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# Time-Space Tradeoffs for Computing Continuous Functions,

1) We use induction to prove that S1(n) = n(n + 1)/2 is true for each n ∈ . DEFINITION 1.3.1 A proof by induction has a predicate P, a basis step, an induction hypothesis, and an inductive step. The basis establishes that P(k) is true for integer k. The induction hypothesis assumes that for some fixed but arbitrary natural number n ≥ k, the statements P(k), P(k + 1), … , P(n) are true. The inductive step is a proof that P(n + 1) is true given the induction hypothesis. It follows from this definition that a proof by induction with the predicate P establishes that P is true for all natural numbers larger than or equal to k because the inductive step establishes the truth of P(n + 1) for arbitrary integer n greater than or equal to k. Also, induction may be used to show that a predicate holds for a subset of the natural numbers. For example, the hypothesis that every even natural number is divisible by 2 is one that would be defined only on the even numbers. The following proof by induction shows that S1(n) = n(n + 1)/2 for n ≥ 0. LEMMA 1.3.1 For all n ≥ 0, S1(n) = n(n + 1)/2. Proof PREDICATE: The value of the predicate P on n, P(n), is True if S1(n) = n(n + 1)/2 and False otherwise. BASIS STEP: Clearly, S1(0) = 0 from both the sum and the closed form given above. INDUCTION HYPOTHESIS: S1(k) = k(k + 1)/2 for k = 0, 1, 2, … , n. INDUCTIVE STEP: By the definition of the sum for S1 given in (1.1), S1(n+1) = S1(n)+ n + 1. Thus, it follows that S1(n + 1) = n(n + 1)/2 + n + 1. Factoring out n + 1 and rewriting the expression, we have that S1(n + 1)=(n + 1)((n + 1) + 1)/2, exactly the desired form. Thus, the statement of the theorem follows for all values of n. We now define proof by contradic