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I am solving a problem from a GRE guide, and stuck at the following problm.

Given that $-1 < a < 0 < \left| a \right| < b < 1$ which of the following quantity is greater?

$$\left(\frac{a^2 \sqrt{b}}{\sqrt{a}}\right)^2$$ or $$\frac{a b^5}{\left(\sqrt{b}\right)^4}$$

I don't know how to simplify first expression. Is it $\frac{a^4 b}{a}$ or $\frac{a^4 b}{-a}$

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  • $\begingroup$ $(\frac{a^2 \sqrt{b}}{\sqrt{a}})^2=\frac{a^4b}{a}=a^3b$? where you keep the sign in a without writing it out? $\endgroup$
    – user614287
    Apr 19, 2020 at 14:12
  • $\begingroup$ I would think $\sqrt{a}$ in the denominator is not even defined since $a<0$ $\endgroup$
    – gt6989b
    Apr 19, 2020 at 14:12
  • $\begingroup$ The problem is taken from Manhattan 5lb. book. In the solution provided in the book, it is assumed that $\left(\sqrt{a}\right)^2$ is $a$, which to me makes no sense. $\endgroup$
    – Ali Baig
    Apr 19, 2020 at 14:33
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    $\begingroup$ I agree with you. We have other questions showing that $\sqrt{a^2}=|a|$, but trying to take the square root first when $a$ is negative is not allowed in the reals. The conditions on $a,b$ indicate we are working in the reals. Demand your money back. $\endgroup$
    – Ross Millikan
    Apr 19, 2020 at 15:02
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    $\begingroup$ If you are in the complex numbers there are two square roots of $a$, but both of them square to $a$. One is $-1$ times the other and squaring that gives $1$. For example, $\sqrt {2i}=\pm(1+i)$. Squaring either of those gives $2i$ $\endgroup$
    – Ross Millikan
    Apr 19, 2020 at 19:01

2 Answers 2

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As written the first is undefined so they cannot be compared. They intend the first to be positive, so the denominator in your expressions should be $-a$. It does not change the answer if you use $a$ because then both are negative but the first is less so.

I believe they intend you to compare $|a^3|b$ with $ab^3$

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  • $\begingroup$ The problem is taken from Manhattan 5lb. book. In the solution provided in the book, it is assumed that $\left(\sqrt{a}\right)^2$ is $a$. However, I didn't notice that assuming $\left(\sqrt{a}\right)^2 = a$ or $\left(\sqrt{a}\right)^2 = -a$ does not affect final answer. Thank you :) $\endgroup$
    – Ali Baig
    Apr 19, 2020 at 14:35
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If you are well versed with basics of complex number, you can simply arrive at the following conclusion:

For some $a<0$, let $b=-a$. Thus $b>0$

Now $\sqrt{a} = i \sqrt{b}$ where $i = \sqrt{-1}$. Now, $$ \left(\sqrt{a}\right)^2 = \left(i\sqrt{b}\right)^2 = i^2 \left(\sqrt{b}\right)^2 = (-1) \cdot b = -b = a $$

Thus, the square of square root of a negative number is the number itself where as the square root of square of a negative number is its positive counterpart.

As an example: $$ \left(\sqrt{-2}\right)^2 = -2 \quad \text{and} \quad \sqrt{(-2)^2} = 2 $$

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  • $\begingroup$ Welcome ot Math.SE! If you are working in $\mathbb{C}$, it makes no sense to compare the elements though... you cannot do algebra in $\mathbb{C}$ and comparisons in $\mathbb{R}$ $\endgroup$
    – gt6989b
    Apr 19, 2020 at 14:48
  • $\begingroup$ At no step did I compare any complex number with any other number. a and b are real numbers to be clear with my solution. I made no other comparisons. Please comment back if I didn't get you right. $\endgroup$
    – SAGALPREET SINGH
    Apr 19, 2020 at 14:55
  • $\begingroup$ To rely on complex interpretation of square roots, you have to say $a,b$ are really complex numbers (with zero imaginary part), otherwise, if you are working in the real numbers, you cannot interpret the roots that way... but once you let $a,b \in \mathbb{C}$, even though you are ending up with numbers having zero imaginary part, comparison is not well defined... $\endgroup$
    – gt6989b
    Apr 19, 2020 at 15:05
  • $\begingroup$ Oh, now I get your point in such case you may take a to be positive real number and takes it complex counterpart as a + 0i. And I think it should not be illegal to say that a = a + i0. Inequalities make no sense in complex numbers but equalities do even across different sets. For example 1 in naturals is equal to 1 in whole numbers, right ? $\endgroup$
    – SAGALPREET SINGH
    Apr 19, 2020 at 16:35

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