Each of the following defines a relation on N :
xy is square of an integer, x, y ∈ N
Determine which of the above relations are reflexive, symmetric and transitive.
Recall that for any binary relation R on set A. We have,
R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.
We have
xy is the square of an integer. x, y ∈ N.
This relation is defined on N (set of Natural Numbers)
The relation can also be defined as
R = {(x, y): xy = a2, a = √(xy), a ∈ N} on N
Check for Reflexivity:
∀ x ∈ N
We should have, (x, x) ∈ R
⇒ xx = a2, where a = √(xx)
⇒ x2 = a2, where a = √(x2)
which is true every time.
Take x = 1 and y = 4
xy = a2
⇒ 1 × 4 = (√(1 × 4))2 [∵ a = √(xy)]
⇒ 4 = (√4)2
⇒ 4 = (2)2
⇒ 4 = 4
So, ∀ x ∈ N, then (x, x) ∈ R
⇒ R is reflexive.
Check for Symmetry:
∀ x, y ∈ N
If (x, y) ∈ R
⇒ xy = a2, where a = √(xy)
Now, replace x by y and y by x. We get
yx = a2, which is as same as xy = a2
where a = √(yx)
⇒ yx = a2
⇒ (y, x) ∈ R
So, if (x, y) ∈ R, and then (y, x) ∈ R ∀ x, y ∈ N
⇒ R is symmetric.
Check for Transitivity:
∀ x, y, z ∈ N
If (x, y) ∈ R and (y, z) ∈ R
⇒ xy = a2 and yz = a2
⇒ xz = a2, may or may not be true.
Let us take x = 8, y = 2 and z = 50
xy = a2, where a = √(xy)
⇒ (8)(2) = (√(8 × 2))2
⇒ 16 = (4)2
⇒ 16 = 16, which is true.
yz = a2
⇒ (2)(50) = (√(2 × 50))2
⇒ 100 = (10)2
⇒ 100 = 100, which is true
xz = a2
⇒ (8)(50) = (√(8 × 50))2
⇒ 400 = (20)2
⇒ 400 = 400
We won’t be able to find a case to show a contradiction.
⇒ xz = a2
⇒ (x, z) ∈ R
So, if (x, y) ∈ R and (y, z) ∈ R, and then (x, z) ∈ R
∀ x, y, z ∈ N
⇒ R is transitive.
Hence, the relation is symmetric and transitivity, but not reflexive.