Is there a difference between $$y=\frac{\sqrt{1-x}}{\sqrt{1+x}}$$ and $$y=\sqrt{\frac{1-x}{1+x}}$$ If there is a difference, why when I give the square for both equations, they will be equal.
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1Not if all quantities are positive, i.e. $-1<x\le 1$. – Mario Carneiro Feb 12 '15 at 09:44
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@Reery. You get it equal after squaring because $1\neq-1$, but $1^2=(-1)^2$. – Feb 12 '15 at 10:26
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Yes. There are differences. In the region where both of $x-1$ and $x+1$ are negative, the upper one will be wrong and lower one is correct. This region is $(-\infty, -1).$
For complex numbers there are no difference because any number is acceptable under the sqrt (positive or negative)
Mojee KD
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You proved that there is no difference on interval $(-1,\infty)$, but what about complex numbers? For which complex numbers $x$ there is no difference? – Feb 12 '15 at 10:15
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In general for real numbers, $ sqrt(\dfrac{a}{b}) = \dfrac{sqrt(|a|)}{sqrt( |b|)} $ for $ (a \wedge b) \epsilon \mathbb{R}$
None
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if this were the case, then you would have $sqrt(-1)$ which is not a member of the set of real numbers. if $a = -2, b = -3$ you would then have $sqrt(\dfrac{2}{3})$ but, that is not equal to $\dfrac{sqrt(-2)}{sqrt(-3)}$ – None Mar 23 '15 at 09:11
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