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$1 = \sqrt{1} = \sqrt{(-1)^2} = \sqrt{-1}\sqrt{-1} = i\cdot i = -1$. I know the mistake is here $\sqrt{(-1)^2} = \sqrt{-1}\sqrt{-1}$ because everything else seems right to me, but I don't understand why?

I guess this is the correct way : $\sqrt{(-1)^2} = |\sqrt{-1}| |\sqrt{-1}|$ ?

Nicola Fattoruso
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  • $\sqrt{(-1)^2} = \sqrt{-1}\sqrt{-1}$ ? –  Nov 10 '18 at 18:26
  • $\sqrt{x^2} \ne \sqrt{x}\sqrt{x}$. – fleablood Nov 10 '18 at 18:26
  • Complex numbers have two square roots each. – Angina Seng Nov 10 '18 at 18:28
  • The think I don't quite get is why people think a proof when they know why something is wrong somehow think a proof with a mistake discredits them. "Is a poptart a sandwich" "No" "why not? A sandwich is filling between two pieces of bread; fruit filling if filling and pastry dough is basically bread" "Gee, you're righ-- a poptart is a sandwich". No, it isnt! The definition wasn't explicit enough. – fleablood Nov 10 '18 at 18:32
  • Think to note is that for $a,b$ there are two $k, -k$ so that $(\pm k)^2 = a$ and two $\pm j$ so that $(\pm j)^2 = b$. So if we have $w^2 =a; v^2=b \implies w^2v^2=(wv)^2 = ab$ and we have $(\pm k)^2(\pm j)^2 = ab$ it doesn't for that all $(\pm k)(\pm j)=\pm (kj) = kj = -kj$ are all equal to each other. [And that IS the assumption we make when we say $\sqrt{ab} = \sqrt{a}\sqrt{b}$. If we restrict ourselves to ONLY possitive values it is mute. But if we allow negative values in but not adjust to allow negative outcomes, we are contradicting ourselves.) – fleablood Nov 10 '18 at 18:56
  • For crying out loud let's just implement the machine learning that prevents this question from being duplicated anymore. We already have like a million datapoints so let's get it rolling. – rschwieb Nov 11 '18 at 03:53

2 Answers2

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In general it is not true that $\sqrt{ab}=\sqrt a \sqrt b$. It is true if a and b are positive reals, but square root behaves strangely for negative numbers (and more generally, complex numbers). This is the error.

The more difficult part of the issue, if you're interested, is figuring out what square root means for complex numbers. And what people do, in the field, is call it a multi valued function. Basically there are two numbers that, when squared, give you your original number. And trying to do algebra objects like that isn't straightforward or predictable, for the purpose of this conversation.

Richard Rast
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  • Alright, didn't remember a and b (in your example) must to be positive reals. Thanks a lot. – Nicola Fattoruso Nov 10 '18 at 18:31
  • "What pepole do in the field is call it a multi valued function" Some people prefer that, not all. There are plenty of reasons to consider only the principal square root since then it is still a function. – JMoravitz Nov 10 '18 at 18:47
  • "square root behaves strangely" I wouldn't say this is strange. There are always two $k$ so that $k^2 = w$ and if we restrict $\sqrt w$ to mean just one of them we are not going to be able to sneak in $(-k)^2$ to replace $k^2$ and expect $\sqrt$ to magically switch to the other. – fleablood Nov 10 '18 at 18:50
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The fact that $x^2 = y^2$ does not imply $x =y$ (as $x = -5$ and $y = 5$ show means that $\sqrt{k}$ means "the number $v$ so that $v^2 = k$" is not an actual well defined statement (because there isn't necessarily one number) and that it is not necessarily $\sqrt{x^2} =x$ nor that $\sqrt{x}^2 = x$.

And it doesn't follow that $\sqrt{ab} =\sqrt a \sqrt b$.

ANd just like $(-1)^2 = (1)^2 = 1$ we also have two numbers $(i)^2 = (-i)^2 = -1$.

If we go through the proof critically:

$1 = \sqrt 1$ sure that's notation. We define $\sqrt{x}$ to be one of the two values so that $v^2 = x$ and in the case of $x > 0$ we arbitrarily choose the positive one.

$\sqrt {1} = \sqrt {(-1)*(-1)}$. Sure. That's just an existential fact.

$\sqrt{(-1)}\sqrt{(-1)}$. Um. No. It is true that there is a $w$ so that $w^2 = -1$ and that therefore $(w*w)^2 = w^2w^2$ but is doesn't follow that $w$ is the one we designate as $\sqrt{-1}$ nor does it for that $w*w$ is the one we desiganta as $\sqrt{(-1)(-1)}$.

fleablood
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