Does $\sqrt{-1\cdot-1}=1$ or $-1$?

Question

Let's define $x=\sqrt{ab}$, where $a=-1$ and $b=-1$.

Does $x=1$, as $-1\cdot-1=1\implies\sqrt{-1\cdot-1}=\sqrt{1}=1$?

Or maybe $x=-1$, as $\sqrt{a^2}=a\implies\sqrt{-1\cdot-1}=\sqrt{(-1)^2}=-1$?

I assume that at least one of the following is true:

1. The former solution is true, and the fallacy is in assuming $\sqrt{a^2}=a$ for a negative base $a$. This doesn't seem to hold, given that this is equivalent to saying that $a^{2^{\frac12}}\not=a^{2\cdot\frac12}=a$ for a negative $a$, where the value of $a$ seemingly has no bearing on the arithmetic performed with the exponents.
2. The former solution is true, and the fallacy is in assuming the positive root is meant in the latter solution; in fact, $\sqrt{-1\cdot-1}=-\sqrt{(-1)^2}=--1=1$. But this implies that, depending on how the radicand is factored, one is forced to take the positive root for one factoring and the negative root for another factoring to get the same answer, which doesn't seem right.
3. The latter solution is true, and the fallacy is in assuming the positive root is meant in the former solution; in fact, $\sqrt{-1\cdot-1}=-\sqrt1=-1$. But, in addition to the problem with the previous solution, this additionally means that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ holds even for negative $a,b$.

Which of my reasonings is incorrect? Or are all of them correct, and there's an additional solution that I'm not seeing?

Answer 1

For $a \in \mathbb{R}^+$, $\sqrt a $ is defined as the positive real $b $ such that $b^2=a $. Thus $\sqrt{x^2} = |x|$ because by definition a square root is positive. Hence in your case $\sqrt {(-1)^2} =1$.

Regarding your other points :

• As you noticed $\sqrt {ab} = \sqrt a \sqrt b $ only applies for $a,b $ positive so it cant be used here

• $\sqrt{a^2}=(a^2)^\frac 12$ is true, but $(a^x)^y = a^{xy} $ for all $x,y \in \Bbb R $ is only true for $a>0$. This is because in general $a^b = e^{b\ln a} $ which only makes sense for $a>0$. With your example, you can see why this will not work : if $(a^x)^y = a^{xy} $ then $((-1)^2)^\frac 12 = (-1)^1 \iff 1=-1$

Answer 2

Based on the definition of the square root which is positive, it should be 1. See here to know more about that.

Answer 3

The following statement is not true:

$$\sqrt{x^2} = x$$

The square root function returns the non-negative square root, or the principal square root, so it’s actually

$$\sqrt{x^2} = \vert x\vert = \begin{cases} x; \quad x\geq 0 \\ -x; \quad x < 0 \end{cases}$$

So, in your case, $x = -1 < 0$, so $\sqrt{(-1)^2} = \vert -1\vert = 1$.