The problem is: \" Let R be the set of all real numbers and consider the followi
ID: 2969007 • Letter: T
Question
The problem is:
" Let R be the set of all real numbers and consider the following subsets of R X R:
(a) R1 = {(x,y) in R X R | xy = 0}
(b) R2 = {(x,y) in R X R | |x - y| < 5} [Note: it reads "such that the absolute value of x - y is less than 5].
(c) R3 = {(x, y) in R X R | x^2 + y^2 = 1}
Which of the above relations is:
(i) Reflexive on R? (ii) Symmetric? (iii) Transitive? "
I am really struggling with how to determine whether a relation is reflexive, symmetric, or transitive, so I'd greatly appreciate anyone's explanation for this problem. I'll be sure to quickly give the points to the best answer. Thanks!
Explanation / Answer
A relation R is reflexive if (x,x) is in R.
Notice that (a) is not reflexive, since (x,x) in R, implies x^2 = 0, but this is not true in general (only true if x =0)
(b) is reflexive since |x-x| = 0 <5.
(c) is not relfexive since x^2 + x^2 = 1 is not true in general, this is 2x^2 =1 is not true in general.
A relation is symmetric if for every (x,y) in R, then (y,x) is in R too.
(a) is symmetric since if (x,y) is in R, then xy=0, so yx=0 so (y,x) is in R.
(b) is symmetric since if (x,y) is in R, then |x-y|<5 so |-(y-x)| < 5 so |y-x| <5, hence (y,x) is in R.
(c) is symmetric since if (x,y) is in R, then x^2 + y^2 =1 , so y^2 + x^2 =1 so (y,x) is in R.
A relation is transitive if for (x,y), (y,z) in R, then (x,z) is in R.
(a) is NOT transitive since (1,0) is in R (this is 1*0=0), (0,2) is in R (this is 0*2=0) , but (1,2) is not in R (this is 1*2 is not 0). Hence the transitive property is not true when x =1, y =0 , z =1.
(b) is NOT transitive since |10-6| = 4 <5, so (10,6) is in R, also |6-2|=4 < 5 (so (6,2) is in R), however |10-2| = 8 which is not less than 5, so (10,2) is not in R. So the transitivity property is not true when x = 10, y = 6 and z = 2.
(c) is NOT transitive. Notice that 0^2 + 1^2 = 1, so (0, 1) is in R. Also, 1^2+0^2 =1 so (0,1) is in R. But (1,1) is not in R since 1^2 + 1^2 is not 1. Hence the transitive property is not true when x =1, y =0 , z =1.
Please rate. It is very well explained. Done by a mathematician.
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