*χ*_{A∪C} (x) + *χ*_{B∪C} (x) = *χ*_{A∪B∪C}(x).

**False**.

Since C is non-empty, we may take x ∈ C.

Then x ∈ A∪C, B∪C, A∪ B ∪ C and so:

*χ*_{A∪C}(x) + *χ*_{B∪C}(x) = 1 + 1 ≠ 1 = *χ*_{A∪B∪C}(x).

*χ*_{A∩C}(x) + *χ*_{A∪C}(x) = 2 · *χ*_{A}(x).

**False**.

Let A := {3}, C := {5}.

Then A ∩ C = ∅, A ∪ C = {3, 5} and so:

*χ*_{A∩C}(5) + χ_{A∪C}(5) = 0 + 1 **≠** 0 = 2 · *χ*_{A}(5).

*χ*_{A×B}(x,y) = *χ*_{A}(x)_{χ}B(y)

**True**.

We need to check four cases:

- x ∈ A and y ∈ B, so (x,y) ∈ A×B ⇒
*χ*_{A×B}(x,y) = 1 = 1·1 =*χ*_{A(x)}χ_{B}(y) - x ∉ A and y∈ B, so(x,y) ∉ A×B ⇒
*χ*_{A×B}(x,y) = 0 = 0·1 =*χ*_{A}(x)*χ*_{B}(y) - x ∈ A and y∉ B, so(x,y) ∉ A×B ⇒
*χ*_{A×B}(x,y) = 0 = 1·0 =*χ*_{A}(x)*χ*_{B}(y) - x ∉ A and y ∉ B, so(x,y) ∉ A×B ⇒
*χ*_{A×B}(x,y) = 0 = 0·0 =*χ*_{A}(x)*χ*_{B}(y)

A×(B∩C) = (A×B) ∩ (A×C).

**True**.

Suppose (x,y) ∈ A × (B∩C), then x ∈ A and y ∈ B∩C.

In particular y ∈ B and y ∈ C, so (x,y) ∈ A×B and (x,y) ∈ A×C.

Hence (x,y) ∈ (A×B) ∩ (A×C).

Conversely, if (x,y) ∈ (A×B) ∩ (A×C), then (x,y) ∈ A × B, so x ∈ A and y ∈ B. Also, (x,y) ∈ A×C, so x ∈ A and y ∈ C.

Since y ∈ B and y ∈ C, we have y ∈ B ∩ C.

Hence (x,y) ∈ A×(B∩C).

Let X, Y and Z be any sets and let f : X → Y and g : Y → Z be two functions.

If f and g are injective, then g ◦ f is injective.

**True**.

Suppose x_{1},x_{2} ∈ X are such that (g◦f)(x_{1}) = (g◦f)(x_{2}).

This means, by definition, that g(f(x_{1})) = g(f(x_{2})).

Since g is injective, we must have f(x_{1}) = f(x_{2}).

Since f is also injective, x_{1} = x_{2}.

This proves that the composition g ◦ f is injective.

For any real numbers x_{1},···x_{n},

**True**.

For each fixed value of i, we may divide the set {1,...,*n*} into

{j : j < i} ∪ {i} ∪ {j : j > n} and hence divide the sum into three parts.

See answer sheet 2 for more detailed answer.

If *n* is odd, then *n* divides Σ*k*.

**True**.

We know (or can prove by induction).

If *n* is odd, then ^{1}/_{2}(*n* + 1) is an integer and hence the sum on the left is an integer multiple 2 of *n*.

Consider the relation ∼ on ℤ where a*∼*b if and only if a−b is a power of 2. Then ∼ is an equivalence relation.

**False**.

The relation is not reflexive since 0 is not a power of 2.

Furthermore, it is not transitive, since 11 ∼ 9 (as 11 − 9 = 2^{1}) and 9 ∼ 5 (as 9 − 5 = 2^{2}), but 11 ∼/∼ 5, since 11 − 5 = 6 is not a power of 2.

Consider the relation ∼ on ℤ where a ∼ b if and only if a−b is divisible by 3. Then ∼ is an equivalence relation.

**True**.

It is reflexive: if *a* ∈ ℤ, then a−a = 0 is divisible by 3.

It is symmetric: if a ∼ b, then a−b is divisible by 3, and so b−a is also divisible by 3, so b ∼ a.

It is transitive: if a ∼ b and b ∼ c, then a−b and b−c are divisible by 3, then so is their sum a−b+b−c = a−c, but this means that a ∼ c.

Let A and B be finite sets with card(A) = card(B). Then there exists a bijection between A and B.

**True**.

Suppose n := card(A) = card(B).

Then we may write out the elements of *A* = {a_{1},...,a_{n}} and *B* = {b_{1},...,b_{n}}, since each set has n elements, there are no repeats in either set.

We define f : A → B by f(a_{i}) := b_{i} for 1 ≤ i ≤ n.

We claim this is a bijection. Given an element b∈B, we know that b=b_{i} for some 1 ≤ i ≤ n, but then f(a_{i}) = b_{i} = b.

This shows that f is surjective. Suppose that x_{1}, x_{2} ∈ A are such that f(x_{1}) = f(x_{2}).

Thenx_{1} =a_{i},x_{2} =a_{j} for some 1 ≤ i, j ≤ n. This means that b_{i} = f(a_{i}) = f(x_{1}) = f(x_{2}) = f(a_{j}) = b_{j}. But since we have listed the elements of B without repeating any elements, we must have i = j and hence x_{1} = a_{i} = a_{j} = x_{2}. This shows that f is injective and hence a bijection.