Kenneth Rosen Edition 6th Exercise 7.5 Question 3 e (Page No. 507)

Question

Which of these relations on the set of all functions from Z to Z are equivalence relations? Determine the properties of an equivalence relation that the others lack.

 

{(f, g) | f(0) = g(1) and f(1) = g(0)}

 

In many places it has been said as not reflexive as f (0) != f (1). But I couldn't understand why this comparison is being made to check the reflexive property.

Why can't we check f(0)=f(0) to confirm the reflexive property. Please help.

Comments

i think its becuz they have said that the relation is equivalence relation...

so {0,1} and {1,0} means that it should also have {0,0} and {1,1} .

from {0,0} we can say that f(0) = g(0).......@1

but given that f(1)=g(0)................@2

but from ......@1 and ........@2

f(0) and f(1) needs to be equal...

which becomes a contradiction. so its not reflexive...and here they mean to say that for given conditions f(0) and f(1) must be equal which is false...

Answer 1

For Reflexivity Property:

Suppose , function $f$ supported this property, $f\left ( x \right )=x$

So, from this property, we can say, 

$f\left ( 0 \right )=0$

$f\left ( 1 \right )=1$

$f\left ( 0 \right )\neq f\left ( 1 \right )$

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Symmetry Property

It is given

$f\left ( 0 \right )= g\left ( 1 \right )$ and $f\left ( 1 \right )= g\left ( 0 \right )$

$f$ and $g$ working on same value in reverse.

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Transitivity Property

Say, $f\left ( 0 \right )=g\left ( 1 \right )=h\left ( 0 \right )=0$

$f\left ( 1 \right )=g\left ( 0 \right )=h\left ( 1 \right )=1$

Now, $f\left ( 0 \right )\neq h\left ( 1 \right )$

So, transitivity property also not satisfied