T={(x,y):x−y∈l} as 0∈IT is a reflexive relation. If x−y∈1⇒y−x∈I ∴T is symmetrical also If x−y=l1 and y−z=l2 Then x−z=(x−y)+(y−z)=l1+I2∈I ∴T is also transitive. Hence T is an equivalence relation. Clearly x=x+1⇒(x,x)∈/S ∴S is not reflexive.
Let R be the real line. Consider the following subsets of the plane R×R. S={(x,y):y=x+1 and 0<x<2},T={(x,y):x−y is an integer }. Which one of the following is true?
Held on 30 Apr 2008 · Verified 6 Jul 2026.
neither S nor T is an equivalence relation on R
both S and T are equivalence relations on R
S is an equivalence relation on R but T is not
T is an equivalence relation on R but S is not
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