This preview shows page 1. Sign up to view the full content.
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...
View
Full
Document
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.
 Fall '08
 STAFF
 Math, Calculus

Click to edit the document details