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Which is the correct value of $\sqrt{(-1)^2}$ ?

  1. $\sqrt{(-1)^2} = \sqrt{(-1)\cdot(-1)} = \sqrt{1} = 1$

  2. $\sqrt{(-1)^2} = ((-1)^2)^{1/2} = (-1)^{2\cdot1/2} = (-1)^1 = -1$

I was surprised to realize that I hadn't even understood this basic math. Are there any exponentiation rules to undertaand to get the correct value?

pdh0710
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2 Answers2

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The first is correct.

In the second, the part where you have $((-1)^2)^{1/2}=(-1)^{2\cdot1/2}$ is not valid. You are citing a property of exponents that is not guaranteed to be correct when the base ($-1$) is negative.

2'5 9'2
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  • Why it "is not guaranteed to be correct when the base is negative" ? – pdh0710 Jul 07 '24 at 03:44
  • Exponentiation is only unique if the exponent is an integer or if the base is a positive real number. By convention, $\sqrt \cdot$ returns a complex number with argument (i.e., angle) in the range $(-\pi/2, \pi/2]$, while $(\cdot)^{1/2}$ is simply not well defined. – Joseph Camacho Jul 07 '24 at 04:14
  • Given some space to write it all down, and given complete definitions for $b^r$ in general, and assuming $b>0$, I could prove that $(b^r)^s=b^{r\cdot s}$. It simply can't be done if $b$ is negative, because it's just not true. $((-1)^2)^{1/2}$ is $1^{1/2}$, which is $1$. And $(-1)^{2\cdot1/2}$ is $(-1)^1$, which is $-1$. They just are not the same. – 2'5 9'2 Jul 07 '24 at 04:15
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It is more appropriate to see it as a precedence issue. Square root and exponential operations are of the same precedence, the rule is then to apply by order, that is determined by the form of the expression. $\sqrt{(-1)^{2}}$ is to do the square (exponential) first, then the square root, so the first answer is correct. $\left(\sqrt{-1}\right)^{2}$ will do the square root first, so it ends with $\mathbf{i}^{2}=-1$. Cancelling these two operations is invalid because depending on the operand, the operation of square root (to produce the principle value) and that of square may not be exactly inverse to each other

Shine
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