The relation R = { ( a , b ) : g c d ( a , b ) = 1 , 2 a b , a , b , Z } is [2023]

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Q.
The relation $R = {(a, b) : g c d (a, b) = 1, 2 a \neq b, a, b, \in Z}$ is [2023]

1 reflexive but not symmetric

2 transitive but not reflexive

3 symmetric but not transitive

4 neither symmetric nor transitive

Ans.
(4)

Reflexive : $\gcd (a, a) = a \neq 1 for a \neq 1 .$

Symmetric: Let $a = 2, b = 1$

$\gcd (a, b) = 1; b \neq 2 a$

Thus, $(2, 1) \in R; \gcd (1, 2) = 1 .$ But $2 = 2 \times 1$

$Therefore (1, 2) \notin R . Hence, R is not symmetric.$

Transitive: $(2, 3) \in R$

$(3, 8) \in R .$ But $(2, 8) \notin R .$

$Hence R is not transitive.$

$R is neither symmetric nor transitive.$

Solution

Q.
The relation $R = {(a, b) : g c d (a, b) = 1, 2 a \neq b, a, b, \in Z}$ is [2023]

1 reflexive but not symmetric

2 transitive but not reflexive

3 symmetric but not transitive

4 neither symmetric nor transitive

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Solution

Q. The relation R={(a,b):gcd(a,b)=1,2a≠b,a,b,∈Z} is [2023]

1 reflexive but not symmetric

2 transitive but not reflexive

3 symmetric but not transitive

4 neither symmetric nor transitive

Ans. (4) Reflexive : gcd(a,a)=a≠1 for a≠1. Symmetric: Let a=2, b=1 gcd(a,b)=1; b≠2a Thus, (2,1)∈R; gcd(1,2)=1. But 2=2×1 Therefore (1,2)∉R. Hence, R is not symmetric. Transitive: (2,3)∈R (3,8)∈R. But (2,8)∉R. Hence R is not transitive. R is neither symmetric nor transitive.

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Q.
The relation $R = {(a, b) : g c d (a, b) = 1, 2 a \neq b, a, b, \in Z}$ is [2023]