# IB DP Maths Topic 8.1 De Morgan’s laws: distributive, associative and commutative laws (for union and intersection) HL Paper 3

## Question

Given the sets $$A$$ and $$B$$, use the properties of sets to prove that $$A \cup (B’ \cup A)’ = A \cup B$$, justifying each step of the proof.

## Markscheme

$$A \cup (B’ \cup A)’ = A \cup (B \cap A’)$$     De Morgan     M1A1

$$= (A \cup B) \cap (A \cup A’)$$     Distributive property     M1A1

$$= (A \cup B) \cap U$$     (Union of set and its complement)     A1

$$= A \cup B$$     (Intersection with the universal set)     AG

Note:     Do not accept proofs using Venn diagrams unless the properties are clearly stated.

Note:     Accept double inclusion proofs: M1A1 for each inclusion, final A1 for conclusion of equality of sets.

[5 marks]

[N/A]

## Question

Consider the sets A = {1, 3, 5, 7, 9} , B = {2, 3, 5, 7, 11} and C = {1, 3, 7, 15, 31} .

Find $$\left( {A \cup B} \right) \cap \left( {A \cup C} \right)$$.

[3]
a.i.

Verify that A \ C ≠ A.

[2]
a.ii.

Let S be a set containing $$n$$ elements where $$n \in \mathbb{N}$$.

Show that S has $${2^n}$$ subsets.

[3]
b.

## Markscheme

EITHER

$$\left( {A \cup B} \right) \cap \left( {A \cup C} \right) = \left\{ {1,\,2,\,3,\,5,\,7,\,9,\,11} \right\} \cap \left\{ {1,\,3,\,5,\,7,\,9,\,15,\,31} \right\}$$    M1A1

OR

$$A \cup \left( {B \cap C} \right) = \left\{ {1,\,3,\,5,\,7,\,9,\,11} \right\} \cup \left\{ {3,\,7} \right\}$$     M1A1

OR

$${B \cap C}$$ is contained within A     (M1)A1

THEN

= {1, 3, 5, 7, 9} (= A)     A1

Note: Accept a Venn diagram representation.

[3 marks]

a.i.

A \ C = {5, 9}     A1

= {15, 31}    A1

so A \ C ≠ A     AG

Note: Accept a Venn diagram representation.

[2 marks]

a.ii.

METHOD 1

if $$S = \emptyset$$ then $$n = 0$$ and the number of subsets of S is given by 20 = 1     A1

if $$n > 0$$

for every subset of S, there are 2 possibilities for each element $$x \in S$$ either $$x$$ will be in the subset or it will not     R1

so for all $$n$$ elements there are $$\left( {2 \times 2 \times \ldots \times 2} \right){2^n}$$ different choices in forming a subset of S      R1

so S has $${2^n}$$ subsets      AG

Note: If candidates attempt induction, award A1 for case $$n = 0$$, R1 for setting up the induction method (assume $$P\left( k \right)$$ and consider $$P\left( {k + 1} \right)$$ and R1 for showing how the $$P\left( k \right)$$ true implies $$P\left( {k + 1} \right)$$ true).

METHOD 2

$$\sum\limits_{k = 0}^n {\left( \begin{gathered} n \hfill \\ k \hfill \\ \end{gathered} \right)}$$ is the number of subsets of S (of all possible sizes from 0 to $$n$$)     R1

$${\left( {1 + 1} \right)^n} = \sum\limits_{k = 0}^n {\left( \begin{gathered} n \hfill \\ k \hfill \\ \end{gathered} \right)} \left( {{1^k}} \right)\left( {{1^{n – k}}} \right)$$      M1

$${2^n} = \sum\limits_{k = 0}^n {\left( \begin{gathered} n \hfill \\ k \hfill \\ \end{gathered} \right)}$$ (= number of subsets of S)     A1

so S has $${2^n}$$ subsets      AG

[3 marks]

b.

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.