A set S of ordered pairs of integers is defined recursively by 2 things: (1,1) i

Question

A set S of ordered pairs of integers is defined recursively by 2 things:

(1,1) is in S

if(m,n) is in S,

then (m+2,n) is in S,

and (m,2+n) is in S, and (m+1,n+1) is in S

Describe the ordered pair set?

The answer n+2l and n+2k I cannot understand. if someone can explain better, I would appreciate so much... . The answer should be a sentence that provides an easy test to check if a given pair is in the set or not.

Applying the "if" to (1, 1), we see that (3, 1), (1, 3) and (2, 2) are in S.

Applying it again, to (2, 2), we will get all pairs natural numbers (n, n) in the set.

Then considering (3, 1) and (1, 3), you can see that you can get (5, 1), (1, 5). Similary, from (2, 2), we'll get (2, 4), (4, 2) and so on. It becomes apparent that we have the general form

(n + 2l, n+2k) for all natural numbers n and integers l, k ? 0. In other words, all pairs of integers (n, m) where n and m differ by a multiple of 2 (including 0).

Explanation / Answer

You got it right mate. The set is all ordered pairs (m,n) where m and n differ by a multiple of 2 including 0 and m,n are natural numbers.

It is given that, if (m,n) is in S then

a) (m+1,n+1) is in S

b) (m+2,n) is in S

c) (m,n+2) is in S

the base case given is (1,1) is in S.

i.e, initially m = n

then from (a), we get all (n,n) where n is a Natural no.

from (b) and (c), we get all (m,n) where |m-n| = 2k which implies S has elements of the form (n+2l,n+2k).

Hope this helps.

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