Writing proofs for MATH 51H Section 2: Set theory, proofs of existential statements, proofs of uniqueness statements, proof by cases September 22, 2018 Recall from last week that the purpose of a proof is to communicate an argument. Your writing alone must convey the argument without you being there to explain what you meant to say. Here are some general reminders/guidelines to follow which will make your proofs clearer and more readable: Write in complete, grammatically correct sentences. Signal to your reader what you are doing in your proof. If you are proving a universal statement about elements in some set (for example, a vector space), say that you are arguing with a general element of that set. Don t just start doing an argument with a vector v without saying what v is. If you are doing a proof by contradiction, say at the beginning of the proof that you are doing a proof by contradiction. If you are doing a proof by induction, say at the beginning of the proof that you are doing a proof by induction and clearly label the base case and induction step somehow. If at some point in the proof you want to break the argument up into cases, say that you are breaking the proof up into cases and label the cases. If your proof has multiple parts (for example, an proof or showing that two sets are equal), then before each of these parts state what you are proving. This week we ll cover some basic notions in set theory that you ll need in this class and your life afterwards. In particular, we ll discuss what it means to show that two sets are equal. We ll also cover proving existence results, proving uniqueness results, and the strategy of division into cases. 1 Basic set theory At this point in the course, you ve encountered the notion of two sets (probably the span of some vectors) being equal. We intuitively know what it means for two sets to be equal: they 1
have exactly the same elements. In contrast to the situation for proving that two numbers are equal, where you start with one and do some manipulations to turn it into the other, it s not 100% obvious how to go about proving set equality. Recall that given two sets A and B, we say that A is a subset of B and write A B if every object in A is contained in B, i.e. if x A, then x B as well. We say that A and B are equal if they have the same elements, i.e. every element in A is also in B and every element in B is also in A. Note that A = B is equivalent to A B and B A. The most common strategy in showing that two sets are equal is to show these two inclusions separately. So to show A = B, show that A B and then that B A. We will shortly see two examples of this sort of proof. Assume, for the purposes of this section, that all of the sets considered are contained in some fixed larger set, so that we can define complements. Recall that given two sets A and B, their union A B is the set of all elements which are in A or B, A B = {x : x A or x B}. Here or is inclusive, so elements which are in both A and B are in A B. Also recall that given A and B, their intersection A B is the set of all elements which are in both A and B, A B = {x : x A and x B}. We can also define the union and intersection of any number of sets, finite or infinite, in the same manner. Given a sequence of sets A 1, A 2, A 3,..., then their union is A n = {x : x A n for some n N} n=1 and their intersection is A n = {x : x A n for all n N}. n=1 Since we have fixed a universe, we define the complement of A, denoted A c, to be all of the elements in the universe which are not in A, A c = {x : x / A}. Here you should be careful: complements of sets depend on the universe, that is, the set with respect to which we take the complement. For example, the complement of {1, 2, 3,... } in Z is different than its complement in Q. We ll now present two examples which demonstrate the strategy discussed above for proving that two sets A and B are equal show A B and then B A. Here is a template showing the general form of one of these proofs: Proof. First we will show that A B. Let x A. 2
proof that x B Thus, since x A was arbitrary, A B. Now we will show that B A. Let x B. proof that x A Thus, since x B was arbitrary, B A. Putting the two inclusions together, we conclude that A = B. 1.1 Examples Our first example relates unions, intersections, and complements. Proposition (De Morgan s laws, part 1). For any sets A and B, (A B) c = A c B c. Proof. First we will show that (A B) c A c B c. Let x (A B) c. Then by definition, x / A B, which means that x / A and x / B. But what this means is that x A c and x B c. Hence, x A c B c. Thus, since x (A B) c was arbitrary, we see that (A B) c A c B c. Now we will show that A c B c (A B) c. Let x A c B c. Then by definition, x A c and x B c, which means that x / A and x / B. Hence, x / A B, which means that x (A B) c. Since x A c B c was arbitrary, we see that A c B c (A B) c. Since (A B) c A c B c and A c B c (A B) c, we conclude that (A B) c = A c B c. This second example is more in line with what you ve done in class. Proposition. Let v 1, v 2 R n. Then span{v 1, v 2 } = span{v 1, 2v 2 }. Proof. First we ll show that span{v 1, 2v 2 } span{v 1, v 2 }. Let v span{v 1, 2v 2 }. Then v = a 1 v 1 + a 2 (2v 2 ) for some a 1, a 2 R. But then v = a 1 v 1 + (2a 2 )v 2, a linear combination of v 1 and v 2. Hence, v span{v 1, v 2 }. Since v was arbitrary, we ve shown that span{v 1, 2v 2 } span{v 1, v 2 }. Now we ll show that span{v 1, v 2 } span{v 1, 2v 2 }. Let v span{v 1, v 2 }. Then v = a 1 v 1 + a 2 v 2 for some a 1, a 2 R. But note that we can write v = a 1 v 1 + a 2 v 2 = a 1 v 1 + a 2 2 (2v 2 ), a linear combination of v 1 and v 2. Hence, v span{v 1, 2v 2 }. Since v was arbitrary, we ve shown that span{v 1, v 2 } span{v 1, 2v 2 }. Hence, since span{v 1, 2v 2 } span{v 1, v 2 } and span{v 1, v 2 } span{v 1, 2v 2 }, we conclude that span{v 1, v 2 } = span{v 1, 2v 2 }. 1.2 Exercises Exercise 1. Prove the second part of De Morgan s laws: for two sets A and B, (A B) c = A c B c. Exercise 2. Prove that for any sets A, B, and C, A (B C) = (A B) (A C). Exercise 3. Show that for any v 1, v 2 R n, span{v 1, v 2 } = span{v 1 + v 2, v 1 v 2 }. 3
Exercise 4. Show that {10n : n Z} = {2n : n Z} {5n : n Z}. Exercise 5. Prove that [n, n + 1) = [0, ). n=0 2 Proving existential statements, Part 1 Suppose that you want to prove that there exists some object X (which may be an integer, a vector, a set, etc.) such that some property P(X) (X satisfies some equation, X is in some span, X has a certain number of elements satisfying some conditions, etc.) holds. There are several strategies for proving these sort of existence statements. The simplest one is to give an explicit example of an X for which P(X) is true. You did one of these proofs on the previous homework, where you had to find a polynomial that took certain values at certain points. The real work in these sort of results is in finding the example, and then the proof itself is usually just verifying that the example does indeed work. The general structure of the proof is to give a candidate object for X and then show that P(X) does, in fact, hold: Proof. Consider A. proof that P(A) is true Thus, there exists an X such that P(X) is true, namely A. 2.1 Examples Proposition. There exist three distinct positive integers a, b, and c such that 1 a + 1 b + 1 c = 1. Proof. Consider a = 2, b = 3, and c = 6. Then 1 a + 1 b + 1 c = 1 2 + 1 3 + 1 6 = 3 + 2 + 1 = 6 6 6 = 1, and a b, b c, c a. Thus, there do indeed exist three such integers, namely a = 2, b = 3, c = 6. Proposition. There exist x, y R that solve the system of equations { 2x y = 1. x + y = 2 Proof. Consider x = 3 and y = 5. Then 2x y = 2 3 5 = 6 5 = 1 and x+y = 3+5 = 2. Thus, there do indeed exist two such real numbers, namely x = 3 and y = 5. 4
2.2 Exercises Exercise 6. Show that if a, b, c, d Z with a c, then there is an r Q such that ar + b cr + d = 1. Exercise 7. Prove that there exist arbitrarily large gaps between prime numbers. That is, for each n N, there exist at least n consecutive positive integers which are not prime. 3 Proving existential statements, Part 2 Another type of existence proof is to show indirectly that an object satisfying the property P(X) exists. There are several ways that you could do this. You could invoke some theorem which tells you that such an object exists, such as the intermediate value theorem or the mean value theorem. Alternatively, you could do a proof by contradiction (which we discussed last week) by showing that if such an object didn t exist, then we d be led to something which is false. Here are examples of both of these types of indirect proofs of existence. 3.1 Examples Proposition. The polynomial p(x) = x 9 + 23x 8 39x 6 + 120x 4 5x 2 10 has a root in [0, 1]. Proof. Note that p(x) is a continuous function on [0, 1], and thus the intermediate value theorem from calculus applies. We have p(0) = 10 and p(1) = 1+23 39+120 5 10 = 90. Thus, by the intermediate value theorem, p(x) = 0 for some x [0, 1]. That is, p(x) has a root in [0, 1]. Recall that an integer p 2 is prime if whenever p = ab, a, b N, then either a = 1 or b = 1. We ll use without proof the fact that any positive integer greater than 1 is divisible by some prime. This is, in fact, an existence statement: given any integer greater than 1, there exists a prime dividing it. The proof of this is left as an exercise. Proposition. There are infinitely many prime numbers. Proof. Suppose by way of contradiction that there were only finitely many prime numbers p 1, p 2,..., p k. Consider the integer n = p 1 p 2 p k + 1. Since there exists a prime number (for example, 2), n > 1, so that n is divisible by some prime p i, i = 1,..., k. So since the difference of two multiples of a number is still a multiple of that number, we have that p i (n p 1 p k ). But n p 1 p k = 1, so this says that p i 1, a contradiction since p i 2 by the definition of a prime number. Thus, there must be infinitely many prime numbers. 5
3.2 Exercises Exercise 8. Prove that there is an x R such that x = 2 x. Exercise 9. Show that every integer greater than 1 is divisible by a prime. Exercise 10. Recall that an integer n 2 which is not prime is called composite. Prove that if n is a composite integer, then n has a prime factor p such that p n. 4 Proving uniqueness, Part 1 One very useful strategy for proving that if there is an object satisfying some property, then it is unique, is to assume that there are two objects with this property and then show that they must be equal. You ve already seen this technique used in class several times, so we ll recap the proofs as examples. The general strategy to show that an object, if it exists, which satisfies some property P(X) is the unique object that satisfies P(X) is to show that if P(A) and P(B) are true for some objects A and B, then we must have A = B. Proof. Suppose that P(A) and P(B) are true. proof that A = B Thus, we have shown that any two objects satisfying P(X) are equal, which implies that the object satisfying P(X), if it exists, is unique. 4.1 Examples Proposition. The identity element in a group G is unique. Proof. Suppose that e 1 and e 2 are such that, for all g G, ge 1 = e 1 g = g and ge 2 = e 2 g = g. Then setting g = e 2 in the first equation, we see that e 1 e 2 = e 2. Similarly, setting g = e 1 in the second equation, we see that e 1 e 2 = e 1. Hence, e 1 = e 1 e 2 = e 2, and we see that any two identity elements are equal, which implies that the identity element in G is unique. Proposition. Let A R. Then the supremum of A, if it exists, is unique. Proof. Suppose that a 1, a 2 are both the supremum of A. By definition, this means that for all a A, a a 1 and a a 2. In particular, specializing a to be a 2, we see that a 2 a 1. Reversing the roles, when a = a 1, we see that a 1 a 2. Now, a 2 a 1 and a 1 a 2 taken together imply that a 1 = a 2. Thus, the supremum, if it exists, is unique, for any supremum of A equals a 1. The following example was not covered in class. Proposition (Division algorithm, uniqueness). Suppose that n, d N and we have found integers q, r Z such that n = dq + r and 0 r < d. Then this pair (q, r) is unique. 6
Proof. Suppose that q 1, r 1, q 2, r 2 Z are such that n = dq 1 + r 1 and n = dq 2 + r 2 with 0 r 1, r 2 < d. Then we have dq 1 + r 1 = dq 2 + r 2, and moving the q terms to one side and r terms to the other, this is equivalent to d(q 1 q 2 ) = r 2 r 1. By the definition of divisibility, we see that d (r 2 r 1 ). However, 0 r 1, r 2 < d, so that r 1 r 2 < d. Since d can t divide any nonzero integers with absolute value less than d, we conclude that r 1 = r 2. Plugging this back in to the equation above, since d > 0, q 1 q 2 = 0, i.e. q 1 = q 2. Hence, (q 1, r 1 ) = (q 2, r 2 ), and thus we see that the pair (q, r) in the division algorithm is unique. 4.2 Exercises Exercise 11. Show that the infimum of a set of real numbers, if it exists, is unique. Exercise 12. For any r R, show that there is at most one integer in the interval (r, r + 1 2 ). 5 Proving uniqueness, Part 2 The next strategy is a way to show that the solution to some equation or system of equations is unique. You start with the equation or system of equations and show through manipulations what the solution must be if it exists. If all of the steps you ve used to get to that point are reversible, as in the case of Gaussian elimination, then this sort of proof also shows existence. The general form of the proof is as follows. If your aim is to prove that some solution to the equation f(x) = 0 is unique, then you should start your proof by assuming that x is such that f(x) = 0 holds, and then try to solve for x. If you get that x must equal some constant, then this shows that the solution in x, if it exists, is unique. Here are two examples demonstrating this strategy. 5.1 Examples Proposition. The solution in x and y to the system of equations { x + y = 0, x 2y = 0 if it exists, is unique. Proof. Suppose that x and y are such that x + y = 0 and x 2y = 0. Then adding two times the first equation to the second, we see that 3x = 0. Thus, x = 0. Now, this implies that since x + y = 0 that y = 0 as well. We have shown that if x and y solve the above system of equations, then x = y = 0. Hence, the solution is indeed unique. It s obvious in the previous example that x = y = 0 is a solution without doing any manipulations. In addition, since every step we did was reversible, our manipulations also gave a proof that x = y = 0 does indeed solve the above system of equations. 7
Proposition. The solution in R of x + 4 = x 2, if it exists, is unique. Proof. Suppose that x R is such that x + 4 = x 2. Then squaring both sides, x + 4 = x 2 4x + 4, and simplifying gives x 2 5x = 0. This equation factors as x(x 5) = 0, and hence we see that either x = 0 or x = 5. However, x = 0 is not a solution to the original equation x + 4 = x 2, for plugging it in to the left hand side yields 4 = 2 and plugging it in to the right hand side yields 2, and 2 2. Hence, x is forced to be 5. Thus, we have shown that any two solutions to the above equation are equal, for we have shown that they are equal to 5. So, the solution to the equation in question is unique. Here, in contrast to the system of linear equations example above, our manipulations do not show that x = 0 or x = 5 is a solution to the equation x + 4 = x 2. This is because squaring both sides of an equation is not a reversible step, for y 2 = ( y) 2 for all y R, but y y whenever y 0, so we d have to consider both positive and negative square roots if we tried to undo the squaring. You have to plug x = 5 into the equation to check that it is a solution. 5.2 Exercise Exercise 13. Show that the solution (x, y, z) to the system of equations x + y z = 1 x + 2y + z = 0, 3x y z = 2 if it exists, is unique. 6 Proof by cases Sometimes it happens that you need to consider two different possibilities in your proof, and these cases need to be treated differently. In this case, you should label your cases as below, or in some other way signal to the reader what you are proving and when. Remember to inform the reader that you are splitting up the proof into cases. Proof. We split the proof up into k cases. Case 1 : Suppose that [conditions defining Case 1] proof of the result in Case 1 Case 2 : Suppose that [conditions defining Case 2] proof of the result in Case 2 Case k: Suppose that [conditions defining Case k]. 8
proof of the result in Case k Since these cases cover all possibilities, the result holds. 6.1 Examples Proposition. For all r R, r r r. Proof. We split the proof up into two cases: r 0 and r < 0. Case 1, r 0: Suppose that r 0. Then r = r. Certainly r r = r, and r 0 r = r. Hence, r r r in this case. Case 2, r < 0: Suppose now that r < 0. Then r = r. We have r < 0 < r = r, so that r r, and r = ( r) = r r. Hence, r r r in this case. So since these cases cover all possibilities for a real number, for all r R, r r r. Proposition. For all r R, r = r. Proof. We split the proof up into two cases: r 0 and r < 0. Case 1, r 0: Suppose that r 0. Then r 0, so that r = ( r) = r. But r = r, so we conclude that indeed r = r in this case. Case 2, r < 0: Suppose now that r < 0. Then r > 0, so that r = r. But r = r, so we conclude that indeed r = r in this case was well. So since these cases cover all possibilities for a real number, for all r R, r = r. 6.2 Exercises Exercise 14. Prove that for all x, y R, x y = xy. Exercise 15. Prove that if n Z is odd, then 8 (n 3 1). Exercise 16. Prove that if n Z is not a multiple of 5, then 5 (n 4 1). 9