Inequality $\frac{1}{x}>\frac{1}{y}$ when $y>0$ has the same solutions as..

Question

Inequality $\frac{1}{x}>\frac{1}{y}$ when $y>0$ has the same solutions as..

a) $x>y$

b) It should say $y>x$

c) $y-x>1$

d) none of three above

Correct answer: d)

My procedure: If x=0,01 and y=0,1 then a) & c) must be false. However, I can not manage to find a counter "example" to b). It seems to me that if the inequality which is posed in the question above should be true then $y>0$ implies that $x>0$. Otherwise the inequality sign does not seem hold. If both variables must be positive then does it not imply that $y>x$?

Answer 1

The most obvious reason why (b) is wrong is that $y=1$ and $x=0$ is a solution to $y>x$, but is not a solution to $1/x>1/y$. Hence they do not have the same set of solutions.

However, other examples include anything with $y>0$ and $x<0$. That is a solution to $y>x$, but not $1/x>1/y$ with $y>0$.

Explained in another way. It is true that $1/x>1/y$ combined with $y>0$ implies that $x>0$. However, $y>x$ and $y>0$ does not imply that $x>0$, therefore the solution sets cannot be equivalent!

Answer 2

Note that for $y>0$

$$\frac1x>\frac1y\iff \frac1x-\frac1y>0\iff \frac{y-x}{xy}>0$$

that is (for $x,y\neq 0$)

• $x>0$ and $x<y$

or

• $x<0$ and $x>y$

Answer 3

$$\frac{1}{x} > \frac{1}{y} > 0$$ along with $y>0$ implies that $ 0<x<y$

For example $ y=5$ and $x=4, 1/4>1/5$ , and $5>4>0$