## 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!!

Click on question number to see the answer

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 ).

• Show that f is onto .
• Show that f is not a one-one function.

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.

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.

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 = .

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.

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.

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.

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.

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.

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).