Question 1. Structural induction (4 marks) The set \( S \) is defined recursively as follows: (1) \( (0,0) \in S,(1,1) \in S \). (base clause) (2) If \( (n-1, x) \in S \) and \( (n-2, y) \in S \) then \( (n, x+y) \in S \). (recursion clause) (3) Membership for \( S \) can always be demonstrated by (finitely many) successive applications of clauses above. (minimality clause) (a) Is \( (2,2) \in S \) ? Explain. [1 mark] (b) Is \( (10,55) \in S \) ? Explain [1 mark] (c) What is \( S \) ? [2 marks]
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Since n = 2, we have (1, x) and (0, y). From the base clause, we know that (1,1) β S and (0,0) β S. Show moreβ¦
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Let $S$ be the subset of the set of ordered pairs of integers defined recursively by Basis step: $(0,0) \in S$ . Recursive step: If $(a, b) \in S,$ then $(a, b+1) \in S$ $(a+1, b+1) \in S,$ and $(a+2, b+1) \in S$ a) List the elements of $S$ produced by the first four ap- plications of the recursive definition. b) Use strong induction on the number of applications of the recursive step of the definition to show that $a \leq 2 b$ whenever $(a, b) \in S .$ c) Use structural induction to show that $a \leq 2 b$ whenever $(a, b) \in S .$
Induction and Recursion
Recursive Definitions and Structural Induction
Let $S$ be the subset of the set of ordered pairs of integers defined recursively by Basis step: $(0,0) \in S .$ Recursive step: If $(a, b) \in S,$ then $(a+2, b+3) \in S$ and $(a+3, b+2) \in S$ a) List the elements of $S$ produced by the first five appli- cations of the recursive definition. b) Use strong induction on the number of applications of the recursive step of the definition to show that $5 | a+b$ when $(a, b) \in S .$ c) Use structural induction to show that $5 | a+b$ when $(a, b) \in S .$
Use generalized induction as was done in Example 13 to show that if $a_{m, n}$ is defined recursively by $a_{1,1}=5$ and $$a_{m, n}=\left\{\begin{array}{ll}{a_{m-1, n}+2} & {\text { if } n=1 \text { and } m>1} \\ {a_{m, n-1}+2} & {\text { if } n>1}\end{array}\right.$$ then $a_{m, n}=2(m+n)+1$ for all $(m, n) \in \mathbf{Z}^{+} \times \mathbf{Z}^{+}$
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