Find the positive divisors of 20.
Let $d$ be a possible positive divisor of 20. According to Theorem 2.2.2, we know that $0 \leqslant d \leqslant 20$. From the equalities

$$
20=1 \times 20=2 \times 10=4 \times 5=5 \times 4=10 \times 2=20 \times 1
$$

we already know that $1, 2, 4, 5, 10$, and 20 are positive divisors of 20.

On the other hand, we know that $d \neq 0$, since 0 does not divide any non-zero integer. Furthermore, since $3 \times 6=18<20<21=3 \times 7$, $7 \times 2=14<20<21=7 \times 3$, and $8 \times 2=16<20<24=8 \times 3$, we know that 3, 7, and 8 are not divisors of 20.

Moreover, 3 divides $6=3 \times 2$ and $9=3 \times 3$, so since 3 does not divide 20, Theorem 2.2.3 shows that 6 and 9 do not divide 20 either. Finally, for any integer $k$ such that $11 \leqslant k \leqslant 19$, we know that $k<20<k \times 2$, so $k$ does not divide 20.

In conclusion, the positive divisors of 20 are 1, 2, 4, 5, 10, and 20.