By Harris Kwong
It is a textual content that covers the normal themes in a sophomore-level path in discrete arithmetic: common sense, units, evidence recommendations, easy quantity idea, features, family members, and straight forward combinatorics, with an emphasis on motivation. It explains and clarifies the unwritten conventions in arithmetic, and courses the scholars via a close dialogue on how an explanation is revised from its draft to a last polished shape. Hands-on routines aid scholars comprehend an idea quickly after studying it. The textual content adopts a spiral procedure: many themes are revisited a number of occasions, occasionally from a unique point of view or at the next point of complexity. The aim is to slowly advance scholars’ problem-solving and writing abilities.
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Additional resources for A Spiral Workbook for Discrete Mathematics
Recall that a biconditional statement p ⇔ q consists of two implications p ⇒ q and q ⇒ p. Hence, to prove p ⇔ q, we need to establish these two “directions” separately. 11 Let n be an integer. Prove that n2 is even if and only if n is even. Solution: (⇒) We first prove that if n2 is even, then n must be even. We shall prove its contrapositive: if n is odd, then n2 is odd. If n is odd, then we can write n = 2t + 1 for some integer t. Then n2 = (2t + 1) = 4t2 + 4t + 1 = 2(2t2 + 2t) + 1, where 2t2 + 2t is an integer.
Since x is positive, we gather that 1 + n + x > 1 + n. We deduce that the sum of the divisors cannot be n + 1. Therefore, if the sum of the divisors of n is precisely n + 1, then n must be prime. 3 Solution: Let x be a real number. Prove that if x3 − 7x2 + x − 7 = 0, then x = 7. Assume x = 7, then x3 − 7x2 + x − 7 = x2 (x − 7) + (x − 7) = (x2 + 1)(x − 7) = 0. Thus, if x3 − 7x2 + x − 7 = 0, then x = 7. 1 Let x be a real number. Prove that if (2x2 + 3)(x + 5)(x − 7) = 0, then either x = −5, or x = 7.
To prove that p ⇒ q, we proceed as follows: 1. Suppose p ⇒ q is false; that is, assume that p is true and q is false. 2. Argue until we obtain a contradiction, which could be any result that we know is false. Proof by contradiction. How does this prove that p ⇒ q? Assuming that the logic used in every step in the argument is correct, yet we still end up with a contradiction, then the only possible flaw must come from the supposition that p ⇒ q is false. Consequently, p ⇒ q must be true. This is what a typical proof by contradiction may look like: Proof: ...
A Spiral Workbook for Discrete Mathematics by Harris Kwong