Explanation
Given R:{(x,y):x∈Ny∈N,2x+y=41}R:{(x,y):x∈Ny∈N,2x+y=41}
2x+y=412x+y=41
y=41−2xy=41−2x
Since y∈Ny>0y∈Ny>0
41−2x>041−2x>0
−2x>−41−2x>−41
2x<412x<41
x<412x<412
Since x∈N.x∈N. x can take values ={1,2,3....20}={1,2,3....20}
Domain of R={1,2,3.....20}R={1,2,3.....20}
Range of R is y=41−2xforx={1,2,3....20}y=41−2xforx={1,2,3....20}
Range of R={1,3,.......,37,39}R={1,3,.......,37,39}
When x=1y=41−2=39x=1y=41−2=39
When x=2y=41−2x×2=39x=2y=41−2x×2=39
When x=20y=40−2×20=1x=20y=40−2×20=1
Domain of f={1,2.....20}f={1,2.....20}
Range of f={1,3......37,39}f={1,3......37,39}
R is not reflexive
Since (1,1) does not satisfy
2×1+1≠412×1+1≠41
R is not symmetric since (1,39)∈R(1,39)∈R satisfies 2x+y=41but(39,1)∉R2x+y=41but(39,1)∉R
Since 2×39+1≠412×39+1≠41
but 2×1+39=412×1+39=41
Solution:R is not transitive since (1,39) ∈∈ but no values for x=39. satisfies the given relation
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