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I was reading about this known fallacy $$ -1 = i^2 = i \cdot i = \sqrt{-1} \cdot \sqrt{-1} = \sqrt{(-1)(-1)} = \sqrt{1} = 1 $$ and according to Wikipedia "The fallacy is that the rule $\sqrt{xy} = \sqrt{x}\sqrt{y} $ is generally valid only if both x and y are positive"

So my question is, how come we can say that $\sqrt{-3} = \sqrt{3}i$ ?. Aren't we applying the same mistake as the fallacy? Like $\sqrt{-3} = \sqrt{(-1)(3)} = \sqrt{-1}\sqrt{3} = \sqrt{3}i$ cannot be since -1 is negative.

Thanks for reading.

Alex T
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    "if both $x$ and $y$ are positive." If $x=-1$ and $y=3$ then if is false that both are positive. – JRN Jun 06 '15 at 01:14
  • Note the key word generally. The identity may be valid in other cases (as in this one), but it gets more complicated when you allow for nonreal or even nonpositive $x$ and $y$. – Zach Effman Jun 06 '15 at 01:24
  • check this link https://www.mathsisfun.com/numbers/imaginary-numbers.html – alkabary Jun 06 '15 at 01:24

1 Answers1

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First: the fact that the identity $\sqrt a\sqrt b=\sqrt{ab}$ is only valid for nonnegative reals $a,b{}$ does not mean that it is always false for other numbers.

But let's see the proof of the identity to see where it's essencial that $a$ and $b$ are nonnegative:

$$(\sqrt a\sqrt b)^2=\left(\sqrt a\right)^2\left(\sqrt b\right)^2=ab$$ Therefore, $\sqrt{ab}=\sqrt a\sqrt b$.

Now, let's do this for $a=b=-1$:

$$(\sqrt {-1}\sqrt {-1})^2=\left(\sqrt {-1}\right)^2\left(\sqrt {-1}\right)^2=(-1)(-1)$$ Therefore, $\sqrt{(-1)(-1)}=\sqrt {-1}\sqrt {-1}$.

The problem arises in the conclusion. The definition of $\sqrt{\:}$ says: $$\sqrt{x}=y\iff y^2=x\text{ and }y\ge 0$$

The last condition is essential. Many books for high school student say that $\sqrt 4$ has "two solutions" but this is totally wrong. What has two solutions is the equation $x^2=4$, but $\sqrt 4$ is $2$ and not, ever, $-2$.

In the fallacy we are saying in essence that $\sqrt{(-1)(-1)}=\sqrt {-1}\sqrt {-1}$, but this is wrong because $\sqrt {-1}\sqrt {-1}<0$.

Teoc
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ajotatxe
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  • I'm not native English speaker: should that word "ever" be "never"? – ajotatxe Jun 06 '15 at 01:30
  • I think "ever" is fine as you put it. I think that the key word of my question is that Wikipedia put the word "only" as in that identity should be used "only" when x and y are positives. – Alex T Jun 06 '15 at 01:37
  • @VladimirLenin Where did they say that? –  Jun 06 '15 at 02:28
  • @ajotatxe Another issue that is important here is that while $\sqrt{4}=2$, the $\sqrt{-4} = \pm 2i$ depending on which side of the Riemann surface the point $-4$ lies. For example, if the plane is cut along the negative real axis, and we are evaluating $\sqrt{-4}$ "infinitesimally" above the branch cut, then we have $\sqrt{-4}=2i$. – Mark Viola Jun 06 '15 at 04:57