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Give a recursive definition of each of these sets of ordered pairs of positive integers. [Hint: Plot the points in the set in the plane and look for lines containing points in the set.

$S = \left\{{(a,b) \mid a \in Z^+, b \in Z^+ \text{ and } a + b \text{ is odd}}\right\}$

Does this work?

$(1,2) \in A$

$(2,1) \in A$

if $(a, b) \in A$ s.t. $a = 2k + 1$ where $k$ is a constant and a and b are integers, then $(a + 2, b) \in A$

if $(a, b) \in A$ s.t. $a = 2k + 1$ then $(a, b + 2) \in A$

I think the above gives me all the pairs where $a$ is odd $b$ is even.

if $(a, b) \in A$ s.t. $a = 2m$ (where $m$ is a constant) then $(a + 2, b) \in A$

if $(a, b) \in A$ s.t. $a = 2m$ then $(a, b + 2) \in A$

I think these two conditionals give me all the pairs where $a$ is even and $b$ is odd.

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3  
\mathbb{Z} to get $\mathbb{Z}$. Also you should type all of your maths inbetween \$ \$, so it looks like this $(1,2)\in S$. If you need to write something in the middle of the maths, you can use the \text{type your text here} command to get $\text{type your text here}$. –  Git Gud Feb 10 '13 at 10:14
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@GitGud: actually the shorter '\Bbb Z' also works: $\Bbb Z$ –  Marc van Leeuwen Feb 10 '13 at 11:12
    
@MarcvanLeeuwen Cool, thanks. –  Git Gud Feb 10 '13 at 11:13
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What do you mean by a recursive definition of a single set? In a sequence you can refer to previous elements, but here that would only work is you define you set as the union of a growing sequence of sets, which sequence you then define recursively; is that what you want? You cannot define a set in terms of its (completed) self; that would be circular. What you are saying (if $(a,b)\in S$ then also $(a+2,b),(a,b+2)\in S$) does not exclude too large solutions like $S=\Bbb Z^+\times\Bbb Z^+$. –  Marc van Leeuwen Feb 10 '13 at 11:18
    
Oh okay I think I understand what you mean. I want to take some element(s) in a set and build a set equivalent to S. I shouldn't have used S again in my attempted solution. I'll fix it. Thanks yo! –  papercuts Feb 10 '13 at 11:30

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