Novice question: what values of $x$ satisfy $\frac{x^2}{x} \le 0 $

Question

Okay I'm embarrassed to even ask this clearly I am not going to win a Nobel prize in my life.

Question

What values of $x$ satisfy:

$$\frac{x^2}{x} \le 0 $$

My attempt

I'm very tempted to simplify the LHS and say the answer is $x \le 0$

But I have this nagging concern that the answer is actually $x<0$

If I start from the other direction, then I have

$$x \le 0$$

$$x*1 \le 0*1$$

$$x\frac{x}{x} \le 0 \frac{x}{x}$$

And the last line only holds if $x$ is not zero and therefore it changes to

$$x\frac{x}{x} < 0 \frac{x}{x}$$

$$\frac{x^2}{x} < 0 $$

And now that there is no threat of dividing by zero I can reduce it to $x<0$

Is this right?

I feel like in general when doing math I multiply top and bottom by $x$ all the time and I never really think about explicitly calling out that $x$ can't be zero.

Thanks for your help/patience.

Answer 1

If $x=0$, $\frac{x^2}{x}$ is undefined.

If $x>0$, $\frac{x^2}{x}=x >0$.

If $x<0$, $\frac{x^2}{x} = x<0$.

Hence the answer is $x<0$.

Answer 2

The top is always non-negative, so you need the bottom to be purely negative to make the whole thing less than $0$. Hence $x <0$.

Note that having it equal to $0$ is out of the question because this could only occur when the top is $0$ but the bottom isn't, which is impossible because the top is $0$ only when the bottom is.

Answer 3

For

$$\frac{x^2}{x} \le 0 $$

we need $x\neq0$ then numerator is always positive and the sign depends by the denominator then

$$\frac{x^2}{x} < 0 \iff x< 0$$

and never $\frac{x^2}{x} =0 $.