Fractions: Reciprocal
If $$x$$ is the reciprocal of $$y$$, then which of the following may be true?
Incorrect.
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In this answer choice, if you replace $$\frac{1}{y}$$ with $$x$$, you get
>$$\displaystyle\frac{x}{\frac{1}{y}}=\frac{1}{2}$$
>$$\displaystyle \frac{x}{x}=\frac{1}{2}$$
This is obviously wrong, as $$\frac{x}{x}$$ equals 1 for all nonzero $$x$$.
Incorrect.
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In this answer choice, if you replace $$\frac{1}{x}$$ with $$y$$, you get
>$$\displaystyle\frac{\frac{1}{x}}{y}=2$$
>$$\displaystyle \frac{y}{y}=2$$
This is obviously wrong, as $$\frac{y}{y}$$ equals 1 for all nonzero $$y$$.
Incorrect.
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In this answer choice, if you replace $$\frac{1}{y}$$ with $$x$$, you get
>$$\displaystyle\frac{\frac{1}{y}}{x}=\frac{1}{2} $$
>$$\displaystyle \frac{x}{x}=\frac{1}{2}$$
This is obviously wrong, as $$\frac{x}{x}$$ equals 1 for all nonzero $$x$$.
You may be tempted to __Plug In__ values for $$x$$ and $$y$$. The problem, however, is that this is a “_may_ be true" question; even if a certain answer choice isn't true for a specific Plug In, you can't eliminate it since it still "may be true" for some other plug-in.
The only way Plugging In will work is if you pick a specific set of values that makes one of the answer choices true. Though that may be the right idea in a question with inequalities, as you can see, it is unlikely that you "guess" $$x = \sqrt{2}$$ as a plug in.
Incorrect.
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In this answer choice, if you replace $$\frac{1}{x}$$ with $$y$$, you get
>$$\displaystyle\frac{y}{\frac{1}{x}}=2$$
>$$\displaystyle \frac{y}{y}=2$$
This is obviously wrong, as $$\frac{y}{y}$$ equals 1 for all nonzero $$y$$.
Correct.
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In all answer choices other than D, the left sides of the equations become 1 when $$\frac{1}{x}$$ is replaced with $$y$$ or $$\frac{1}{y}$$ is replaced with $$x$$. That makes all of the answer choices other than D necessarily false. At this point, you should select D and move on.
On the other hand, you may be interested to see that D can, in fact, be true. It is easier to find the solutions if $$\frac{1}{x}$$ is replaced with $$y$$ or $$\frac{1}{y}$$ is replaced with $$x$$.
> $$\displaystyle \frac{\frac{1}{x}}{\frac{1}{y}}$$ $$=\frac{1}{2}$$
> $$\displaystyle \frac{y}{\frac{1}{y}}$$ $$=\frac{1}{2}$$
> $$y \div \frac{1}{y} = \frac{1}{2}$$
> $$y \times \frac{y}{1} = \frac{1}{2}$$
> $$y^2=\frac{1}{2}$$
Now square rooting both sides gives possible values for $$y$$.
> $$y = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}}$$
That makes $$x = \pm \sqrt{2}$$. So there are two possible sets of values that work for D. Of course, the solutions are definitely not needed in order to answer this question.
$$\displaystyle\frac{x}{\frac{1}{y}}$$ $$=\frac{1}{2}$$
$$\displaystyle\frac{\frac{1}{x}}{y}=2$$
$$\displaystyle\frac{\frac{1}{y}}{x}$$ $$=\frac{1}{2} $$
$$\displaystyle \frac{\frac{1}{x}}{\frac{1}{y}}$$ $$=\frac{1}{2}$$
$$\displaystyle{\frac{y}{\frac{1}{x}}}=2$$
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