$\dfrac{2+3}{4+6}=\dfrac5{10}=\dfrac{1}{2}$

$\dfrac{2\div3}{4\div6}=\dfrac{\frac23}{\frac46}=\dfrac{\frac23}{\frac23}=1$

$\dfrac{23}{46}=\dfrac12\\$

$\dfrac{2-3}{4-6}=\dfrac{-1}{-2}=\dfrac12\\$

$\dfrac{2\times3}{4\times6}=\dfrac{6}{4\times6}=\dfrac14$

So $\dfrac{2\times3}{4\times6}$ is the smallest.

The numerators all contain $2$ and $3$ and the denominators all contain $4$ and $6$, which are double $2$ and $3$.

In $\dfrac{2+3}{4+6}$ the denominator is $2\times2+2\times3=2(2+3)$ which is twice the numerator.

In $\dfrac{2\div3}{4\div6}$ the denominator is $(2\times2)\div(2\times3)$ which is the same as the numerator. Imagine sharing twice as much money between twice as many people.

In $\dfrac{23}{46}$, the denominator is twice the numerator.

In $\dfrac{2-3}{4-6}$ the denominator is $2\times2-2\times3=2(2-3)$ which is twice the numerator.

In $\dfrac{2\times3}{4\times6}$ the denominator is $(2\times2)\times(2\times3)$, which is $4$ times the numerator.

So $\dfrac{2\times3}{4\times6}$ is the smallest, as its denominator is largest relative to its numerator.

You can find more short problems, arranged by curriculum topic, in our short problems collection.