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i = √(-1)

Therefore:
i² = i·i

Therefore:
i² = √(-1) · √(-1)

By some rule of radicals
i² = √[ (-1) · (-1)]

i² = √[ 1 ]

i² = ±1

Where is the flaw?

2007-10-20 01:33:21 · 5 answers · asked by Anonymous in Science & Mathematics Mathematics

5 answers

You asked "why cant i² = 1 ?"

But you asked exclusively +1... not -1... not ±1... you just asked +1.

By your own proof, i² = ±1... not exclusively +1

So... the real question is why isnt "i² = ±1 ?"

That is because your proof resulted in an extraneous solution.

I think the extraneous solution occurred because you applied the radical after you applied the exponent. Had you applied the radical first, then squared it, the solution would prove just -1.

But that is also the definition expression of the imaginary unit.
i² = -1 and i = √-1
These expressions define the value for i... no proof really necessary.

If i = √-1 then
i² = (√-1)² = -1

Going this route ended up being more restrictive to the solution set.

Perhaps others will have more to add...

2007-10-20 01:39:51 · answer #1 · answered by Anonymous · 0 0

The flaw is in the step:

"by some rule of radicals, √(-1)√(-1) = √((-1)*(-1))"

The problem is, there is no rule of radicals that says you can do this in general. Think about it -- why should we believe that √x√y = √(xy)? The only thing we know is that:

(√x√y)² = √x√y√x√y = √x√x√y√y = (√x)²(√y)² = xy = (√(xy))²

But you know that a²=b² does not imply that a=b. It could be that case that, for instance, a=-b. So the only thing we can deduce about products of roots in general is that √x√y = ±√(xy).

Now, in the special case where x and y are both positive and real, we also know that the principal branch of √(xy), √x, and √y will be positive real numbers, so √x√y and √(xy) are both positive real numbers. Now, if √x√y = -√(xy), then at least one of them would have to be a negative number, a contradiction, so in this special case, we know that the sign must be positive, and so √x√y = √(xy).

The problem is, that -1 is NOT positive and real, so the reasoning applied to the case of positive real numbers simply is inapplicable, and the only thing we know is that √(-1)√(-1) = ±√((-1)*(-1)). And in this case, a manual calculation reveals that the negative sign holds, i.e. √(-1)√(-1) = -√((-1)*(-1)).

One more point: The square root function is a function, and thus has only ONE value, not 2. Whenever the radical sign is used, it returns only the principal square root of the number (which for a complex number re^(iθ) is √r e^(iθ/2), where -π < θ ≤ π), not both of them. So √1 = 1, not ±1. Nonetheless, it is interesting to note that if we do interpret √x as yielding both square roots of x, that your derivation is correct -- from the assumption that i is one of the square roots of -1, it follows that i² is one of the square roots of 1, so i² = ±1 (in particular, i²=-1). Under that interpretation, your error would only have been at the very end, where you interpreted i²=±1. That symbol means i² = 1 OR i² = -1 (which is true), but from the title of your question, you seem to have interpreted it to mean i² = 1 AND i² = -1, which is false.

2007-10-20 08:31:44 · answer #2 · answered by Pascal 7 · 0 0

i or

.2
1 = 1

2007-10-20 01:39:28 · answer #3 · answered by Anonymous · 0 1

There is a problem here due to the fact that the function, sqrt(.), is not well defined as it is not one to one.
Firstly note that sqrt(-1) = +/- i
So when you have written sqrt(-1).sqrt(-1) which is ambiguous as we do not know which is either root, we could have
sqrt(-1)sqrt(-1) = -i . i = 1
So dur to the fact the function sqrt() is is multivaluable the statement i^2= sqrt(-1)sqrt(-1) is not totally correct

2007-10-20 02:32:46 · answer #4 · answered by Anonymous · 0 0

it's not equal to 1. it's equal to 1 and -1. just like you proved. it's gotta be both though.

2007-10-20 01:36:44 · answer #5 · answered by toadstoolcouch 2 · 0 1

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