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I am having a hard time showing this simple relation. Here, all the letters below are in $\mathbb{N} \cup \{0\}$. $$A = \{(a,b) : a + 2b \leq n+2\}$$ $$B = \{(a,c+1) : a + 2c \leq n\}$$

How to show that $B \subset A$?

This is something I got from a sum over the above indices. I wanted a lower bound on the sum (which is where $B$ came in). Obviously I set $b = c+1$ in $A$ to get $B$ but how to show that it's a subset?

Thanks

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up vote 4 down vote accepted

I take it you mean for $n$ to be some fixed positive integer.

To show $B \subset A$, you want to show that any ordered pair of $B$ can also be found in $A$. To that end, let $(a, c+1) \in B$. That is, $a + 2c \leq n$.

We want to know whether $(a,c+1) \in A$. In other words, is $a + 2(c+1) \leq n + 2$? Just follow your nose: $$ \begin{align*} a + 2(c+1) &= [a + 2c] + 2\\ &\leq [n] + 2, \end{align*} $$ where the last line comes from our assumption about $(a,c+1) \in B$ (the brackets are just for emphasis).

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Thanks......... –  matt Aug 6 '12 at 21:23
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If $(a,c+1)$ is an element of $B$ and you set $b=c+1$, then $a+2b=a+2(c+1)=(a+2c)+2\leq n+2$.

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Yeah that's what i did, but is it obvious from this that it's a subset? Am I just being stupid? –  matt Aug 6 '12 at 20:41
    
Then you have shown that $(a,c+1)$ is an element of $A$ since it satisfies the required condition. That's all you need to show $B$ is a subset. –  AMPerrine Aug 6 '12 at 20:42
    
Thanks a lot... –  matt Aug 6 '12 at 21:23
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