Traditionally, underflow is said to occur when the exact result of an operation is nonzero but with an absolute value that is smaller than the smallest normalized floating point number. In the days before IEEE arithmetic, the response to underflow was typically, though not always, flush to zero: return the result $0$. In IEEE arithmetic, the standard response to underflow is to return the correctly rounded value, which may be a subnormal number, $\pm0$ or $±N_{min}$, where $N_{min}$ stands for the smallest normalized floating point number. This is known as gradual underflow.
A subnormal number is a number has a magnitude greater than $0$ but less than $N_{min}$.
I encountered the following question:
If $x$ and $y$ are floating point numbers with the property that $1/2 \leq \frac{x}{y}\leq 2$, then the exact difference $x-y$ is also a floating point number, so that $x\ominus y=x-y$, if gradual underflow is used. Show that this is not always the case if flush to zero is used. Here $x\ominus y$ is equal to round$(x-y)$.
I guess the rounding mode is rounding to the nearest representable floating point number.
I understand the inequality means the exponents of $x$ and $y$ can differ at most by 1. I also know if their exponents are equal, then since they are already in floating point representation, we can just subtract them, and the result is gonna be exact.
But this is the only progress I have made. I also have difficulties understanding the subtraction between two floating numbers in such a complicated scenario. For example, how do we shift the digits to align them since they do not have the same exponents.
I'm not sure how gradual underflow is contributing since how I am supposed to know when the difference will be rounded to a subnormal, $0$, or $N_{min}$.
Any helpful explanation is extremely appreciated.