3 votes 3 votes Let $\text{N}^{+}$ denote the non-zero positive integers. Define a binary relation $\text{R}$ on $\text{N}^{+} \times \text{N}^{+}$ by $(m, n)\text{R}(s, t)$ if $\gcd(m, n) = \gcd(s, t).$ The binary relation $\text{R}$ is Reflexive Symmetric Transitive Not Transitive Set Theory & Algebra goclasses_wq7 goclasses set-theory&algebra relations multiple-selects 2-marks + – GO Classes asked Apr 14, 2022 edited Apr 15, 2022 by Lakshman Bhaiya GO Classes 367 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes Reflexive because $(a,b) \text{R} (a,b)$ Symmetric because $\gcd(m,n) = \gcd(s,t)$ implies that $\gcd(s,t) = \gcd(m,n)$ Transitive because $\gcd(m,n) = \gcd(s,t)$ and $\gcd(s,t) = \gcd(a,b)$ implies that $\gcd(m,n) = \gcd(a,b)$ GO Classes answered Apr 14, 2022 GO Classes comment Share Follow See all 2 Comments See all 2 2 Comments reply Amlan Kumar Majumdar commented Apr 15, 2022 reply Follow Share If m= 2 and n=3 , s=4 and t =5 , a= 10 and b=12 then how it is transitive? 1 votes 1 votes Kabir5454 commented Apr 16, 2022 reply Follow Share GCD(m,n)=GCD(2,3)=1 GCD(s,t)=GCD(4,5)=1 so , GCD(m,n) R GCD(s,t). now, GCD(a,b)=GCD(10,12)=2 so , GCD(s,t) is not related to G(a,b) because two pair are related if their GCD are same. so it is trivially transitive. 1 votes 1 votes Please log in or register to add a comment.