Math 13, Spring 2013, Lecture B: Midterm Name Signature UCI ID # E-mail address Each numbered problem is worth 12 points, for a total of 84 points. Present your work, especially proofs, as clearly as possible. If you find yourself stuck somewhere, move on and come back to the problem later. For problems with multiple parts it is often possible to answer one part without having solved the previous ones, so be sure to give everything a try. You are not permitted the use of a calculator, phone, or other electronic aid. Academic dishonesty in any form will result in a score of zero.
1. Define f : Z Z by f : n 2n. (a) What is the domain of f? What is the codomain (target) of f? The domain of f is Z and the target of f is Z. (b) What is the range of f? Is f surjective? Explain your answer. (You don t need to give a thorough proof, but your answer should include the definition of surjective!) The range of f is 2Z, the set of even integers. Since the target is the set Z of all integers, the function is not surjective. Surjectivity of f would mean that for every element m of the target Z there would exist an element n of the domain Z such that m = f(n), but given any odd m, there is no such n. (c) Is f injective? Prove your answer. Yes, f is injective. Proof. Let n 1, n 2 Z. Then f(n 1 ) = f(n 2 ) 2n 1 = 2n 2, by definition of f Thus f satisfies the definition of injectivity. n 1 = n 2, dividing both sides by 2.
2. Prove by induction that for every natural number n 1 1 2 + 1 2 3 + 1 3 4 + + 1 n(n + 1) = n n + 1. Proof. We will prove the claim (as directed) by induction on n. Base case. Clearly 1 = 1, so the equation above holds when n = 1. 1 2 1+1 Inductive step. Now assume that for a given n N the above equation holds. Then 1 1 2 + 1 2 3 + 1 3 4 + + 1 n(n + 1) = n n + 1 1 according to the inductive hypothesis we ve just made, so, adding to both sides, (n+1)(n+2) we obtain 1 1 2 + 1 2 3 + 1 3 4 + + 1 n(n + 1) + 1 (n + 1)(n + 2) = n n + 1 + 1 (n + 1)(n + 2) n(n + 2) = (n + 1)(n + 2) + 1 (n + 1)(n + 2) = n2 + 2n + 1 (n + 1)(n + 2) (n + 1) 2 = (n + 1)(n + 2) = n + 1 n + 2, proving that the equation above holds with n replaced by n + 1, which completes the inductive step and the proof. Remarks. It is essential to make your proofs as clear and easy to read as you can. In particular, in an inductive proof you need to indicate that you are making and using an inductive hypothesis. Generally, you should make all of your assumptions explicit, you should introduce all notation you use that is not completely standard (so, for example, don t start writing P (n) without defining what you mean), you should write in complete sentences (although it s fine to use symbols and abbreviations in place of English words), and you should explain the logical connection between the various sentences and equations you write down. You should think of the proof as a short essay, which you re writing in order to explain why the proposition you re proving is true.
3. Let A = {x R x 2 1 = 0} and let B = {x 2 1 x R}. (a) Write A in roster notation. Evidently x A x 2 1 = 0 x = ±1, so (b) Write B in interval notation. A = {1, 1}. We have y B ( x R)(y = x 2 1) y [ 1, ) (which you can see, for instance, by thinking of the graph of the parabola given by y = x 2 1), so (c) Write B\A as a union of two intervals. B = [ 1, ). By definition x B\A (x B x A), so we cut out 1 and 1 from [ 1, ) to get B\A = ( 1, 1) (1, ). (d) Evaluate A\B. Since A B, the complement of B in A is empty: A\B =.
4. Prove that a, b Z (a + b) 3 (a 3 + b 3 ) (mod 3). Proof. Let a, b Z. Then (a + b) 3 (a 3 + b 3 ) = (a + b)(a 2 + 2ab + b 2 ) a 3 b 3 = 3(a 2 b + ab 2 ), but since a, b Z, we have a 2 b + ab 2 Z as well, so the left-hand side of the above equation is divisible by 3. Equivalently, (a + b) 3 (a 3 + b 3 ) (mod 3).
5. We have studied the logical or operator, which is sometimes called the inclusive disjunction. For some purposes it is convenient to introduce also xor, the exclusive disjunction, which can be defined by the following truth table. P Q P Q (P Q) T T F T T F T F F T T F F F F T (a) Complete the truth table for (P Q) above. See above. (b) Fill in the blank ( ) below with one of the logical symbols we have studied (,,,,,, or ) to make the following propositional form a tautology: [ (P Q)] [P Q] This can be seen from the above truth table, since P Q means that P and Q have the same truth value as one another. (c) Using just the familiar operators,, and, write a propositional form logically equivalent to P Q. Looking at the truth table above, or using the result of (b), we can see that P Q is logically equivalent to (P Q) (P Q). [Of course there are many equivalent correct answers, such as to name one more.] (P Q) ( P Q)
6. Recall that two integers a and b are called relatively prime if their greatest common divisor is 1. Prove that n and n + 1 are relatively prime for every natural number n. (Hint: do not attempt an inductive proof.) Proof. Let n N. Suppose d Z divides n and n + 1. Then by definition of divisibility there exist k, l Z such that n = kd and n + 1 = ld. Subtracting the first equation from the second, we get 1 = (l k)d, but the only divisors of 1 are 1 and 1, so (since 1 < 1) the greatest common divisor of n and n + 1 is 1. Remarks. If you wanted to use the Euclidean algorithm in this problem, that s a fine approach, but in this case you absolutely must mention that you re using the Euclidean algorithm and explain (briefly) what you re doing (rather than just writing down a couple equations).
7. Some problems with quantifiers. (a) Is the proposition ( y R)( x R)(y = 2x) true or false? Prove your answer. The proposition is true. Proof. Given y R, let x = 1 y. Then x R and y = 2x. 2 (b) Is the proposition ( x R)( y R)(y = 2x) true or false? Prove your answer. The proposition is false. Proof. We will prove the negation of the proposition, namely ( x R)( y R)(y 2x). Indeed, given x R, let y = 2x + 1. Then y R and y 2x, because 2x + 1 = 2x 1 = 0. Here s an alternative approach. Proof. We will prove the proposition is false by assuming it is true and deriving a contradiction. Suppose then that x R satisfies y = 2x for every y R. Then in particular, taking y = 0 yields 0 = 2x so that x = 0, but taking y = 1 yields 1 = 2x so that x = 1. Since 0 1, the original proposition must be false. 2 Remarks on (a) and (b). Some students wanted to prove (a) or the negation of (b) by giving examples, but it is impossible to prove a statement by giving an example (though an example can be helpful for understanding what s going on). On the other hand, one can (and often does) prove an statement by giving an example. Similarly a good method to prove that a statement is false is to give a counterexample, but one cannot give a counterexample to prove that a statement is false (nor is the term counterexample even applicable in this context). (c) Consider the proposition ( G N)( M N)( primes p, q [M, ))( p q < G). Suppose you wanted to try to prove this proposition by contradiciton. What would be your starting assumption? Write your final answer so that it does not contain any negation symbols ( or ). (Aside: Actually a complete proof would be very difficult; a mathematician at the University of New Hampshire, Yitang Zhang, proved the above proposition, related to the twin prime conjecture, about two years ago and has consequently become very famous within the mathematics community.) To prove by contradiction that a given proposition is true, one starts by assuming the negation of that proposition. Here then we have to negate the proposition. Just repeatedly apply the logical identities ( x)(p (x)) ( x)( P (x)) and ( x)(p (x)) ( x)( P (x)):
( G N)( M N)( primes p, q [M, ))( p q G)
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