Question

Prove that if \({A_1},{A_2}, \ldots ,{A_n}\)and \(B\) are sets then,.\(\left( {{A_1} - B} \right) \cup \left( {{A_2} - B} \right) \cup \ldots \cup \left( {{A_n} - B} \right) = \left( {{A_1} \cup {A_2} \cup \ldots \cup {A_n}} \right) - B\)

Short answer

A

It is proved that if \({A_1},{A_2}, \ldots ,{A_n}\) and \(B\) are sets then, \(\left( {{A_1} - B} \right) \cup \left( {{A_2} - B} \right) \cup \ldots \cup \left( {{A_n} - B} \right) = \left( {{A_1} \cup {A_2} \cup \ldots \cup {A_n}} \right) - B\).

Step-by-step solution

Mathematical Induction

The principle of mathematical induction is to prove that \(P\left( n \right)\) is true for all positive integer n in two steps.

1. Basic step : To verify that \(P\left( 1 \right)\) is true.

2. Inductive step : To prove the conditional statement if \(P\left( k \right)\) is true then \(P\left( {k + 1} \right)\) is true.

Proof by induction

Let \(P\left( n \right)\) be the statementas\(\left( {{A_1} - B} \right) \cup \left( {{A_2} - B} \right) \cup \ldots \cup \left( {{A_n} - B} \right) = \left( {{A_1} \cup {A_2} \cup \ldots \cup {A_n}} \right) - B\). Let us prove by induction on n.

Base Case:

For \(n = 1\), the value of LHS is \({A_1} - B\) and RHS is \({A_1} - B\). Since, both are equal the statement is true for \(P\left( 1 \right)\).

Induction Case:

Assume that the statement is true for \(P\left( k \right)\) then prove for \(P\left( {k + 1} \right)\).

Let \(P\left( k \right)\) be true then\(\left( {{A_1} - B} \right) \cup \left( {{A_2} - B} \right) \cup \ldots \cup \left( {{A_k} - B} \right) = \left( {{A_1} \cup {A_2} \cup \ldots \cup {A_k}} \right) - B\). Let us prove the statement is true for \(P\left( {k + 1} \right)\) as follows:

\(\begin{array}{c}\left( {{A_1} - B} \right) \cup \left( {{A_2} - B} \right) \cup \ldots \cup \left( {{A_k} - B} \right) \cup \left( {{A_{k + 1}} - B} \right) = \left( {{A_1} \cup {A_2} \cup \ldots \cup {A_k}} \right) - B \cup \left( {{A_{k + 1}} - B} \right)\\ = \left( {\left( {{A_1} \cup {A_2} \cup \ldots \cup {A_k}} \right) \cup {A_{k + 1}}} \right) - B\\ = \left( {{A_1} \cup {A_2} \cup \ldots \cup {A_k} \cup {A_{k + 1}}} \right) - B\end{array}\)

Thus, \(P\left( {k + 1} \right)\) is also true.

Therefore, the statement \(P\left( n \right)\) is true for every positive integer n.