1873_solutions

Q x are statements that contain an unknown x and that

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Unformatted text preview: true. Then write a proof that Q x must be true. 5. Given that P x is a statement that contains an unknown x and that S is a set, write down an opening line of a proof of the assertion that P x is true for every x S. Solution: Suppose that x S. 6. You are given that P x and Q x are statements that contain an unknown x, that S is a set, that P x is true for every x S and that P x  Q x . Is it possible to deduce that Q x is true for every member x of the set S? Solution: Yes it is possible. For each x we know that P x is true and that P x  Q x and so we know that Q x is true. 7. You are given that P x and Q x are statements that contain an unknown x, that S is a set, that P x is true for every x S and that P x  Q x . Is it possible to deduce that Q x is true for at least one member x of the set S? Solution: No it isn’t possible to deduce that Q x is true for at least one member x of the set S because we have no information to the effect that the set S is nonempty. 8. Write down the contrapositive form of the statement that for every member x of a given set S we have Px Qx. Solution: For every member x of the set S we have Qx  Px . 9. Write down the denial of the statement that for every x we have P x  Q x . Solution: There is at least one x for which P x 10. Prove that for every number x in the interval u 3 v is true and Q x is false. 2, 2 , if we define 3 3 x  2x 2  1 3 4 x2 4 x2 63 and 2x 2  1 3 3x 63 then u 2  v 2  1  uv. Hint: With an eye on the proof shown earlier, show that u 3  v 3  u  v. That earlier proof shows that if x is any number in the interval 3 3 3 x  2x 2  1 4 x2 63 13  3 3x 3 2, 2 then 2x 2  1 63 4 x2 x which we can express as u  v  1. On the other hand, u3  v3  3 3 x  2x 2  1 x2 4  63 and we conclude that u 3  v 3  x. Thus 2x 2  1 3 3x 4 x2 63 u3  v3  u  v and, by factorizing the left side, we obtain the desired result u 2  v 2  1  uv without difficulty. Some Exercises on Solution of Cubic Equations 1. Solve the following equations: a. x 3 6x  6  0 Solution: We look for two numbers r and s such that r3  s3  6 and 3rs  6 Thus r3  3 2 r 6 and so 6r 3  8  0 r6 Which gives us r3 So there are two possibilities: Either r  The equation 3 4 0 2 r3 2 , in which case s  x3 3 4 or r  6x  6  0 becomes x3  r3  s3 3rsx  0 which we can write as x  r  s x2  r2  s2 xr rs  0. xs and so either x 3 3 2 4 or x 3 2  34 3i 2 3 2 b. x 3 9x  12  0 We look for numbers r and s such that r 3  s 3  12 3rs  9 which gives us r3  14 3 r 3  12 3 4 . 3 4 , in which case s  3 2. and so r3 9 0 3 r3 and so we have the possibilities r  3 3 or r  3 9 that boil down to the one possibility that r  3 3 and s  3 9 . So the equation x 3 9x  12  0 becomes x  r  s x 2  r 2  s 2 xr xs rs  0. which gives us the real solution x  3 3 3 9 and also the possibility 33  39 33  39 3i 33 39 i 39 x  2 2 3 c. x 3 3x  2  0 One might notice that x  1 is an obvious solution of this equation. That obse...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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