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Answer:R is an equivalence relation and the set of all elements related to 1 is 1.
Hint: To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.
$\inline Given: \! Set A=\left \{ x \: \epsilon \: Z ;0 \leq x \leq 12 \right \}$
$\inline Also\: given\: that\: relation\: R=\left \{ \left ( a, b \right )\! :\! a=b \right \} is \: defined\: on \: set\: A$
Explanation:
Reflexivity:
Let a be an arbitrary element of A.
Then,
$a=a \; \; \; \; \; \; [Since, every\: element \: is\: equal\: to\: itsel\! f]$
$\inline (a, a) \: \epsilon \: R\: f\! or\: all\: a \: \epsilon \: A$
So, R is reflexive on A
Symmetry:
$\inline Let (a, b)\: \epsilon \: R$
$\inline a=b$
$\inline b=a$
$\inline (b, a) \: \epsilon \: R\: f\! or\: all\: a,b \: \epsilon \: A$
So, R is symmetric on A.
Transitivity:
$\inline Let\: a, b\: and\: (b, c) \: \epsilon \: R$
$\inline a=b ...(i)\: and \: b=c ...(ii)$
multiplying eqn (i) and (ii), we get
$\inline ab=bc$
$\inline a=c$
$\inline there\! f\! ore, (a, c) \: \epsilon \: R$
So, R is transitive on A.
Therefore, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on A.
$R=\left \{ \left ( a, b \right ):a=b\: \right \} and\: 1 \: is\: on\: element\: o\! f \: A.$
$R=\! \left \{\left ( 1, 1 \right ):1=1 \right \}$
Thus, the set of all elements related to 1 is 1.

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