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# 11_Transcript - 1 Uniqueness 1.1 Welcome Welcome to the...

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Unformatted text preview: 1 Uniqueness 1.1 Welcome Welcome to the eleventh module in our course on proofs. Before going on, please do the assigned reading. 1.2 Introduction You will encounter statements of the form There is a unique object with a certain property such that something happens Instead of the word “unique” you may see “one and only one” or “exactly one”. Here are two examples that deal with uniqueness. Proposition 1. Two lines in the plane which are not parallel will intersect in one and only one point. Proposition 2. Up to addition by a constant, there is a unique function f ( x ) such that f ( x ) = f ( x ) . Uniqueness may occur either in the forward or backward processes, and we must proceed differ- ently in each case. 1.3 Uniqueness in the Forward Method If a statement of the form There is a unique object with a certain property such that something happens. appears in the forward process we assume not only that There is an object with a certain property such that something happens. but that the object is unique How does this help us? We can assume that there are two such objects, X and Y , and equate them. Using the statement X = Y in the forward process is called the forward uniqueness method . Here is an example. Proposition 3. If there is a unique root of ax 2 + bx + c = 0 when a 6 = 0 , then b 2- 4 ac = 0 . As usual, we begin by explicitly identifying the hypothesis and conclusion. Hypothesis A: There is a unique root of ax 2 + bx + c = 0 when a 6 = 0. Conclusion B: b 2- 4 ac = 0. The appearance of “unique” in the hypothesis tells us to use the forwards uniqueness method. Of course, we could just use the quadratic formula but here we will do it without using this information. In keeping with the plan mentioned above, we assume that there are two objects with the desired property and then equate them. 1 A1 Let us suppose that r 1 and r 2 are the roots of ax 2 + bx + c = 0. Here we are making implicit use of prior knowledge, that a quadratic polynomial can be factored into two linear factors. A2 Since the root is unique we may also suppose r 1 = r 2 . There are two natural things to do when dealing with roots, factor and substitute. We will factor. A3 Dividing the equation by a and recognizing that r 1 and r 2 are roots, we get ( x- r 1 )( x- r 2 ) = x 2 + ( b/a ) x + c/a . A4 Expanding the left side gives x 2- ( r 1 + r 2 ) x + r 1 r 2 = x 2 + ( b/a ) x + c/a ....
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11_Transcript - 1 Uniqueness 1.1 Welcome Welcome to the...

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