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If $d$ is the greatest common divisor of all the numbers in set $B$, let $A=\{b/d:b\in B\}$. Then for each $a,b\in A$ ($a>b$) we have $$\frac{a-b}{d(a,b)}\in A.(\ast )$$ Observe that the greatest common divisor of the set $A$ equals $1$, therefore we can find a finite subset $A_1\subset A$ for which $\gcd A_1=1$. We may think that the sum of elements of $A_1$ is minimal possible. Choose numbers $a,b\in A_1$ ($a>b$) and replace $a$ in the set $A_1$ with $\frac{a-b}{d(a,b)}$. The greatest common divisor of the obtained set equals $1$. But the sum of numbers decreases by this operation, which contradicts the minimality of $A_1$.

Thus, $A_1=\{1\}$. Therefore all the numbers in the set $A$ have residue $1$ modulo $d$. Take an arbitrary $a=kd+1\in A$ and $b=1$. Then $k\in A$ by $(\ast )$ and hence $k=ds+1$. But $(k,kd+1)=1$, therefore $\frac{kd+1-ds-1}{d}=k-s=(d-1)s+1\in A$, so $s$ is divisible by $d$. But $s\in A$, therefore $s-1$ is also divisible by $d$, hence $d=1$ (that means that $B=A$). Thus we have checked that if $a=kd+1=k+1\in A$ then $a-1=k\in A$. Then all non-negative integers belong to $A$ because it is infinite.