w16_206_hwk01_solns

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Math 206, Spring 2016 Optional Assignment 1 – Solutions Due: January 29, 2016

(1) In set theory, DeMorgan’s Laws say the following: if A and B are subsets of some set S , then

• (A ∪ B) = Ā ∩ B̄

• (A ∩ B) = Ā ∪ B̄Prove one of these equalities.

Solution. We’ll prove the first result. Since we’re asked to prove an equality of sets, we’ll achieve thisgoal by proving two statements:(a) (A ∪ B) ⊆ Ā ∩ B̄ and

(b) Ā ∩ B̄ ⊆ (A ∪ B).To give some variety in style, we’ll prove the first containment by using the logical rules that dictate theinteraction of negation and conjunctions; for the second containment, we’ll use a “proof by contradiction”approach.

To prove the first containment, we need to argue that if x ∈ (A ∪ B), then x ∈ Ā ∩ B̄. So suppose

that x ∈ (A ∪ B). By definition of complement, this means that x ∈ A ∪ B. Now recall that z ∈ A ∪ Bis equivalent to the truth of the statement “z ∈ A or z ∈ B”. Hence x ∈ A ∪ B means that it is not thecase that x ∈ A or x ∈ B. Hence we conclude that x ∈ A and x ∈ B.∗ By the definition of complement,this means that x ∈ Ā and x ∈ B̄. The definition of intersection then gives x ∈ Ā ∩ B̄.

Next, we’ll prove that Ā ∩ B̄ ⊆ (A ∪ B). So let x ∈ Ā ∩ B̄. We want to show that x ∈ (A ∪ B). Forthe sake of reaching a contradiction, suppose instead that this desire result were false. In other words,suppose that x ∈ A ∪ B. We’ll show that this assumption leads to a contradiction, from which we canconclude that our desired conclusion must indeed hold. Now if x ∈ A ∪ B is true, then this means thateither x ∈ A or x ∈ B .

Suppose, then, that x ∈ A is true. Then we’d have x ∈ A and — by our original assumption — thatx ∈ Ā ∩ B̄. Now the statement x ∈ Ā ∩ B̄ tells us (among other things) that x ∈ Ā. But then we wouldhave x ∈ A and x ∈ Ā. This cannot happen, so it must not be the case that x ∈ A.

Since we’re assuming x ∈ A ∪ B, however, to say that x ∈ A is false implies that x ∈ B must betrue. But then the original assumption x ∈ Ā ∩ B̄ would imply that x ∈ B̄, and so we’d have x ∈ Band x ∈ B̄. Again: this leads to a contradiction.

We see that assuming x ∈ A ∪ B leads to a contradiction, so we must conclude that this statement

is false. Hence x ∈ A ∪ B, which is the same as saying x ∈ (A ∪ B), as desired.

(2) In set theory, the distributive laws say the following: if A,B and C are sets, then• A ∩ (B ∪ C ) = (A ∩ B) ∪ (A ∩ C )• A ∪ (B ∩ C ) = (A ∪ B) ∩ (A ∪ C )

Prove one of these equalities.

Solution. In this solution we’ll prove the first equality. You are encouraged to give a proof of thesecond statement yourself. If you get stuck or want to check your work, feel free to contact the instructor.

To prove the first equality, we’ll follow the standard method for checking that two sets are equal:verifying a dual containment. We begin by proving that

A ∩ (B ∪ C ) ⊆ (A ∩ B) ∪ (A ∩ C ).

So let x ∈ A ∩ (B ∪ C ) be given. By the definitions of intersection and union, this means that x ∈ A,and also that either x ∈ B or x ∈ C . From this we conclude that either x ∈ A and x ∈ B , or that x ∈ Aand x ∈ C .† Now if x ∈ A and x ∈ B , then we have x ∈ A ∩ B; on the other hand, if x ∈ A and x ∈ C ,then we have x ∈ A ∩ C . Hence our preceding statement gives x ∈ (A ∩ B) ∪ (A ∩ C ), and we concludethat A ∩ (B ∪ C ) ⊆ (A ∩ B) ∪ (A ∩ C ).

∗Below we’ll see (using truth tables) that ¬(P ∨ Q) = ¬P ∧ ¬Q. In this case we take P to be the statement x ∈ A, and then ¬P becomes x ∈ A.†Using truth tables one can verify that if P ,Q and R are statements, then P ∧ (Q∨R) is equivalent to the statement (P ∧ Q)∨

(P ∧ R). But don’t take my word for it; write out the truth table and check for yourself!

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For the reverse containment, suppose that x ∈ (A ∩ B) ∪ (A ∩ C ). By the definitions of union andintersection, this means that we have either x ∈ A ∩ B , or that x ∈ A ∪ B . If the former is true,then the definition of intersection gives x ∈ A and x ∈ C ; if the latter is true, then we get x ∈ A andx ∈ B. Notice that in either case we have x ∈ A, and additionally that x ∈ B ∪ C (since in the firstcase x ∈ B ⊆ B ∪ C , whereas in the second x ∈ C ⊆ B ∪ C ). By the definition of intersection, we havex ∈ A ∩ (B ∪ C ) as desired.

(3) In logic, DeMorgan’s Laws say the following: if P and Q are statements, then• ¬(P ∧ Q) = (¬P ) ∨ (¬Q)• ¬(P ∨ Q) = (¬P ) ∧ (¬Q)

Use truth tables to prove these assertions.

Solution. We’ll prove the first result. Our method is to write out a truth table. We’ll show thatwhatever truth values P and Q take, the two statements ¬(P ∧ Q) and (¬P ) ∨ (¬Q) take the same truthvalue. Hence these two statements will be equivalent.

P Q P ∧ Q ¬(P ∧ Q) ¬P ¬Q (¬P ) ∨ (¬Q)T T T F F F FT F F T F T TF T F T T F TF F F T T T T

(4) Suppose that P ,Q and R are statements. Use truth tables to prove that the statement ”P ⇒ (Q ∨ R)”is equivalent to the statement ”(P ∧ ¬Q) ⇒ R”.

Solution. This time we have three initial statements, and so there are 23 8 possible combinations of truth values they can take on. While our truth table will then be a little longer, the basic strategy is thesame: we show that the two given statements take on the same truth values, and hence are equivalent.

P Q R Q ∨ R P ⇒ (Q ∨ R) ¬Q P ∧ ¬Q (P ∧ ¬Q) ⇒ RT T T T T F F T

T T F T T F F TT F T T T T T TT F F F F T T FF T T T T F F TF T F T T F F TF F T T T T F TF F F F T T F T

(5) Suppose that P ,Q and R are statements. Use truth tables to prove that the statement ”(P ∨ Q) ⇒ R”is equivalent to the statement ”(P ⇒ R) ∧ (Q ⇒ R)”. [Note: this is the logical foundation for ”proof by cases.”]

Solution. As in the previous problem, we simply list all possible combinations of truth values for our

three initial statements, then make a truth table to verify the desired result.



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P Q R P ⇒ R Q ⇒ R) (P ⇒ R) ∧ (Q ⇒ R) P ∨ Q (P ∨ Q) ⇒ RT T T T T T T TT T F F F F T FT F T T T T T T

T F F F T F T FF T T T T T T TF T F T F F T FF F T T T T F TF F F T T T F T

(6) Suppose you have a sequence {a1, a2, a3, · · · }, where each ai ∈ R. In calculus you (probably) learnedthat the formal definition of limn→∞ an = L is the statement: for every ε > 0 there exists N ∈ N suchthat for all n ≥ N we have |an − L| < .(a) Write the definition of the limit in the form of quantified variables followed by a predicate.(b) What does it mean — as a quantified statement — to say that limn→∞ an = L?

Solution. For the first part, the definition in quantified form is

∀ε ∈ R ∃N ∈ N ∀n ∈ N : ε ≤ 0 ∨ (n < N ∨ |an − L| < ε).

This can be simplified slightly if we allow ourselves to use the notation R+ to denote the set of positivereal numbers; we’d then have

∀ε ∈ R+ ∃N ∈ N ∀n ∈ N : n < N ∨ |an − L| < ε.

Finally, it’s worth noting that it’s pretty common practice in mathematics to include any relevantrestrictive qualities on variables immediately after quantifying (but before expressing the predicate); soyou’d more commonly see this written as

∀ε > 0 ∃N ∈ N ∀n ≥ N : |an − L| < ε.

(This latter expression is pretty sloppy since it doesn’t even describe the set that n is meant to be drawnfrom; for mathematicians who know what a sequence represents this is supposed to be “obvious,” but

this goes to show you that mathematicians aren’t nearly as careful in applying logic as they could be.)For the second part: to say that limn→∞ an = L means that the negation of the definition holds,

namely that there exists some ε > 0 so that for all N ∈ N there exists some n ≥ N with |an − L| ≥ ε.

(7) In number theory, Goldbach’s conjecture states that every positive even number can be written as thesum of two prime numbers.(a) Write Goldbach’s conjecture in the form of a quantified statement. (You might choose to use 2Z

to denote the set of even numbers, and P to denote the set of primes.)(b) Many mathematicians believe that Goldbach’s conjecture is true. What would you need to do in

order to prove that Goldbach’s conjecture is false?

Solution. Goldbach’s conjecture says:

∀x ∈ 2Z ∃ p, q ∈ P : x = p + q.

In order to prove Goldbach’s conjecture is false, we’d need to prove the following statement is true:

∃x ∈ 2Z ∀ p, q ∈ P : x = p + q.

In words, we’d have to argue that there is some even number which cannot be written as the sum of (any) two primes.


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