Mathematical proof 15.pdf - cannot think of a constructive proof or that a constructive proof is very long and tedious In any case existential proofs

# Mathematical proof 15.pdf - cannot think of a constructive...

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Unformatted text preview: cannot think of a constructive proof, or that a constructive proof is very long and tedious. In any case, existential proofs are another valuble technique in proofs. Let's look at a familiar example from the calculus. An Example from Calculus First let's recall the Intermedaite Value and Mean Value Theorems: Intermediate Value Theorem. If a real valued funtion f is continuous on the closed interval [a, b] and if N is a number strictly between f(a) and f(b), then there t here exists a number c in (a, b) such that f(c) = N. Mean Value Theorem. If a real valued function f is continuous on the closed interval [a, b] and f is differentiable diff erentiable on the open interval (a, b) then there is a number c in (a, b) such that f'(c) = (f(b) (f( b) - f(a))/(b-a). We can use the t he Mean Value Theorem to prove that certain polynomials do not have more than one real root. (A root of a polynomial p(x) is a number c such that p(c) = 0.) Theorem. The polynomial p(x) = x3 + x - 1 has exactly one real root. Proof. The proof is in two parts. Part 1. (Direct Existential Proof.) First we will prove p(x) has one real root. We appeal to the Intermediate Value Theorem with a = 0 and b = 1: p(0) = - 1 < 0 and p(1) = 1 > 0. Since 0 (N = 0) is between -1 (=p(0)) and 1 (=p(1)), we may conclude that there is a real number c, between 0 and 1, for which p(c) = 0. Part 2. (Proof by Contradiction.) Now we will prove that p(x) has only one root. Assume to the contrary that p(x) has more than one root. Let's suppose two distinct roots c1 and c2, so p(c1) = p(c2) = 0. Then by appealing to the Mean Value Theorem, there must be a number c between c1 and c2 for which p'(c) = (p(c2) p(c1))/(c2 - c1) = 0. But a direct calculation shows that p'(x) = 3x2 + 1, which can never be zero since x2 >= 0 for all real numbers x. The contradiction completes the proof. q Example: Continuous Motion Here is an example where we appeal to the Mean Value Theorem to obtain the existence of something. Theorem. If an object is traveling in a straight line li ne with a differentiable position function s(t), where t denotes the time variable, for t between a and b, then there is a time t0, between a and b where the instantanious velocity at t = to is equal to the average velocity over the entire path. Proof. The velocity function is the derivative of the position function v(t) = s'(t). According to the Mean Value Theorem, there is a value t = t0 between a and b where v(t0) = s'(t0) = (s(b) ( s(b) - s(a))/(b-a) = average velocity over the path. ...
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