Solve $\frac{2}{x}<3$ over the reals.

Question

I was using the method of intervals to solve this Question. I seemed to miss $0$ on the number line while using the method of intervals. Why is $0$ included here in the method of intervals?

Answer 1

Because $\frac2x$ is not continuous at $x=0$. Note that just to the left of $0$, $\frac2x$ is large in magnitude but negative; while just to the right of $0$, it is large and positive.

Answer 2

$$\frac{2}{x}-3<0$$ or $$\frac{2-3x}{x}<0,$$ which gives the answer: $x>\frac{2}{3}$ or $x<0$.

Answer 3

For an inequality $f(x) <0$ or $f(x)>0$, the method of intervals finds all the points where the graph of $f(x)$ crosses the $x$-axis (some people call these "split points".) Once you find them, they cut the $x$-axis into intervals. Each interval is either part of the solution set or not.

The big point is this: If one interval had values of $x$ that made $f(x) >0$ and other values of $x$ that made $f(x)<0$ then somehow the graph would have to cross the $x$-axis in that interval. But you know that it doesn't, because you have all the split points.

Finally, there are two ways for a graph to get to the other side of the $x$-axis. One is to intersect it (at such points $f(x) =0.$ The other way is to jump over it, (at such points $f(x)$ is discontinuous. So the collection of split points is the set of all points where either $f(x)=0$ or $f(x)$ is discontinuous.

For your problem we have $$\frac{2}{x}-3 <0.$$

So we solve $$\frac{2}{x}-3 = 0$$

and get $x=2/3$. Then we look for discontinuities and find $x=0$ is a problem.

The set of split points is $\{0, 2/3\}$ which cut the real line into the three intervals $(-\infty, 0)$, $(0,2/3)$, and $(2/3,\infty).$ I think you know the rest of the story.

Answer 4

First suppose $x>0$. Then $$ \frac{2}{x}<3\iff3x>2\iff x>2/3. $$ If $x>0$ and $x>2/3$, then $x>2/3$. Now suppose that $x<0$. Now when we multiply by $x$ the inequality sign flips. Hence $$ \frac{2}{x}<3\iff3x<2\iff x<2/3. $$ If $x<0$ and $x<2/3$, then $x<0$. Thus the solution set is $(-\infty, 0)\cup(2/3,\infty)$.