How to prove that the g.c.d is equal to the prime factorization raised to the minimum of two powers for the prime factorization of $a$ and $b$ as $$a = p_1^{\alpha_1}p_2^{\alpha_2}\cdots{p_t}^{\alpha_t}$$ and $$b = p_1^{\beta_1} p_2^{\beta_2} \cdots p_s^{\beta_s}.$$ I want to prove that $d = gcd(a,b)$ is equal to $${p_1}^{\min(\alpha_1, \beta_1)}{p_2}^{\min(\alpha_2, \beta_2)}\cdots {p_r}^{\min(\alpha_r\beta_r)}$$ so I begin by proving that $d$ is a common divisor of $a$ and $b$.
$$\frac{a}{d} = \frac{p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_t^{\alpha_t}}{p_1^{\min(\alpha_1\beta_1)}p_2^{\min(\alpha_2\beta_2)} \cdots {p_s}^{\min(\alpha_s\beta_s)}}$$ and
$$\frac{b}{d} = \frac{p_1^{\beta_1}p_2^{\beta_2} \cdots p_s^{\beta_s}}{p_1^{\min(\alpha_1\beta_1)}{p_2}^{\min(\alpha_2\beta_2)} \cdots p_s^{\min(\alpha_s\beta_s)}}$$
I don't know if $a$ is always divisible by $d$ in $\mathbb{Z^+}$ I'm also very new to proofs can I have some guidance?

In $a / d$ you have factors
$$
F_i = \frac{p_i^{\alpha_i}}{p_i^{\min(\alpha_i, \beta_i)}}
$$
and two cases: 


*

*$\min(\alpha_i, \beta_i) = \alpha_i$ then $F_i = 1$ and

*$\min(\alpha_i, \beta_i) = \beta_i$ then $\alpha_i \ge \beta_i$ and
$\alpha_i - \beta_i \ge 0$ so $F_i = p_i^{\alpha_i - \beta_i} \in \mathbb{Z}$.


This gives $F_i \in \mathbb{Z}$ and for the product $a/d \in \mathbb{Z}$.
A similar argument yields $b/d \in \mathbb{Z}$ and we see that $d$ is a common divisor of $a$ and $b$.
