1) Define a relation R on the set of real numbers by: (x , y) R if and only if |
ID: 3006458 • Letter: 1
Question
1) Define a relation R on the set of real numbers by: (x , y) R if and only if |x y| < 4 . Is the relation R reflexive? symmetric? transitive? Is R an equivalence relation? Justify your answers.
2) Define a relation R on R as follows: (x , y) R if and only if (x y)(x^2 + y^2 1) = 0 . Is R reflexive? Is R symmetric? Is R transitive? Is R an equivalence relation? Justify your arguments.
3) Define a relation R on the set Z (a,b) element of R if and only if 3a+5b is divisible by 8. prove that R is equvlance relation.
4) Define a relation R on the set Z (a,b) element of R if and only if 2a+3b is divisible by 5. prove that R is equvlance relation. what is the equvlance class of {0]}
5) R-{0} define a relation if xy< or equal to zero Is R reflexive? Is R symmetric? Is R transitive?
6) R on Z (a,b) element if and only if R is a+b is even. prove that R is equvlance relation. what is class of 1?
Explanation / Answer
Ans(1):
Given condition is |x y| < 4
reflexive
Yes it is reflexive because |x x| < 4 or |0|<4 or 0<4 is clearly True.
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symmetric
Yes it is symmetric because
|x y| < 4
or |- (-x + y)| < 4
or | (-x + y)| < 4 (because we know that |-x|=x )
or | (y -x)| < 4
or | y -x| < 4 is True
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transitive
No, it is not transitive because. I will prove that using an counter example
take x=1, y=4, z=7
clearly |x y| < 4 or |1-4| < 4 or |-3|<4 or 3<4 is True.
similarly |y z| < 4 or |4-7| < 4 or |-3|<4 or 3<4 is True.
But |x z| < 4 or |1-7| < 4 or |-6|<4 or 6<4 is False.
Hence it is not transitive.
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Since it fails transitive property, so we can say it is Not an equivalence relation.
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