In lecture 6 of Harvard online videos on abstract algebra (https://www.youtube.com/watch?v=lTM7TSlDij4&list=PLA58AC5CABC1321A3&index=6&t=2280s, minute 37), the lecturer claims the following:
Let $m,n \in \mathbb{Z}$ (not both zero). Then $\mathrm{gcd}(m, n)$ is equal to the unique positive integer $d$, such that $d|m,n$, and that if $e|m,n$ then $e|d$.
I have taken a simple example with $m=16=2\cdot8$ and $n=48=2 \cdot 3 \cdot 8$. There is indeed a $d=16=\mathrm{gcd}(m, n)$ that divides $m$ and $n$, and for example $e=8$ that divides $m$ and $n$ and also divides $d$. However, I can also find another $d^{\prime}=8\neq\mathrm{gcd}(m, n)$ that divides $m$ and $n$, and an $e^{\prime}=2$ that divides $m$ and $n$ and which also divides $d^{\prime}$.
Is something missing in that definition? In this example those conditions do not define a unique number equal to the gcd.