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Expert-verified Found in: Page 155 ### Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095 # Let S be a subset of a universal set U. The characteristic function of S is the function from U to the set {0,1} such that fs(x) = 1 if x belongs to S and if x does not belong to S and fs(x) = 0. Let A and B be sets. Show that for all $x\in U$ $\begin{array}{r}\left(a\right){f}_{A\cap B}\left(x\right)={f}_{A}\left(x\right)\cdot {f}_{B}\left(x\right)\\ \left(b\right){f}_{A\cup B}\left(x\right)={f}_{A}\left(x\right)+{f}_{B}\left(x\right)-{f}_{A}\left(x\right)\cdot {f}_{B}\left(x\right)\\ \left(c\right){f}_{\overline{A}}\left(x\right)=1-{f}_{A}\left(x\right)\\ \left(d\right){f}_{A\oplus B}\left(x\right)={f}_{A}\left(x\right)+{f}_{B}\left(x\right)-2{f}_{A}\left(x\right)\cdot {f}_{B}\left(x\right)\end{array}$

$\begin{array}{r}\left(a\right){f}_{A\cap B}\left(x\right)={f}_{A}\left(x\right)\cdot {f}_{B}\left(x\right)\\ \left(b\right){f}_{A\cup B}\left(x\right)={f}_{A}\left(x\right)+{f}_{B}\left(x\right)-{f}_{A}\left(x\right)\cdot {f}_{B}\left(x\right)\\ \left(c\right){f}_{\overline{A}}\left(x\right)=1-{f}_{A}\left(x\right)\\ \left(d\right){f}_{A\oplus B}\left(x\right)={f}_{A}\left(x\right)+{f}_{B}\left(x\right)-2{f}_{A}\left(x\right)\cdot {f}_{B}\left(x\right)\end{array}$

See the step by step solution

## Step: 1

a)

$\begin{array}{r}{f}_{A\cap B}\left(x\right)=1\\ ↔x\in A\cap B\\ ↔x\in \text{Aand}x\in B\\ ↔{f}_{A}\left(x\right)=1\text{and}{f}_{B}\left(x\right)=1\\ ↔{f}_{A}\left(x\right)\cdot {f}_{B}\left(x\right)=1\end{array}$

## Step: 2

b)

$\begin{array}{r}{f}_{A\cup B}\left(x\right)=1\\ ↔x\in \text{Aor}x\in B\\ ↔{f}_{A}\left(x\right)={\mathrm{lorf}}_{B}\left(x\right)=1\\ ↔{f}_{A}\left(x\right)+{f}_{B}\left(x\right)-{f}_{A}\left(x\right)\cdot {f}_{B}\left(x\right)=1\end{array}$

## Step: 3

c)

$\begin{array}{r}{f}_{\overline{A}}\left(x\right)=1\\ ↔x\in \overline{A}\\ ↔x\notin A\\ ↔{f}_{A}\left(x\right)=0\\ ↔1-{f}_{A}\left(x\right)=1\end{array}$

d)

$\begin{array}{r}{f}_{A\oplus B}\left(x\right)=1\\ ↔x\in A\oplus B\\ ↔\left(x\in \text{Aand}x\notin B\right)\text{or}\left(x\notin \text{Aand}x\in B\right)\\ ↔{f}_{A}\left(x\right)+{f}_{B}\left(x\right)-2{f}_{A}\left(x\right)\cdot {f}_{B}\left(x\right)=1\end{array}$ ### Want to see more solutions like these? 