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Consider this expression $$\sqrt{1-x^2}$$ We could factor the negative one out like this $$\sqrt{-1(x^2-1)}$$ Now we could use take the two factors and separate them $$\sqrt{-1}\sqrt{(x^2-1)}$$

But the root of negative one is not real, I feel really dumb asking this and I don't know anyone who can help me with this at the current moment. Thank you.

egreg
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Shayan
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    $\sqrt{ab}=\sqrt{a}\sqrt{b}$ is valid only if $a,b\geq 0$ – kingW3 Jun 20 '17 at 14:53
  • $\sqrt{ab} = \sqrt a \sqrt b$ only when $a, b \ge 0$. – Gregory Jun 20 '17 at 14:53
  • @kingW3 beat me to it – Gregory Jun 20 '17 at 14:54
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    Possible duplicate of Why $\sqrt{-1 \times {-1}} \neq \sqrt{-1}^2$? – kingW3 Jun 20 '17 at 14:55
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    In order to have a real value for $\sqrt{1-x^2}$, you need $-1\le x\le 1$, so $x^2-1\le0$. So the last expression is the product of two imaginary numbers (for $x=\pm1$ it is zero). On the other hand, you should never use $\sqrt{-1}$. Avoid it with great care: in the complex numbers there is $i$ with the property that $i^2=-1$; but also $-i$ satisfies the property and square roots in the complex numbers are not single valued, whereas square roots of nonnegative real numbers are well defined. – egreg Jun 20 '17 at 14:58
  • Actually, $\sqrt{z_1z_2}=\sqrt{z_1}\sqrt{z_2}$ is true in terms of set equivalence. That is that any value of $\sqrt{z_1z_2}$ can be expressed as the product of some value of $\sqrt{z_1}$ and some value of $\sqrt{z_2}$. And conversely, the product of any value of $\sqrt{z_1}$ and any value of $\sqrt{z_2}$ can be expressed by some value of $\sqrt{z_1z_2}$. – Mark Viola Jun 20 '17 at 15:34

2 Answers2

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The property :

$$\sqrt{ab} = \sqrt a \sqrt b$$

is valid, only when $a,b\geq 0$.

Thus, you cannot handle the expression $\sqrt{1-x^2}$ this way.

Take also in account that this expression is valid in $\mathbb R$ only when $1-x^2 \geq 0$.

Rebellos
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You are actually making two mistakes without knowing.

If the expression $\sqrt{1-x^2}$ is defined, then certainly $1-x^2\ge0$. Now when you write

$\sqrt{1-x^2}=\sqrt{(-1)(x^2-1)}$, you have the product of two negatives. And of course, factoring as a product of two square roots is not allowed.

If you want to switch to the complex numbers, then you could write

$$\sqrt{-1}\sqrt{x^2-1}=i\sqrt{x^2-1},$$ where the square root is that of a negative.

Then

$$i\sqrt{x^2-1}=i\cdot i\sqrt{1-x^2}=-\sqrt{1-x^2}\ !$$

The apparent paradox is due to the fact that the rule "the square root of a product is the product of the square roots" doesn't hold in the complex.