Virat Batch 2027
Champion Batch 2028
Divine JEE 1-on-1
a, b ∈ z, GCD(a, b)=1, b ≠ 2 a a, b \in z, \operatorname{GCD}(a, b)=1, b \neq 2 a \\ a, b ∈ z, GCD(a, b)=1, b = 2 a
GCD(a, a) ≠ 1 ⇒ Not Reflexive \operatorname{GCD}(a, a) \neq 1 \Rightarrow \text { Not Reflexive } \\ GCD(a, a) = 1 ⇒ Not Reflexive
GCD(3,7)=1 & GCD(7,15)=1 \operatorname{GCD}(3,7)=1 \& \operatorname{GCD}(7,15)=1\\ GCD(3,7)=1 & GCD(7,15)=1
But GCD(3,15) ≠ 1 ⇒\operatorname{GCD}(3,15) \neq 1 \Rightarrow\\GCD(3,15) = 1 ⇒ Not Transitive
Take a=2, b=1 ⇒ gcd(2,1)=1a=2, b=1 \Rightarrow \operatorname{gcd}(2,1)=1\\a=2, b=1 ⇒ gcd(2,1)=1
Also 2 a=4 ≠ b2 a=4 \neq b\\2 a=4 = b
Now when a=1, b=2 ⇒ gcd(1,2)=1a=1, b=2 \Rightarrow \operatorname{gcd}(1,2)=1\\a=1, b=2 ⇒ gcd(1,2)=1
Also now 2 a=2=b2 \mathrm{a}=2=\mathrm{b}\\2 a=2=b
Hence a=2 ba=2 ba=2 b
⇒ R\Rightarrow R⇒ R is not Symmetric
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