Name: Student number: METU MATH 111, Midterm 2 ^{1 2 3 4 5 Σ}

Thursday, December 22, 2011, at 17:40 (100 minutes), totally 60 points

Instructors: E.Emelyanov, G.Ercan, S.Finashin, E.Solak Instructions: Please, show clearly the logic of your solutions.

Problem 1. (13 pts.) Let f : {1, 2, 3, 4} → {5, 6, 7} be the function defined by the following graph.

1 •  • 5
2 •  • 6
3 •  • 7

4 •

a) Find the following images and preimages: f(2) = f({2, 4}) = f^{−1}(5) = f^{−1}({5, 7}) =

b) Find a subset A ⊆ {1, 2, 3, 4} such that the function f_{A} : A → {5, 6, 7} (the restriction function of f to A) is invertible. Find the inverse function f_{A}^{−1}.

Problem 2. (13 pts.) Consider the function f : ℝ − \{2\} ⟶ ℝ defined by f(x) = \frac{3x}{x−2} .

(a) Show that f is one-to-one.

Solution. Let x_{1}, x_{2} ∈ ℝ − \{2\} such that f(x_{1}) = f(x_{2}). Then \frac{3x_{1}}{x_{1} − 2} = \frac{3x_{2}}{x_{2} − 2} and hence

3x_{1}x_{2} − 6x_{1} = 3x_{2}x_{1} − 6x_{2}. It follows that x_{1} = x_{2}. Thus f is one-to-one.

(b) Find the range of f.

Solution. For every x ∈ ℝ − \{2\}, \frac{3x}{x − 2} ≠ 3, because otherwise 0 = −6, which is impossible.

Hence the range of f is a subset of ℝ − {3}.

Conversely if y ∈ ℝ − \{3\}, then f(\frac{−2y}{3 − y}) = y, that is, y belongs to the range of f.

As a result the range of f is ℝ − {3}.

(c) For x which belongs to the range of f, find f^{−1}(x).

Solution. For x which belongs to the range of f, f(x) = y ⟺ x = f^{−1}(y).

By a similar argument as in part (b), we see that f^{−1}(x) = \frac{−2x}{3 − x}.

Problem 3. (8 pts.) Let f : X → Y . Prove that f(A ∪ B) = f(A) ∪ f(B) for any subsets A and B of X.

Solution. \underline{f(A ∪ B) ⊂ f(A) ∪ f(B)}

For any y ∈ f(A ∪ B), there exists x ∈ A ∪ B such that y = f(x). Since x ∈ A ∪ B, we have either x ∈ A, or x ∈ B.

\begin{matrix}x ∈ A ⟹ y ∈ f(A) \\ x ∈ B ⟹ y ∈ f(B) \end{matrix} Thus, y ∈ f(A) ∪ f(B).

\underline{f(A ∪ B) ⊃ f(A) ∪ f(B)}

Let us take any y ∈ f(A) ∪ f(B), then either y ∈ f(A) or y ∈ f(B). If y ∈ f(A), then there exists x ∈ A such that y = f(x). If y ∈ f(B), then there exists x ∈ B such that y = f(x). So, there exists x ∈ A ∪ B such that y = f(x). Thus, y ∈ f(A ∪ B).

Problem 4. (13 pts.) Consider the partial order relation “⊆” on the power set P(A) of A = {1, 2, 3, 4, 5}. Let B = {{1, 2}, {2, 3}, {2, 3, 4}, {4}}.

(a) Does B have a smallest element with respect to “⊆”?

Solution. No. A smallest element should be a subset of all other elements of B, but none of the elements of B is a subset of all the others.

(b) Find the minimal elements of B with respect to “⊆”, if any.

Solution. There are three minimal elements in B: {1, 2}, {2, 3}, and {4}. They do not contain any other element of B. The remaining element {2, 3, 4} is not minimal because it contains {2, 3} (and {4}).

(c) Find the least upper bound of B in P(A) with respect to “⊆”, if any.

Solution. {1, 2, 3, 4} is the least upper bound of B in P(A). This set contains all the elements of B as subsets, and it is the smallest set with such a property.

Problem 5. (13 pts.) Let A = {1, 2, 3, 4, 5} and let R be the relation defined on the set B = A × A as follows:

(x_{1}, y_{1})R(x_{2}, y_{2}) iff x_{1} + y_{1} = x_{2} + y_{2}, for all (x_{1}, y_{1}), (x_{2}, y_{2}) in B.

Solution. (a) Show that R is an equivalence relation on B. We have to show that R is reflexive, symmetric and transitive.

(i) Since for all (x, y) ∈ B, x + y = x + y, the relation R is reflexive.

(ii) Let (x_{1}, y_{1}) and (x_{2}, y_{2}) be arbitrary elements of B. Then since ((x_{1}, y_{1}), (x_{2}, y_{2})) ∈ R → x_{1} + y_{1} = x_{2} + y_{2} → x_{2} + y_{2} = x_{1} + y_{1} → ((x_{2}, y_{2}), (x_{1}, y_{1})) ∈ R. So R is symmetric. (iii)Let (x_{1}, y_{1}) , (x_{2}, y_{2}) and (x_{3}, y_{3}) be arbitrary elements of B. Then since ((x_{1}, y_{1}), (x_{2}, y_{2})) ∈ R → x_{1} + y_{1} = x_{2} + y_{2} and ((x_{2}, y_{2}), (x_{3}, y_{3})) ∈ R → x_{2} + y_{2} = x_{3} + y_{3} → x_{1} + y_{1} = x_{3} + y_{3} → ((x_{1}, y_{1}), (x_{3}, y_{3})) ∈ R. So R is transitive. Since R is reflexive, symmetric and transitive, R is an equivalence relation.

(b) Determine the equivalence classes [(1, 3)], [(2, 4)] and [(1, 1)] of elements (1, 3), (2, 4) and (1, 1) in B, respectively.

[(1, 3)] = {(1, 3), (2, 2), (3, 1)}

[(2, 4)] = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}

[(1, 1)] = {(1, 1)}

(c) Determine the partition of B induced by R.

A/R = {(1, 1)} ⋃{(1, 2), (2, 1)} ⋃{(1, 3), (2, 2), (3, 1)} ⋃{(1, 4), (2, 3), (3, 2), (4, 1)} ⋃{(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)} ⋃{(2, 5), (3, 4), (4, 3), (5, 2)} ⋃{(3, 5), (4, 4), (5, 3)} ⋃{(4, 5), (5, 4)} ⋃{(5, 5)}.