Operations on rational numbers

A drop of $\frac{2}{3}$°C per hour for $\frac{3}{2}$ hours means a total change of $-\frac{2}{3}\times \frac{3}{2}$. The negative sign represents a drop in temperature.

$-\frac{2}{3}\times \frac{3}{2}$ The 2 in the numerator and 2 in the denominator cancel, as do the 3s: $-\frac{1}{1}\times \frac{1}{1}=-1$

Negative × positive = negative. The result is $-1$°C, meaning the temperature dropped by 1 degree Celsius.

The total temperature change is $-1$°C.

This is a multiplication problem: rate × time = total change. We compute $-\frac{2}{3}\times \frac{3}{2}$.

Since the temperature is dropping, the answer must be negative. $+1$°C implies a rise, which is physically incorrect here.

$-\frac{4}{6}$ would come from incorrectly adding numerators and denominators instead of multiplying: $\frac{2+2}{3+3}$. This is not how multiplication works.

$-\frac{1}{3}$ might come from subtracting: $\frac{2}{3}-\frac{3}{2}$... but even that gives a different result. It doesn't arise from correct multiplication.

$-\frac{2}{3}\times \frac{3}{2}=-1$ after cross-cancelling. The answer is $-1$°C.