74-hw7s - MATH 74 HOMEWORK 7 SOLUTIONS 1 It is true that because R ∩ S is nonempty it must contain an element and because S ∩ T is nonempty it

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Unformatted text preview: MATH 74 HOMEWORK 7 SOLUTIONS 1. It is true that because R ∩ S is nonempty, it must contain an element, and because S ∩ T is nonempty, it must contain an element. But these elements could, in principle, be different. The proof doesn’t show that they are the same, it just calls them by the same letter. This doesn’t make them the same, and misses the point of the theorem ( to show that you can find an element in both sets). [The argument fails because the statement in the theorem is false, incidentally.] 2. I would give this under half credit. The person writing it understands how the proof should be structured, but not much else (or they have done a poor job of conveying their understanding.) I don’t understand the vague phrase “combining these two.” My guess is that “these two” refers to the conditions x ( x- 1) ≥ 0 and- ( x- 1)( x- 3) ≤ 0. But this is only a guess (and if it’s right I’m not sure how these are to be “combined” to yield the result). There is also the fact that the condition x ( x- 1) ≥ 0 (from x ∈ R ) has little to do with the conclusion— it isn’t even necessary to establish the conclusion! Reading the proof, I do not get the feeling that the writer noticed this. The point, as I see it, is that S = T , so you can conclude x ∈ T from the fact that x ∈ S alone. (A 10/10 proof would have begun by proving this, or at least stating it, or at least noticing it along the way.) 3. The first displayed equation reverses an inequality. It reads f ( x * ) ≤ (something), when what is known is that...
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This homework help was uploaded on 04/02/2008 for the course MATH 74 taught by Professor Courtney during the Fall '07 term at University of California, Berkeley.

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74-hw7s - MATH 74 HOMEWORK 7 SOLUTIONS 1 It is true that because R ∩ S is nonempty it must contain an element and because S ∩ T is nonempty it

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