Download Math 248, Methods of Proof, Winter 2015

Math 248, Methods of Proof, Winter 2015
Homework 4, due Friday, 1/23
1. Prove that there are m, n ∈ Z such that 2m + 7n = 1.
2. Prove that there is an odd number M such that for all real r ≥ M ,
1
< 0.01
2r
3. Prove (by contradiction) that there does not exists a smallest positive real number (that is there does not
exists an r ∈ R such that r > 0 and, if s ∈ R and s > 0 then r ≤ s).
Sometimes we will want to prove that a statement of the form (∀x)(P (x)) is false. If we do this by giving
a constructive proof of the negation ¬(∀x)(P (x)) ⇐⇒ (∃x)(¬P (x)), then the x which we determine
satisfies ¬P (x) is referred to as a counter-example.
So, for instance, consider the following statement: for all x ∈ N, x2 + x + 41 is prime. This statement
is false, because for x = 41, (41)2 + 41 + 41 = (41)(41 + 1 + 1) is not prime (curiously, for all numbers
x = 1, 2, 3, . . . , 39, x2 + x + 41 does turn out to be prime!). We refer to x = 41 as a counter-example to
the statement for all x ∈ N, x2 + x + 41 is prime because it is a specific number showing that the negation
(there is x ∈ N such that x2 + x + 41 is not prime) is true.
4. Find a counter example to the statement: if x ∈ R such that x > 0, then x2 − x > 0.
5. Prove or give a counter-example: for all integers a, b, c, if a | bc, then a | b or a | c.
Sometimes we will want to prove theorems of the following form (P ∨ Q) =⇒ R. We can do this directly,
and thus begin by assuming P ∨ Q and attempt to establish R. Assuming P ∨ Q, of course, means that
P is true or Q is true (or both). So we proceed by breaking into more than one case. First we assume
that P is true and work to show that R is true. Then we assume instead that Q is true and work to show
that R is true. The form looks like the following:
Theorem 1. (P ∨ Q) =⇒ R.
Proof. Assume P is true
Work to show that R is true
Assume instead that Q is true
Work to show that R is true
Conclude if P ∨ Q is true then R is true, and hence
(P ∨ Q) =⇒ R
Here’s an example of this in action.
(
r
Definition 1. Given r ∈ R, we define |r| =
−r
if r ≥ 0
if r < 0
Theorem 2. Suppose that r, x ∈ R are specified real numbers and r > 0. If x > r or x < −r, then
|x| > r.
Proof. Suppose first that x > r. Since r > 0 and x > r it follows that x > 0 and thus |x| = x by
definition. Thus |x| = x > r as required. Suppose on the other hand that x < −r. Then x < −r < 0
whence |x| = −x by definition. Furthermore x < −r implies −x > r and thus |x| = −x > r as required.
Thus if x > r or x < −r it follows that |x| > r.
Of course, frequently the ∨ which requires the use of cases in the proof will be buried in the problem.
Here’s an example:
Theorem 3. Let x ∈ R. Then −|x| ≤ x ≤ |x|.
Proof. Let x ∈ R be arbitrary. Then either x ≥ 0 or x < 0.
Suppose first that x ≥ 0. Then |x| = x be definition and since x ≥ 0, −|x| = −x ≤ 0. Thus −|x| = −x ≤
0 ≤ x = |x| or simplifying −|x| ≤ x ≤ |x| as required.
Suppose on the other hand that x < 0 so that −x > 0. Now by definition |x| = −x, or −|x| = x, and so
|x| = −x > 0. We conclude that −|x| = x < 0 < −x = |x|, or simplifying, −|x| ≤ x ≤ |x| as required.
Since x ∈ R was arbitrary, this completes the proof.
Another way to state the theorem above, having specified x, is: if x ≥ 0 or x < 0, then −|x| ≤ x ≤ |x|.
That makes the (P ∨ Q) =⇒ R form a bit more apparent.
6. Prove that for all x ∈ R, |x| ≥ 0. Prove this as though you know how to do usual arithmetic on R (and
in particular that −a > −b if and only if a < b), but are just learning about |x| for the first time.
Of course, when encountered in the wild, a proof might require more than just two cases.
7. Prove that |xy| = |x||y| for all x, y ∈ R.
Now onto sets.
8. Let A and B be sets. Prove that if x 6∈ B and A ⊆ B, then x 6∈ A.
9. Give examples of sets A, B, and C such that the following are true (or write NOT POSSIBLE if, well, it
isn’t possible):
(a) A ⊆ B, B 6⊆ C, and A ⊆ C.
(b) A 6⊆ B, B 6⊆ C, and A ⊆ C.
(c) A ⊆ B, B 6⊆ C, and A 6⊆ C.
Recall that ∈ and ⊆ have different meanings. So we can say
{1, 2, 3} ⊆ {1, 2, 3, 4}
but not that
{1, 2, 3} ∈ {1, 2, 3, 4}.
Similarly, we can say that
3 ∈ {1, 2, 3}
but not that
3 ⊆ {1, 2, 3}.
Of course things get interesting when some of the elements in a set are themselves sets. So, for instance
{1, 2, 3} ∈ {1, 2, 3}, 4, {3, 4}
and
{1, 2, 3}, 4 ⊆ {1, 2, 3}, 4, {3, 4} ,
but one cannot say
{1, 2, 3} ⊆ {1, 2, 3}, 4, {3, 4} .
One could say
{1, 2, 3} ⊆ {1, 2, 3}, 1, 2, 3, {3, 4}
and
{1, 2, 3} ∈ {1, 2, 3}, 1, 2, 3, {3, 4} ,
because {1, 2, 3} is both an element and a subset of
{1, 2, 3}, 1, 2, 3, {3, 4} .
10. Let A and B be sets. Prove that A ⊆ B ⇐⇒ A − B = ∅.
11. Let A, B, and C be sets. Prove that (A − B) − C = (A − C) − (B − C).
Having discussed the idea of subsets of a set A, one wonders whether the collection of all subsets of A is
itself a set. The answer turns out to be yes (axiomatically) and leads to the following notation.
Definition 2. Given a set A, we say that the powerset of A is the set P(A) = {B | B ⊆ A}.
So, for example, if A = {1, 2, 3}, then P(A) = ∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3} . Note that
the elements of the powerset of A are themselves sets.
12. Let A = {1, ∅, {2, 3}}. Write out P(A).
13. Let A and B be sets. Prove that P(A) ∪ P(B) ⊆ P(A ∪ B).
14. Give a counterexample which shows that P(A ∪ B) 6= P(A) ∪ P(B).
The grader will carefully consider 7 and 11.