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Here are the propositions:

$$i=\sqrt{-1}$$ $$i^2=-1$$ $$(i)(i)=-1$$ $$\sqrt{-1}\sqrt{-1}=-1$$ $$\sqrt{(-1)(-1)}=-1$$ $$\sqrt{1}=-1$$

There's an error in the propositions above. I think it's in the fifth line where $\sqrt{(-1)(-1)}=-1$.

Are $\sqrt{ab}=\sqrt{a}\sqrt{b}$ and $a=(\sqrt{a})(\sqrt{a})$ different from each other?

Jyrki Lahtonen
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DarkArrow29
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3 Answers3

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1) The rule "$\sqrt{a}\sqrt{b}= \sqrt{ab}$" is not true for a and b non-real.

2) As Elio Joseph said, defining "i" as "$\sqrt{-1}$ is not rigorous. Better is to define the complex numbers to b pairs of real numbers with addition defined by (a, b)+ (c, d)= (a+ c, b+ d) and multiplication by (a, b)x(c, d)= (ac- bd, ad+ bc). We think of the real numbers as "embedded" in the complex numbers by a<==> (a, 0). The "imaginary numbers" are the set of pairs (0, b). And, of course, "i" is (0, 1): (0, 1)x(0, 1)= (0x0- 1x1, 0x1+ 1x0)= (-1, 0) which is what we associate with "-1".

user247327
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Both 5 and 6 are wrong, assuming the usual choice of positive square root. In order to make $$f(x)=\sqrt{x} $$ a bonafide function, mathematicians have defined it to output the positive square root. By this logic, both 5 and 6 are wrong, as they select the negative square root.

operatorerror
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The error is that you can't say that $$i=\sqrt{-1}$$

because the function $\sqrt.$ is not defined on $\mathbb C$ properly.

You can't assigned a unique value for $\sqrt{z}$ if $z\in \mathbb C$ like you do for real positive numbers.

Therefore, writing $\sqrt{-1}$ instead of $i$ may and will lead to wrong results.

E. Joseph
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    You have to be a bit careful with stating this. Often $\sqrt{\cdot}$ is defined to be single-valued so that it is indeed a function. It is the difference between $\sqrt{\cdot}$ and $\cdot^{\frac{1}{2}}$. – Cameron Williams Jun 21 '16 at 17:50
  • $i$ can be defined as one of the roots of $\sqrt{-1}$ – operatorerror Jun 21 '16 at 17:50