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I'm trying to find a proof that shows if $a$, $b$ are in the natural numbers, then the sum of the additive semigroups $\mathbb N a + \mathbb N b$ is a subset of $\mathbb N d$ where $d = \gcd(a,b)$.

Michael Hardy
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user130272
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  • This is obvious since for all $n,m$ you have $$na+mb = d \left( n \frac{a}{d} + m \frac{b}{d} \right) \in d \Bbb{N}$$ – Crostul Mar 04 '16 at 18:38

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$d$ is a divisor of both $a$ and $b$. So \begin{align} a & = jd \\ b & = kd \end{align} for some $j,k\in\mathbb N$.

So $ma+nb = mjd + nkd = (mj+nk)d \in d\mathbb N$.

Michael Hardy
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