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I am sure you are aware of the dillema of 1=-1 when we assume √-1=i. To refresh your memory it goes like this:

1= √1= √((-1)(-1))= √-1√-1= i*i= -1.

We must realize that saying √-1=i is only half true, therefore not true. √-1= +/-i. This is because i×i=-1 and so does -i×-i. Therefore we need to change our above statement to:

1= √1= √((-1)(-1))= √-1√-1= +/-i*+/-i= (+/-)-1=(+/-)1.

But then, wait, now 1 is equal to +/-1. It is equal to both itself and its opposite all of the sudden. Doesnt this make absolutely no sense? I think √-x should equal the set of no numbers. What do all of the educated mathematicians say?

jacobj
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The "laws of exponents" need modification when dealing with complex numbers. In particular, it is not true that $\sqrt{xy} = \sqrt{x} \sqrt{y}$.

Robert Israel
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  • Why not? Doesnt √x√y mean that x gets multiplied by the y before we take the square root of it? – jacobj Aug 02 '16 at 14:45
  • No $\sqrt{x} \sqrt{y}$ means the product of the square root of $x$ and the square root of $y$. $\sqrt{xy}$ means the square root of the product of $x$ and $y$. – Robert Israel Aug 02 '16 at 16:56
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Let's examine your chain of equalities:

$1 =\sqrt 1 = \sqrt{(-1)(-1)} = i*i = -1$.

At this junction, it is important to make an interpretation of what $1 = \sqrt{1}$ means. What does it mean? For this statement to be well defined, the square root of any value must be unique. Do you see why? For otherwise, I could very well say the following: $1 = \sqrt 1 = -1$ !

Hence, the square root when used in this expression is a well defined function, that is to say that the square root of a number is unique. However, this is only possible if the square root is restricted to the positive numbers (which is done, precisely to avoid the situation you have brought up!). Which renders $\sqrt{-1}$ meaningless.

The moral of the story is that the square root may have the same symbol, but is a different demon when you change domains, as you have done, from the natural numbers to the whole set of integers.

That should answer your question, but as an aside I think I should add the precise meaning.

Let $\mathbb{R}^+$ be the positive real line. Define $\sqrt{x} : \mathbb{R}^+ \rightarrow \mathbb{R}^+$ by the rule that $(\sqrt{x})^2=x$. Well-definition is easy to check.

Now, what do we do when we have negative numbers? Worse, what do we do when we have negative powers? Read up on how the logarithm and exponential are implemented on the complex plane, because these functions are the key to every question you will have about powers of numbers.

Sarvesh Ravichandran Iyer
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  • We only define √1 to equal 1? That means the √1 does not equal -1. And in that case -1 times -1 does not equal 1. We both know this is untrue. – jacobj Aug 02 '16 at 14:47
  • It is logically sound to say that √1 equals 1 and -1 together. This means that 1 does not equal √1 and -1 does not equal √1, but +/-1 equals √1. If we pick and choose we destroy previous axioms in math, but then convienently still use them as if we havent. – jacobj Aug 02 '16 at 14:50
  • @jacobj Yes, $-1$ is not the positive square root of $1$. The square function is the inverse of the positive square root function, and not the general square root function. Do you see why? Because the general square root function is not a function at all! If I call $f$ the general square root, then can I say that $f(1) = 1$ and $f(1) = -1$? This would mean $f$ is not a function! And only functions can have defined inverses! – Sarvesh Ravichandran Iyer Aug 04 '16 at 03:18
  • It so happens that the positive square root function is one-one and onto, and hence invertible. The function so inverted is the square function, $g(y) = y^2$. The square function is the inverse of the square root function only on the positive numbers. Of course, we can extend, or continue the square function to the negative numbers, but we notassume that the continuation is also invertible. And that, is the fallacy of this argument. – Sarvesh Ravichandran Iyer Aug 04 '16 at 03:20