Tardigrade
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Tardigrade
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Mathematics
The relation R = (1, 1), (2, 2) (3, 3), (1, 2), (2, 3), (1, 3) on set A = 1, 2, 3 is
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Q. The relation $R = \{(1$, $1)$, $(2$, $2) (3$, $3)$, $(1$, $2)$, $(2$, $3)$, $(1$, $3)\}$ on set $A = \{1$, $2$, $3\}$ is
Relations and Functions - Part 2
A
Reflexive but not symmetric
69%
B
Reflexive but not transitive
9%
C
Symmetric and transitive
14%
D
Neither symmetric nor transitive
8%
Solution:
Reflexive :
$(1$, $1)$, $(2$, $2)$, $(3$, $3) \in R$, $R$ is reflexive
Symmetric :
$(1$, $2) \in R$ but $(2$, $1) \notin R$, $R$ is not symmetric.
Transitive :
$(1$, $2) \in R$ and $(2$, $3) \in R$
$ \Rightarrow (1$, $3) \in R$, $R$ is transitive.