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I don't have any experience working with radicals, but I'm working on a function that requires products of nth roots to be positive or negative, depending on the number of negative factors.

I've done some initial research, and reviews these Stack questions: Square roots — positive and negative and The Product Rule of Square Roots with Negative Numbers but I couldn't find the information I was seeking (or am not fully understanding the answers.)

Are the following expressions true? If not, how can I produce the those results?

$\sqrt[2]{1*-1} = -1$

$\sqrt[3]{1*1*-1} = -1$

$\sqrt[3]{1*-1*-1} = 1$

[update] This is what the function does:

$\sqrt[n]{\overline{\Delta_1}*\overline{\Delta_2} *...*\overline{\Delta_n}} \text{ }*\text{ } \frac{\overline{\Delta_1}*\overline{\Delta_2} *...*\overline{\Delta_n}}{\Delta_1*\Delta_2*...*\Delta_n}$

such that if there are an odd number of negative factors, the product is negative, otherwise positive.

  • Is there a more compact way to express this?

also, any tips on notation are appreciated.

mnp
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2 Answers2

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We have that

  • $\sqrt[2]{1\times(-1)} = \sqrt{-1} \neq -1\quad \color{red}\checkmark$ indeed $-1\times -1=1$
  • $\sqrt[3]{1\times1\times(-1)} = \sqrt[3]{-1} = -1\quad \color{green}\checkmark$ indeed $-1\times -1\times -1=-1$
  • $\sqrt[3]{1\times(-1)\times(-1)} = \sqrt[3]{1}=1\quad \color{green}\checkmark$ indeed $1\times 1\times 1=1$

As a general rule

  • for $n\in \mathbb{N}$ even and $a\ge 0$ we have

$$\sqrt[n] a=b \iff b\ge 0 \quad b^n=a$$

  • for $n\in \mathbb{N}$ odd and we have

$$\sqrt[n] a=b \iff b^n=a$$

user
  • 162,563
1

I am not sure if this is the method you want.

We have Euler's formula $$e^{i\theta} = i\sin\theta+\cos\theta$$ We can take $n^{th}$ root of both sides to obtain $$\sqrt[n]{e^{i\theta}}=\sqrt[n]{i\sin\theta+\cos\theta}\tag{1}$$

It seems like you are only asking for the cases for $1$ and $-1$, so let's do the following.

It is clear that $\sqrt[n]{1}$ is $1$ or $-1$ regardless of whether $n$ is odd or even. The problem is how to figure out $\sqrt[n]{-1}$. For $n$ is odd, $\sqrt[n]{-1}=-1$.

For $n$ is even, let's suppose $\theta = \pi$. Then, from $(1)$ we have $$\sqrt[n]{e^{i\pi}}= \sqrt[n]{-1}$$ $$e^{i\frac{\pi}{n}} = \sqrt[n]{-1}$$ $$\sqrt[n]{-1}= i\sin\frac{\pi}{n}+\cos\frac{\pi}{n}$$ I apologize if this is not what you are looking for.

Larry
  • 5,140