Book edition 7th

Author(s) Kenneth H. Rosen

Pages 808 pages

ISBN 9780073383095

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Short Answer

Question:
a. Give an example to show that the inclusion in part (b) in exercise 40 may be proper.
b. Show that if f is one-to-one, the inclusion in part(b) in exercise 40 is an equality.

Answer:

The function is

$a) f (S) \cup f (T) = f (S \cup T) b) f (S \cap T) \subseteq f (S) \cap f (T)$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1:

Union: $A \cup B$ all elements that are either in A or in B.

Intersection: $A \cap B$ all elements that are both in A and in B.

X is a subset of Y if every element of X is also an element of Y.

Notation: $X \subseteq Y$

Step 2:

Given: (a)

$f : A \to B S \subseteq A and T \subseteq A f (S \cap T) = f (S) \cap f (T)$

For example:

Let $A = R, B = R, S = {0, 1} and T = { 1,2}$

$f(x) = \{1 if x is even 0 if x is odd$

Let us determine the image of every element is S and T.

$f (0) = 1 f (1) = 0 f (2) = 1$

$f (S)$ contains all elements that are

$f (S) \cap f (T) = {0, 1}$

$S \cap T$ contains all elements in both sets.

$f (S \cap T) = {0}$

We then note in this case $f (S \cap T) \subseteq f (S) \cap f (T)$ , but not $f (S \cap T) = f (S) \cap f (T)$

Step 3:

Given: (a)

$f : A \to B S \subseteq A and T \subseteq A$

is one-to-one

To proof: $f (S \cap T) \subseteq f (S) \cap f (T)$

PROOF

FIRST PART

Let $x \in f (S \cap T)$ . then there exists an element $y \in S \cap T$ such that $f (y) = x$ .

By the definition of the union

$y \in S \land y \in T$

$f (S)$ contains all elements that are the image of all an element in S.

$f (T)$ contains all elements that are the image of all an element in S.

$f (y) \in f (S) \land f (y) \in f (T)$

Since

$f (y) = x$

$x \in f (S) \land x \in f (T)$

By the definition of union:

role="math" localid="1668423347151" $x \in f (S) \cap f (T)$

By the definition of subset:

$f (S \cap T) \subseteq f (S) \cap f (T)$

Step 4:

FIRST PART

Let $x \in f (S \cap T)$ , then there exists an element $y \in S \cap T$ such that $f (y) = x$ .

By the definition of the union

$y \in S \land y \in T$

$f (S)$ contains all elements that are the image of all an element in S.

$f (T)$ contains all elements that are the image of all an element in S.

$f (y) \in f (S) \land f (y) \in f (T)$

Since $f (y) = x$

$x \in f (S) \land x \in f (T)$

By the definition of union:

$x \in f (S) \cap f (T)$

By the definition of subset:

$f (S \cap T) \subseteq f (S) \cap f (T)$

Hence, the solution is

Since $f (S) \cap f (T) \subseteq f (S \cap T)$ and $f (S \cap T) \subseteq f (S) \cap f (T)$ the two sets have to be equal:

$f (S) \cap f (T) = f (S \cap T)$

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Discrete Mathematics and its Applications

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Short Answer

Question:
a. Give an example to show that the inclusion in part (b) in exercise 40 may be proper.
b. Show that if f is one-to-one, the inclusion in part(b) in exercise 40 is an equality.

Step by Step Solution

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Step 2:

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Discrete Mathematics and its Applications

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Short Answer

Question: a. Give an example to show that the inclusion in part (b) in exercise 40 may be proper. b. Show that if f is one-to-one, the inclusion in part(b) in exercise 40 is an equality.

Step by Step Solution

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b. Show that if f is one-to-one, the inclusion in part(b) in exercise 40 is an equality.