Discrete Mathematics MATH 302 - 501 & 502, Spring 2014

Assignment 4due Feb 17, 2014

Exercise 1 (10 points).Prove or disprove that every nonnegative integer can be written as the sum of three (notnecessarily distinct) square numbers (0, 1, 4, 9, 16, 25, . . .).

Solution to Exercise 1.7 cannot be written as the sum of 3 squares! Assume for the sake of contradiction that 7 canbe written as the sum of three squares. The sum does clearly not involve 9, 16, 25, and largernumbers. If the sum would not involve at least one 4, then its value would bounded from aboveby 1 + 1 + 1 = 3 < 7. Hence the sum must involve at least on 4. Clearly, since 4 + 4 = 8 > 7,the sum has exactly one 4. But then its value is bounded from above by 4 + 1 + 1 = 6 < 7,which is a contradiction.

Exercise 2 (40 points).Prove or disprove that an 8 × 8 chessboard from which all 4 corner cells are removed can becovered by 15 so-called T-tetrominos, which are pieces of the following form:

As in the lecture, the pieces can be rotated.

Solution to Exercise 2.The center cell of a T-tetromino is either black or white. If it is black, then the T-tetrominocovers 1 black and 3 white cells. If it is white, then the T-tetromino covers 1 white and 3 blackcells. Let n denote the number of T-tetrominos with 1 black and 3 white cells and let m denotethe number of T-tetrominos with 1 white and 3 black cells. The chessboard has 64 − 4 = 60cells, 30 of each color. For a tiling we must have 30 = 3x+ y and 30 = 3y +x. But y = 30− 3ximplies 30 = 3(30−3x)+x = 90−8x. Therefore x = 60

8 = 7.25, which is not a natural number.Therefore the shape cannot be covered.

Exercise 3 (50 points).We use P(A) to denote the powerset of A. Prove or disprove the following propositions:

1. {∅} ⊆ {{∅}, ∅, {{∅}, ∅}}

2. |{{∅}, ∅, {{∅}, ∅}} − {{∅, {∅}}, {{∅}}}| = 2.

3. P({∅})− {∅} = {∅}

4. {∅} × ∅ = ∅

Discrete Mathematics MATH 302 - 501 & 502, Spring 2014

5. For every pair of sets (A,B) we have P(A×B) = P(A)×P(B)

Solution to Exercise 3.

1. Yes, because {∅} has only one element ∅ and ∅ ∈ {{∅}, ∅, {{∅}, ∅}} by roster notation.

2. The set on the left hand side has 3 elements and the set on the right hand side has 2elements. Only the element {∅, {∅}} appears on both sides. Therefore the set differencehas 3− 1 = 2 elements.

3. This is wrong, because and P({∅}) = {∅, {∅}} hence P({∅})− {∅} = {{∅}}

4. This is true, because there is no pair of elements (a, b) such that b ∈ ∅.

5. For example, this can be seen by comparing cardinalities. Let A = B = {a}. Then

|P(A×B)| = |P({(a, a)})| = 2 6= 4 = |{∅, {(a, a)}} × {∅, {(a, a)}}| = |P(A)×P(B)|.