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Unformatted text preview: Math 6a  Solution Set 3 Problem # 1 Story problem 1 ”My question for you, dear Carib, is this: Can you find another truthfunctional connective that is associative? That is, is there some well defined way of taking two truth values, p and q, and producing a thirdcall it p ⊕ qso that ( p ⊕ q ) ⊕ r = p ⊕ ( q ⊕ r ) for any choice of truth values (true or false) p,q, and r ?” Solution. We can simply define p ⊕ q = T for all possible truth values of p and q. Then, trivially, both ( p ⊕ q ) ⊕ r and p ⊕ ( q ⊕ r ) both are always equal to T no matter what value p, q, and r have. Problem # 2 Story problem 2 ”Here’s another question for you: Which of these connectives have inverses ? (Every real number p other than 0 has a multiplicative inverse, that is, a number p 1 such that p · p 1 = 1 . ) In other words, if p is an arbitrary truthvalue (i.e., true or false), can we always find a truth valuecall it p 1such that p ∨ p 1 = F Or p & p 1 = T Or p ≡ p 1 = T ?” Solution. In the first case, we cannot find an inverse in general, since if p = T there is no truth value q such that p ∨ q = T ∨ q = F. Similarly, in the second case, we again cannot find an inverse in general, since if p = F there is no truth value q such that p & q = F & q = T. In the third case, we can find an inverse in general. Simply let p 1 = p. Then, if p = T we have p ≡ p 1 = p ≡ p = T ≡ T = T and similarly, if p = F, we have p ≡ p 1 = p ≡ p = F ≡ F = T. Problem # 3 Story problem 3 Prove that ’ ∨ ’distributes over ’ ≡ : i.e., that p ∨ ( q ≡ r ) = ( p ∨ q ) ≡ ( p ∨ r ) 1 2 for all truthvalues p,q, and r. Solution. First, note that if p = T, the p ∨ ( q ≡ r ) = T no matter what q and r are, and similarly, p ∨ q and p ∨ r are both T and thus ( p ∨ q ) ≡ ( p ∨ r ) = T ≡ T = T and thus the left side equals the right side. So consider the case when p = F. Then, p ∨ x is true if x is true, and false if x is false, and thus, in either case p ∨ x = x. Thus, we can simplify the right hand side p ∨ ( q ≡ r ) = q ≡ r and the left hand side ( p ∨ q ) ≡ ( p ∨ r ) = q ≡ r and thus, again, both sides are equal, no matter what q and r are. This covers all cases. Problem # 4 Story problem 4 ”Well,” his majesty said at last, ”tell me this: for what truthvalues x is ( x ∨ x ) = F true?” Solution. There are only two possible truth values, T or F. When x = T, x ∨ x = T ∨ T = T so this is not a solution to the above equation. When x = F, x ∨ x = F ∨ F = F so x = F is a solution to the above equation. Thus, the only solution is x = F. Problem # 5 Story problem 5 ”That’s an interesting question,” Abubakari said (aloud), once he’d taken it all in. He was quite enjoying this. ”Can we find a polynomial f ( x ) with coefficients in Lthat is, with coefficients that are 0 or 1which has no root? That is, such that f ( x ) is never zero, no matter what x is?” Solution. Consider the polynomial...
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 Fall '07
 Wilson
 Math, Addition, Elementary algebra, idempotent, story problem

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