on with complex exponents. In particular, they indicate that complex exponents (such as roots) are multivalued.

Yeah, similarly, by naively taking the natural logarithm of both sides of equation e^{0}=e^{2\pi i} one gets the result 0=2\pi i.
 
ranjitnepal said:
please tell me the solution i did is right or wrong and why?

we know,
I=√-1
Well, there are your first two errors! To begin with, it is "i", not "I"!
(I suspect your editor automatically changed your "i" to "I". So did mine- I had to "fool" it by typing "ia", then going back and deleting the "a"!)
More importantly, it is a mistake to write "a= \sqrt{-1} because there is no such number before we define "i" and you can't define a new number to be something that doesn't exist to begin with! Defining "i" to be "the number whose square is -1" is better because -1, at least, does exist before we define the complex numbers. But has the difficulty that once we start working with the complex numbers we find that every number, except 0, has two square roots and this does not tell us which of the two roots of -1 "i" is.

Better is to define the complex numbers to be the set of orderd pairs of real numbers, (a, b), with addition defined by (a, b)+ (c, d)= (a+ b, c+ d) and multiplication by (a, b)*(c, d)= (ac- bc, ad+ bc). We can then identify the real numbers with pairs of the form (x, 0) and "i" with (0, 1).

I2=√-1*√-1 =√(-1)2 = √1 = 1
With the "ordered pairs" definition, above, i^2= (0, 1)(0, 1)= (0(0)-(1)(1), 0(1)+ 0(1))= (-1, 0)which we have already identified with the real number -1.
 

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