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# Why isn't the imaginary number i = 1? (Disprove me please) watch

1. So, we are given by definition that i = sqrt(-1).
Lets make that i = (-1)^(1/2).

Next, using equivalent fractions;
i = (-1)^(2/4)

Then, using indicie laws that (a^b)^c = a^(bc);
i = ( (-1)^(2) ) ^ (1/4)

Expanding;
i = (1) ^ (1/4)
or i = fourth root of 1
Hence i = 1, -1, i and -i (since x^4 - 1 = 0 has 4 solutions; 2 real, 2 complex)

What I need disproving is that i = 1, -1 and -i, since that's clearly false.
I'm not sure which step has incorrect procedures. My maths teacher said that the step in which I square out the -1 in turn generates extra solutions that aren't true.

Just need to ensure my teacher is correct and I haven't just broken all of maths, because it still seems to me that I have. This thought occurred to me when I woke up last Friday morning and I've been trying to disprove it ever since, but to no

Thanks!
2. (Original post by NathCave)
x
Firstly, because that's how it is defined.

Secondly, when you write then you must first undo this step before evaluating the expression because otherwise you come to illogical conclusions.

And what your teacher said is also correct. When we square an equation which only has one solution, we introduce an extraneous solution, because we now allow both positive and negative values to enter the fold.

Consider the following example

Here we have x positive. Not suppose we square this equation.

Now here we have two values, although in reality x only takes on one of them. This is similar to what you did in your 'proof'.

This concept of squaring producing extraneous results is especially important when it comes to complex numbers because squaring takes you from the complex plane to the real plane.

For example, but
3. Complex exponentiation is multivalued, and if you want to make sense of your operations, you will have to define branch cuts and work there.

Also, bear in mind that your index laws that apply for real exponential don't necessarily apply to complex exponentiation.
4. (Original post by Desmos)
Secondly, when you write then you must first undo this step before evaluating the expression because otherwise you come to illogical conclusions.
I'm confused - what illogical conclusions would be made if this step is not undone? It works for all real numbers, is it just the fact that this process doesn't carry over to the complex side of maths?

(Original post by Desmos)
Consider the following example

Here we have x positive. Not suppose we square this equation.

Now here we have two values, although in reality x only takes on one of them. This is similar to what you did in your 'proof'.

This concept of squaring producing extraneous results is especially important when it comes to complex numbers because squaring takes you from the complex plane to the real plane.

For example, but
This makes it so clear as to why I was wrong, thank you! ;D
5. (Original post by NathCave)
I'm confused - what illogical conclusions would be made if this step is not undone? It works for all real numbers, is it just the fact that this process doesn't carry over to the complex side of maths?
Yes, those rules aren't necessarily valid over C.
6. Also, you might want to ask yourself why you're not similarly concerned that

, so , so ?
7. (Original post by DFranklin)
Also, you might want to ask yourself why you're not similarly concerned that

, so , so ?
Ah, I see why the logic doesn't work with all real numbers. Yeah, this thought was a random one I had just after waking up - I didn't really consider other examples. Thanks for all the feedback, much appreciated

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