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If

1) $\sqrt{-a} = \sqrt{a} \times \sqrt{-1}$

2) $\sqrt{a} = \sqrt{a}\times \sqrt{1}$

Then, according to me $\sqrt{a}$ could also be written as $\sqrt{-a}\times\sqrt{-1}$.

Am I correct?

egreg
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An Engineering aspirant
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    This is in spirit a duplicate of https://math.stackexchange.com/questions/438/why-sqrt-1-times-1-neq-sqrt-12. – Travis Willse Jun 13 '18 at 17:25
  • you cannot because the operation $\sqrt{a\cdot b} = \sqrt a +\sqrt b$ is defined only for $a,b\gt 0$. If not youll open up the gate to a lot of messy things . – The Integrator Jun 13 '18 at 17:26
  • If you multiply a square root of $a$ by a square root of $b$, you get a square root of $ab$, but not necessarily the one you first thought of. – Angina Seng Jun 13 '18 at 17:29
  • Be careful of your definitions. The usual use of the √ function takes a positive real number to a positive real number. You can extend this to work in the complex plane, but which of the two square roots are you going to use? If you are going to use both, then you can write that, but note you aren't working with functions at that point. – Kaynex Jun 13 '18 at 17:30
  • If you admit this, you have $-1=\sqrt{-1}\sqrt{-1}=\sqrt{1}=1$, not really good for doing maths. – egreg Jun 13 '18 at 17:31
  • Okay I understand now that the square root function is only defined for all positive real numbers but I saw something like this in my textbook: √-2 = √2 x √-1 = √2 x i. If the function isn't defined for negative real number then why can we consider 'iota' or √-1. – An Engineering aspirant Jun 13 '18 at 17:34
  • Maybe you should add to your question a quoted paragraph or two from your textbook so we can see exactly what it said, not just “something like this.” Also let us know the author’s name and the title and edition of the book (in case someone has access to a copy). – David K Jun 13 '18 at 18:29

4 Answers4

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You must define what you mean by the square root.

It is usually understood to mean a function $\sqrt{\cdot}\,:[0,+\infty)\longrightarrow[0,+\infty)$, meaning it's only defined on nonnegative real numbers, and its result is always a nonnegative real number. In other words, we always take the positive square root: $\sqrt{a^2}=|a|$ $($which is not necessarily $a)$.
This function is multiplicative: $\sqrt{a\cdot b} = \sqrt{a} \cdot \sqrt{b}$. However, the domain of the function must be respected, and trying to apply this relation to numbers outside the domain does not make sense (because the function isn't even defined at those points!).

In other contexts (for instance, complex analysis), one generally understands that square roots (and other $n$-th roots) are multivalued functions: any nonzero complex number has $n$ distinct $n$-th roots. In this context, we don't say that '$2$ is the square root of $4$', but rather that '$2$ is a square root of $4$', the other being $-2$.

Fimpellizzeri
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$\pm i$ are square roots of $-1,$ and $\pm i\sqrt 35$ are square roots of $-35.$ \begin{align} (i) \times (i\sqrt{35}) & = -\sqrt{35} \\ (-i) \times (i\sqrt{35}) & = +\sqrt{35} \\ (i) \times (-i\sqrt{35}) & = +\sqrt{35} \\ (-i) \times (-i\sqrt{35}) & = -\sqrt{35} \end{align} So some care about what is meant by $\text{“ }\sqrt{-35\,} \text{ ''}$ is in order.

Michael Hardy
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$$\sqrt{-a}\times\sqrt{-1}$$ is not well-defined, because there are more than one possible answers to that.

As you know a complex number has two square roots, and that is the problem with expressions such as $\sqrt {-5} \sqrt {-1} = \sqrt 5$ because it could as well be $- \sqrt 5$

Mohammad Riazi-Kermani
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By allowing negative and positive values to be solutions to $\sqrt{x}$, we have

\begin{align} \sqrt{36} & = \pm 6 \\ -6 & = -1 \cdot 6 \\ & = (\sqrt{-1})^2 \cdot (\sqrt{6})^2 \\ & = \sqrt{-1} \cdot \sqrt{-1} \cdot \sqrt{6}\cdot \sqrt{6} \\ & = \sqrt{-1} \cdot \sqrt{6} \cdot \sqrt{-1}\cdot \sqrt{6} \\ & = \sqrt{-6} \cdot \sqrt{-6} \\ & = \sqrt{-6\cdot-6} \\ & = \sqrt{36} \end{align}

However, the square function $f(x) = x^2$ is not one to one, and thus does not have a well-defined inverse. Which is essentially how we can "prove" that $x \not= \sqrt{x^2}$. By presuming unique factorization in a field $(\mathbb{C})$ that is not a unique factorization domain, we get a result of limited utility.

Michael Hardy
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David Diaz
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    You can write $3\times 5$ or $3\cdot5;$ there's no need to use an asterisk for that. Using an asterisk for ordinary multiplication is a workaround for situations in which one is limited to the characters on the keyboard. – Michael Hardy Jun 13 '18 at 17:55