CS 1050: Constructing Proofs
Solutions to Problem Set 1 Problem 1
Direct Proofs (10 points) 1. Prove that for every real number x, x2 2x + 2 > 0. Proof. x2 2x + 2 = (x 1)2 + 1 (x 1)2 0 (x 1)2 + 1 > 0 2. Suppose a and b are integers such that a b is divisible a3 b3 is also divisible by 7.
by 7. Prove that a2 b2 is also divisible by 7. Prove that Proof. a2 b2 = (a b)(a + b) Since (a b) is divisible by 7, therefore (a2 b2 ) is also divisible by 7. a3 b3 = (a b)(a2 + ab + b2 ) Since (a b) is divisible by 7, therefore (a3 b3 ) is also divisible by 7.
Problem 2
Examples/Counter-examples (20 points) Prove or disprove each of the following propositions by giving an example or a counter-example. 1. There is a natural number p such that p, p + 2, p + 6 and p + 8 are all primes. Proof. [True] Let p = 5. Then, p = 5, p + 2 = 7, p + 6 = 11 and p + 8 = 13 are all primes.
2. If x and y are positive irrational numbers, then x y is also an irrational number. Proof. [False] Let x = 2, y = 2. Then x y = 2 2 = 2 is rational. 3. If x and y are positive rational numbers, then xy is also a rational number. Proof. [False] Let x = 2, y = 1 . Then xy = 2 2 = 2
1
2 is irrational.
4. If A and B are two nite sets, then |A B| = |A| + |B|. Here |A| denotes the number of elements in the set A. Proof. [False] Let A = {1, 2, 3, 4}, B = {3, 4, 5}. Then, A B = {1, 2, 3, 4, 5}. In this case, RHS = |A| + |B| = 4 + 3 = 7 and LHS = |A B| = 5.
Problem 3
Proofs by Contrapositive (10 points) 1. Let x and y be real numbers. Prove that if x + y is irrational, then either x or y is irrational. Proof. We prove the contrapositive, i.e. we prove that if x and y both are rational then x + y is also rational. Let x and y be both rational. Then we can write: a where b = 0 b c where d = 0 d a c ad + bc + = where b d bd
x = y = x+y = So, x + y is rational.
bd = 0
2. Let a and b be integers. Prove that if a b is odd, then a and b both are odd. Proof. We prove the contrapositive, i.e. if either a or b is even, then a b is even. Let a be even (the case when b is even is similar). Then, a = 2 x for some integer x. Now, a b = (2 x) b = 2 (b x). Therefore, a b is even.
Problem 4
Proofs by Contradiction (10 points) 1. Let x1 , x2 , . . . , xn be positive real numbers. Let y 1 be their average, i.e. y = n ( n xi ). Prove that y xi for some i {1, 2, . . . , n}. i=1 Proof. Suppose on the contrary that y > xi i {1, 2, . . . , n}. n
i=1
ny > y >
xi
1 n
n
i=1
xi
CONTRADICTION as
1 y= n
n
i=1
xi
Therefore, y xi for some i {1, 2, . . . , n}. 2. Prove that for any integer n, at least one of the three integers n, 2n + 1, 4n + 3 is not divisible by 7. Proof. Let all of n, 2n + 1, 4n + 3 be divisible by 7. n = 7a for some integer a. 2n + 1 = 7b for some integer b. 4n + 3 = 7c for some integer c. Therefore, adding the three equations we get 7n + 4 = 7(a + b + c) which implies 4 = 7(a + b + c n) which is a contradiction since 4 is not divisible by 7.
Problem 5
Proofs by Cases (10 points) 1. For any real numbers x and y, prove that |x + y| |x| + |y|. Here |x| denotes the absolute value of x which equals x if x 0 and equals x if x < 0. (Hint: There are four compared to zero). Proof. Case 1 : x 0, y 0 |x| = x as x 0 |y| = y as y 0 |x + y| = x + y as x 0, y 0 x + y 0. Therefore, |x + y| = |x| + |y|. Case 2 : x < 0, y < 0 |x| = x as x < 0 |y| = y as y < 0 |x + y| = (x + y) as x < 0, y < 0 x + y < 0. Therefore, |x + y| = |x| + |y|. Case 3 : x 0, y < 0 |x| = x as x 0 |y| = y as y < 0 y x + y < x as x 0, y < 0 If x + y 0 then |x + y| = x + y < x = |x|. Therefore, |x + y| < |x| + |y| (just adding a positive number to RHS). cases depending on whether x and y are greater-or-eqaul/less
If x + y < 0 then |x + y| = (x + y) y = |y|. Therefore, |x + y| |x| + |y| (just adding a positive number to RHS). Case 4 : x < 0, y 0 This is similar to Case 3 and can be proved similarly.
2. If n is a natural number (written in usual decimal notation), prove that the last digit of n4 is either 0, 1, 5 or 6. Proof. If n is a natural number, then it can be written as 10 a + i where i {0, 1, 2, . . . , 9}. Therefore, n4 = (10 a + i)4 = 10 k + i4 where k is rest of the expansion. So last digit of n4 is same as that of i4 . Case 0 : i = 0 i4 = 0 last digit of n4 = 0 Case 1 : i = 1 i4 = 1 last digit of n4 = 1 Case 2 : i = 2 i4 = 16 last digit of n4 = 6 Case 3 : i = 3 i4 = 81 last digit of n4 = 1 Case 4 : i = 4 i4 = 256 last digit of n4 = 6 Case 5 : i = 5 i4 = 625 last digit of n4 = 5 Case 6 : i = 6 i4 = 1296 last digit of n4 = 6 Case 7 : i = 7 i4 = 2401 last digit of n4 = 1 Case 8 : i = 8 i4 = 4096 last digit of n4 = 6 Case 9 : i = 9 i4 = 6561 last digit of n4 = 1
Problem 6
Puzzle (10 points) 5
There are seven jars with seven coins in each jar. In one of the jars, all coins are fake. All coins in all the remaining six jars are true coins. A fake coin looks exactly like a true coin but weighs less. A fake coin weighs 9 grams whereas a true coin weighs 10 grams. You are given a weighing scale. How will you nd out which jar has fake coins with only one weighing? Select 1 coin from rst jar, 2 coins from second jar, 3 coins from third jar, 4 coins from fourth jar, 5 coins from fth jar, 6 coins from sixth jar, and all 7 coins from seventh jar. Total number of coins selected is 1 + 2 + . . . + 7 = 28. If all coins were true then the total weight of selected coins would have been 280 grams. However coins in the k th jar are fake (k is unknown). Since we selected k coins from k th jar, and each fake coin weighs 1 gram less, the total weight of selected coins will be 280 k grams. Thus, we can weigh the selected coins and if their total weight is 280 i grams for some i, then declare that the ith jar has fake coins.
