Problem 2.

(a) let p(z) be a statement whose truth value depends on z. an ezample is a value of z that makes p(x) true, and a counterezample is a value of x that makes p(x) false. fill in the blank spaces with "is true", "is false'", or "nothing" as appropriate: "3z s.t. p(x)" an example proves a counterexample proves

in (b) - (h), determine whether the given statement is true or false, and briefly justify your answer.

(b) every integer is even or odd
(c) every integer is even or every integer is odd.
(d) some rational numbers are real numbers
(e) every complex number is real or imaginary.
(f) there is x € r such that for every r, y 〈 z.
(g) for every y r there is x r such that y 〈 x.

a) "An example proves P" False

b) True

c) False

d) True

e) False

f) False

g) True

Step-by-step explanation:

a) If we want to prove P, we have to verify that for all z, P(z) is true. Then, an example is not enough to prove P, it could happen that P(x) is false for some x that is not an example. To elaborate, denote the set of real numbers of R, and define P(z):="z is positive", for z∈R. An example is z=2. However, P(z) is not true in general, a counterexample is z=-1.

If you take P as a proposition, you have to decide the truth value of P, and it must be the same for all possible values of z. In this case, examples and counterexamples can be used sometimes to determine whether P is true or false. We apply this idea below.

b) This is true. Every integer is either even or odd because when you divide by 2, you always get a remainder of 0 (then the integer is even) or 1 (then the integer is odd).

c) This is false. It is not true that every integer is even, a counterexample is the integer 3. Then the proposition "Every integer is even" is false. Similarly, the proposition "Every integer is odd" is false (2 is a counterexample). Both propositions are false, thus the compound proposition is false.

d) This is true. An example is enough: rake r=2. r is rational, and real. Thus some (just one is enough) rational numbers are real.

e) This is false. A counterexample proves this: take z=1+i. z is not imaginary because its real part (1) is not zero. Also, z is not real (i≠0) so not all complex numbers are real or imaginary.

f) False. This would imply that for all real numbers y, y<x. In particular with y=x+1, x+1<x hence 1<0, which is a contradiction.

g) True. Given y∈R, take x=y+1. Then y<x=y+1 as required.

Step-by-step explanation:

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