If $x/y = 0$, does $x$ have to equal $0$? I'm a college student in Calculus I. This question came up on a quiz. The professor's answer implied that if $x/y = 0$, $x$ does not have to equal zero. I am curious if this is true. Thanks!
 A: For a well defined expression we need $y\neq 0$ and therefore
$$\frac x y =0 \iff y\cdot \frac x y =y\cdot 0 \iff x=0$$
Are you sure your professor answer's implies that $x\neq 0$?
In case we are dealing with limits, what is true is that
$$\lim_{y\to \infty} \frac x y =0$$
that is $\frac x y\to 0$ which is different concept with respect to $\frac x y= 0$.
A: Fundamentally,
$$
\frac{x}{y}
$$
means 'what number, when multiplied by $y$, gives you $x$?'. If $x/y=0$, then this implies that
$$
y\times0=x
$$
Anything multiplied by $0$ is $0$. Hence, $x=0$.
Note that $y$ can't equal infinity because infinity is not a number. The following expression does not make sense
$$
\frac{x}{\infty}=0
$$
However, we can say that
$$
\lim_{n \to \infty}\frac{x}{n}=0
$$
This expression means 'as $n$ gets larger and larger, $x/n$ gets smaller and smaller'. It does not mean that if $n=\infty$, then $x/n=0$.
A: If $x,y$ are numbers then automatically $\frac xy=0\implies x=0y=0$
Now being a bit informal it is also true that $\frac x\infty=0$ regardless the value of number $x$, but $\infty$ is not a number $y$.
Maybe what your professor meant was an informal way of looking at $x,y$ as potential limits, understand do we need $f(t)\to 0$ if we have $\frac{f(t)}{g(t)}\to 0$, not necessarily since $g(t)\to\infty$ is also a solution (assuming $f$ bounded for instance).
We often do informal calculus when talking about limits, for instance $\frac{3+x}{x^2}\to \frac 30=\infty$ or $|x\sin(x)|^{\frac 1x}\to 0^\infty=0$. But keep in mind this is still informal (these final equal signs, are abuse of notation), even if this helps figuring quickly what the final result should be.
