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$\cfrac{-2}{(2x-1)}$?

I want to see if I'm not forgetting reciprocals. And that I'm correct and not misremembering.

It's the negative in front of the parentheses that's throwing me off.

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    Welcome to Mathematics Stack Exchange. The reciprocal of $-1$ is $-1$ – J. W. Tanner Sep 24 '20 at 20:49
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    Multiplying a number (except $0$, of course) with it's reciprocal gives $1$. You can easily check if the reciprocal is correct or not. (it is) – Devansh Kamra Sep 24 '20 at 20:51
  • Thank you. This is because I can think of this as $\cfrac {-1}{1} \cdot \cfrac{(2x-1)}{2}$ correct? so the reciprocal would be $\cfrac{1}{-1} \cdot \cfrac{2}{(2x-1)}$ which is ultimately $\cfrac{-2}{(2x-1)}$ correct? Is my way of thinking wrong? –  Sep 24 '20 at 20:53
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    @TotomaLover2312 Generally $\dfrac{a}{-b}$ is the same as $\dfrac{-a}{b}$, but the proper way is to keep the denominator positive. Your thinking is correct though. – Devansh Kamra Sep 24 '20 at 20:56

2 Answers2

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The definition of reciprocal is, "the reciprocal of a number $ u $ is a number $ v $ with the property that $ u \cdot v = 1 $." So if $ u = \frac{-(2x-1)}{2} $, then does your proposed $ v = \frac{-2}{2x-1} $ satisfy this definition? Try multiplying it out and decide for yourself if it does.

Jake Mirra
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The reciprocal of a fraction is simply found by swapping the denominator with the numerator, and also by swapping the numerator for the denominator. eg the reciprocal of
$$\frac{a}{b}=\frac{b}{a}$$

A-Level Student
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