Math 3000     Exam II     Sample

Exam totals 100 points. You must do problem 1 and then any 4 problems from the remaining 8. Show all work on these sheets. Clearly mark the questions you want to have graded. Good Luck!!

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1. Define the following terms (do not just indicate notation):

  1. A function f: A B is a surjection if ...
  2. The binomial coefficient C(n,k) counts the number of ...
  3. A set of natural numbers, S, is inductive if ...
  4. A function f: A B is a bijection if ....
  5. The binomial coefficient C(7,4) has numerical value equal to ...

2. Use induction to prove that if S is a set with n elements, then Ã(S), the power set of S, has 2n elements.

3. Let f: be given by f(x) = [5x -4] . ( [ ..] denotes the floor function, round down to the nearest integer, so for example: [4.325] = 4, [3.999] = 3, [ -2.34] = -3, [16] = 16 ).

4. Use induction to prove that for all natural numbers n,

(1/3)n3 + (1/5)n5 + (7/15)n     is an integer.

5. Consider the set Nm = {1,2,...,m}. If 1/4 of all the 3 element subsets of Nm contain the element 5, what is m ?

6. Given that f: A B is a bijection and that g: B C is a bijection, prove that g o f : A C is a bijection.

7. Suppose that there are 25 points in the plane and a certain number of lines which meet each other only at these points. If there are 6 lines through each of the points and each line contains exactly 5 of the points, how many lines are there in total?

8. Give a combinatorial proof of the binomial identity:

9. Use the Well Ordering Principle to prove that every natural number n > 1 has a prime factor.



Answer to question 1:

  1. A function f: A B is a surjection if ...the range of f equals the codomain (B).
  2. The binomial coefficient C(n,k) counts the number of ... subsets of size k in a set of size n.
  3. A set of natural numbers, S, is inductive if ... whenever i S then i+1 S.
  4. A function f: A B is a bijection if ....whenever f(x) = f(y) then x = y.
  5. The binomial coefficient C(7,4) has numerical value equal to ... 35.
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Answer to question 2:

Let K = {n : if |S| = n then |Ã(S)| = 2n}. If |S| = 1, then S = {x} and the power set of S contains just two elements, the empty set and the set S itself. Since 2 = 21, 1 K. Now suppose that m K and consider a set S with |S| = m+1. Remove an element x from S. The new set S' = S - {x} has 2m subsets since |S'| = m. All other subsets of S' contain the element x and are of the form {x} a subset of S'. There are 2m of these as well. So, the total number of subsets of S is 2m + 2m = 2(2m) = 2m+1. Thus, m+1 K and K is an inductive set. By the PMI, K = .

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Answer to question 3:

a) Let y. For x = (y+4)/5 we have f(x) = f((y+4)/5) = [ 5(y+4)/5 -4] = [y] = y (since y is an integer.) Thus, any element in the codomain is in the range and f is onto.

b) f(1) = 1 and f(1.1) = [5.5 - 4] = [1.5] = 1, so f is not a one-one function.

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Answer to question 4.

Let A be the set of natural numbers for which the statement is true. The statement, with n = 1, is 1/3 + 1/5 + 7/15 is an integer. As the sum evaluates to 1, the statement is true in this case, and so, 1 is in A. Now, assume that the statement is true for the integer m, and consider n = m + 1. We have:
(m+1)3/3 + (m+1)5/5 + 7(m+1)/15 = m3/3 + (3m2 + 3m + 1)/3 + m5/5 + (5m4 + 10m3 + 10m2 + 5m + 1)/5 + 7m/15 + 7/15 = an integer + (15m4 + 30m3 + 45m2 + 30m + 15)/15 = an integer + m4 + 2m3 + 3m2 + 2m + 1 which is an integer, showing that the statement is true for n = m + 1. Thus, A is an inductive set and by PMI the statement is true for all n.

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Answer to question 5.

There are C(m,3) 3 element subsets. A 3 element subset containing 5 consists of 5 and two other elements from the remaining m-1 elements. Thus, there are C(m-1,2) of these. Therefore, we have ¼ C(m,3) = C(m-1,2). So, m(m-1)(m-2)/4(3)(2)(1) = (m-1)(m-2)/(2)(1) and so m = 12.

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Answer to question 6:

We must show that g o f is onto and one-to-one.
Suppose that g o f(x) = g o f(y). This means that g(f(x)) = g(f(y)). Since g is one-to-one, we have that f(x) = f(y). Since f is one-to-one this implies that x = y. Therefore, g o f is one-to-one.

Now, let c be an arbitrary element of C. Since g is onto, there is an element b in B with g(b) = c. Since f is onto, there is an element a in A so that g(a) = b. Now, consider g o f (a) = g(f(a)) = g(b) = c. Therefore, c is in the image of g o f and since c was arbitrary, image (g o f) = C, i.e., g o f is onto.

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Answer to question 7.

Consider the set of ordered pairs (P,L) where P is one of the points and L is one of the lines that contains the point P. We can count the size of this set in two ways. If we let b = the total number of lines, then the number of ordered pairs is 5b (for each possible line L there are 5 choices for the point P). On the other hand, the number of ordered pairs is 25(6), since for each possible point P there are 6 choices for L. Since we are counting the same thing in different ways, we must have 5b = 25(6) or b = 30.

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Answer to question 8:

Use the binomial identity C(n,k) = C(n-1,k) + C(n-1,k-1) to get

C(2n,n) = C(2n-1,n) + C(2n-1,n-1).

Since C(2n-1,n) = C(2n-1, 2n-1 -n) = C(2n-1,n-1), we have C(2n,n) = 2 C(2n-1,n-1).

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Answer to question 9.

Let n be a natural number greater than 1. If n is prime, then it divides itself and we are done. Suppose that n is not prime. Let S be the set of all divisors of n which are greater than 1. S is non-empty since n is not prime. By the WOP, S has a smallest element. Call this smallest element d. If d is not prime, then it has a divisor k with 1 < k < d. But, since k divides d and d divides n, k would be a factor of n which is less than d. This contradiction shows that d is a prime and so n has a prime divisor.

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