The given relation is defined on the set of real numbers. `a R b iff |a| = |b|`
. Find whether these relations are reflexive, symmetric or transitive.

Answer

(i) Not reflexive, symmetric, not transitive
(ii) Reflexive, symmetric, transitive
(iii) Reflexive, symmetric, not transitive
(v) Not reflexive, not symmetric, transitive

Explanation

(i) `aRb iff |a-b|gt0`
Reflexivity a - a = 0
`therefore (a, a) cancelinR`
`therefore` R is not reflexive
Symmetry `(a, b) in Rimplies|a-b|gt0`
`implies |-(b-a)gt0|`
`implies |b-a|gt0`
`implies (b, a) in R`
`therefore` R is symmetric relation.
Transitivity `(a, b) in R and (b, c) in R`
`implies |a-b|gt0and|b-c|gt0`
`implies |a-b|+|b-c|gt0` [by addition]
Now, let `a gt b` and `b gt c`, then `a gt c`
`|a-b|+|b-c|=a-b+b-c=a-cgt0`
implies `|a-c|gt0`
If `a lt b` and `b gt c`, then
`|a-b|+|b-c|=-(a-b)+(b-c)=-a+2b-c`
`cancelimplies |a-c|gt0`
`therefore ` R is not transitivity relation.
(ii) `aRbiff|a|=|b|`
Reflexivity We have, |a| = |a|
`implies aRaAAa`
`therefore` R is symmetric relation.
Transitivity `(a, b) in R and (b, c) in R`
implies |a| = |b| and |b| = |c|
implies |a| = |c|
implies `(a, c) in R`
`therefore R` is transitive relation.
(iii) `aRb iff |a| ge |b|`
Reflexivity For any `a in R`, we have `|a| ge |a|`
So, `aRa AA a`
`therefore` R is not symmetric relation.
Symmetry `aRb implies |a| ge |b|`
implies `|b| le |a|`
`therefore` R is not symmetric relation.
Transitivity aRb and bRc implies |a| `ge` |b| and `|b| ge |c|`
implies `|a| ge |c|`
implies aRc
`therefore` R is reflexive on R.
Symmetry Let `(a, b) in R`, then (a, b) `in` R
implies `1 + ab gt 0`
implies 1+ ba `gt` 0
implies (b, a) `in` R
`therefore` R is symmetric on R.
Transitivity We observe that `(1, (1)/(2)) in R` and `((1)/(2), -1) in R` but `(1, -1) cancelin R` because
1 + (1) (-1) = 0 `cancelgt` 0
`therefore` R is not transitive on R.
(v) `aRb iff |a| le b`
Reflexivity Let -1 `in R`, then `|-1|cancelle(-1)`
`therefore R` is not reflexive relation
Symmetry Now, let - 3 R 4, then `|4| cancelle -3`
`therefore` R is not symmetric relation
Transitivity aRb and bRc implies `|a| le b and |b| le c`
Then, `|a| le c implies aRc`
`therefore R` is transitive relation.