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Can i write $1/4a^{-2}$ as $4a^2$ ? Or is the right answer to do it like:

$$1/4a^{-2} = 1/4 \cdot 1/a^{-2} = 1/4 \cdot a^2 = a^2/4$$

In the problem there is no parenthesis around $4a$ but assuming there were parenthesis like $1/(4a)^{-2}$ would it be correct to write $4a^2$ ?

Michael Hardy
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  • To the $\LaTeX$ifiers: Please do not do stack the fractions in this post. If you stack the fraction, the question will get lost. – Ross Millikan Jul 05 '16 at 01:22
  • @RossMillikan : I fear that calling MathJax $\text{“}\LaTeX\text{''}$ could lead to unpleasant consequences. Someone could master MathJax and think they've mastered $\LaTeX$, and then encounter actual $\LaTeX$. They'd get a considerable shock. $\qquad$ – Michael Hardy Jul 05 '16 at 01:34
  • My guess, like Ross, is that $(1/4) a^{-2}$ is intended. The other possibility would be $1/(4a^{-2})$. Definitely not $1/(4a)^{-2}$. Because it could be ambiguous, it should be written with parentheses. – GEdgar Jul 05 '16 at 01:37

2 Answers2

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Officially, you should parse $1/4a^{-2}$ as $(1/4) \cdot a^{-2}$. The exponential is evaluated first, then the multiplies and divides from left to right. Certainly the $4$ belongs in the denominator. The question is whether the $a^{-2}$ belongs in the denominator. Given that the exponent is negative, I would think it likely that the author meant that. On the other hand, if you see $1/4a^2$ it is likely that the author meant $1/(4a^2)$ instead of $a^2/4$ Parentheses are your friend.

Ross Millikan
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If you have $\dfrac{1}{(4a)^{-2}}$ then it would be $(4a)^2 = 16a^2$. Let me know if you have further questions..

DeepSea
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