0

Question is from my lecture notes. What will I show? It is clear, isn't it?

So, $5\mathbb{N}+3\mathbb{N}$={$5n_{1}+3n_{2}$:$n_{1},n_{2} \in\mathbb{N}$}={$0,3,5,6,8,9,10,11,12,...$}.

  • 5
    A statement with "$\ldots$" can never be clear ... Try to show $5\Bbb N+3\Bbb N=\Bbb N\setminus{1,2,4,7}$ – Hagen von Eitzen May 19 '16 at 11:30
  • 4
    This is a special case of Frobenius Coin problem. Given $p,q \in \mathbb{Z}_{+}$ with $\gcd(p,q) = 1$, the largest integer that cannot be represented as $p\mathbb{N} + q\mathbb{N}$ is $pq - p -q$. If you can establish this, there are finitely many integers remain to check whether they are in $p\mathbb{N} + q\mathbb{N}$ or not. – achille hui May 19 '16 at 11:37

2 Answers2

0

$$8=5+3$$ $$9=3+3+3$$ $$10=5+5$$ Add $\color{red}{3}$. Then $$11=8+3=(5+3)+3$$ $$12=9+3=(3+3+3)+3$$ $$13=10+3=(5+5)+3$$ Add $\color{red}{3}$. Then $$14=11+3$$ $$15=12+3$$ $$16=13+3$$ and so on.

Roman83
  • 18,308
0

I assume that the dots mean “and all natural numbers $>12$”.

First you prove that, for every $n\ge8$, you have $n\in 5\mathbb{N}+3\mathbb{N}$.

The base step is obvious: $8=5+3$.

Assume $n>8$ and that the statement is true for every $m$ with $8\le m<n$. If $n-3>8$, we are done, because by the induction hypothesis $n-3=5a+3b$ for some $a,b\in\mathbb{N}$ and so $n=5a+3(b+1)$. Otherwise $n=9$, $n=10$ or $n=11$; but $9=3\cdot 3$, $10=5\cdot 2$ and $11=5\cdot1+2\cdot 3$.

Now you just have to check the cases $0,1,2,3,4,5,6,7$ manually for deciding whether they belong to the set or not.

You may want to generalize the result to

If $r,s\in\mathbb{N}$ and $\gcd(r,s)=1$, then there exists $k\in\mathbb{N}$ such that, for every $n\ge k$, $n\in r\mathbb{N}+s\mathbb{N}$.

What would be the obvious candidate for $k$?

egreg
  • 244,946