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Exercise 8b Given that X = Whole numbers less than 24, P = Prime numbers less than 24, Q = Even numbers less than 24. Find (a) P ∩ Q, (b) P ∪ Q, (c) P' ∩ Q, (d) (P ∪ Q)'

Mathematics

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Step-by-step solution

4 steps
  1. List the elements of sets Q and P.

    • Q={all perfect squares less than 30}Q = \{\text{all perfect squares less than } 30\} The perfect squares are 12=11^2=1, 22=42^2=4, 32=93^2=9, 42=164^2=16, 52=255^2=25. The next is 62=366^2=36, which is not less than 30. So, Q={1,4,9,16,25}Q = \{1, 4, 9, 16, 25\}.
    • P={all odd numbers from 1 to 10}P = \{\text{all odd numbers from } 1 \text{ to } 10\} The odd numbers between 1 and 10 (inclusive) are 1, 3, 5, 7, 9. So, P={1,3,5,7,9}P = \{1, 3, 5, 7, 9\}.
  2. Find (i) Q∩PQ \cap P.

    This is the set of elements common to both Q and P.

    Q∩P={1,4,9,16,25}∩{1,3,5,7,9}Q \cap P = \{1, 4, 9, 16, 25\} \cap \{1, 3, 5, 7, 9\} Q∩P={1,9}Q \cap P = \{1, 9\}

    The answer for (i) is {1,9}\boxed{\{1, 9\}}.

  3. Find (ii) Q∪PQ \cup P.

    This is the set of all elements in Q or P (or both).

    Q∪P={1,4,9,16,25}∪{1,3,5,7,9}Q \cup P = \{1, 4, 9, 16, 25\} \cup \{1, 3, 5, 7, 9\} Q∪P={1,3,4,5,7,9,16,25}Q \cup P = \{1, 3, 4, 5, 7, 9, 16, 25\}

    The answer for (ii) is {1,3,4,5,7,9,16,25}\boxed{\{1, 3, 4, 5, 7, 9, 16, 25\}}.

  4. Find (iii) (P∪Q)∩ϕ(P \cup Q) \cap \phi.

    First, we use the result for P∪QP \cup Q from Step 3:
    P∪Q={1,3,4,5,7,9,16,25}P \cup Q = \{1, 3, 4, 5, 7, 9, 16, 25\}.
    Now, we find the intersection of this set with the empty set ϕ\phi. The intersection of any set with the empty set is always the empty set.

    (P∪Q)∩ϕ={1,3,4,5,7,9,16,25}∩ϕ(P \cup Q) \cap \phi = \{1, 3, 4, 5, 7, 9, 16, 25\} \cap \phi (P∪Q)∩ϕ=ϕ(P \cup Q) \cap \phi = \phi

    The answer for (iii) is ϕ\boxed{\phi}.

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