$\sqrt{(-1)^2}$: Is it -1 or 1?

  • Thread starter footmath
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In summary, the value of $\sqrt{(-1)^2}$ is always 1 because the square root function returns the positive root. The imaginary number $i$ is related to $\sqrt{(-1)^2}$ as $i = \sqrt{(-1)^2} = \sqrt{-1}$. However, the square root function only returns the positive root, so $\sqrt{-1} = i$ instead of -i. This expression is not undefined or indeterminate and has a clearly defined value of 1.
  • #1
footmath
26
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Is it correct ?
$ \sqrt{(-1)^2} $ = $ \sqrt{(-1)}\sqrt{(-1)} $= $ i*i $=$ i^2 $ =$ -1 $
or
$ \sqrt{(-1)^2} $ = $ \sqrt{(1)} $=1
 
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  • #2
You mean this?
[itex]\text{1) }\sqrt{(-1)^2} = \sqrt{(-1)} \cdot \sqrt{(-1)} = i \cdot i = i^2 = -1[/itex]
[itex]\text{2) }\sqrt{(-1)^2} = \sqrt{1} = 1[/itex]

Note that the product property of radicals,
[itex]\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}[/itex]
holds only if both a and b are non-negative.
 

FAQ: $\sqrt{(-1)^2}$: Is it -1 or 1?

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

The value of $\sqrt{(-1)^2}$ is 1.

Why is $\sqrt{(-1)^2}$ equal to 1 instead of -1?

This is because the square root function always returns the positive root. In this case, the positive root of (-1)^2 is 1.

Can the value of $\sqrt{(-1)^2}$ be -1 in any context?

No, the value of $\sqrt{(-1)^2}$ will always be 1 regardless of the context.

How does the imaginary number $i$ relate to $\sqrt{(-1)^2}$?

The imaginary number $i$ is defined as the square root of -1. So, $i = \sqrt{(-1)^2} = \sqrt{-1}$. However, the square root function only returns the positive root, so $\sqrt{-1} = i$ instead of -i.

Is $\sqrt{(-1)^2}$ an undefined or indeterminate expression?

No, $\sqrt{(-1)^2}$ is a well-defined expression with a clearly defined value of 1.

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