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Question
Defines a relation on N:
x + 4y = 10, x, y ∈ N
Determine the above relation is reflexive, symmetric and transitive.
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Solution
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, if xRy and yRz, then xRz.
We have
x + 4y = 10, x, y ∈ N
This relation is defined on N (set of Natural Numbers)
The relation can also be defined as
R = {(x, y) : x + 4y = 10} on N
Check for Reflexivity:
∀ x ∈ N
We should have, (x, x) ∈ R.
4x + x = 10, which is obviously not true everytime.
Take x = 4,
4x + x = 10
⇒ 16 + 4 = 10
⇒ 20 = 10, which is not true.
This is 20 ≠ 10.
So, ∀ x ∈ N, then (x, x) ∉ R.
R is not reflexive.
Check for Symmetry:
∀ x, y ∈ N
If (x, y) ∈ R
4x + y = 10
Now, replace x by y and y by x. we get,
4y + x = 10, which may or may not be true.
Take x = 1 and y = 6
4x + y = 10
⇒ 4(1) + 6 = 10
⇒ 10 = 10
4y + x = 10
⇒ 4(6) + 1 = 10
⇒ 24 + 1 = 10
⇒ 25 = 10, which is not true.
⇒ 4y + x ≠ 10
⇒ (x, y) ∉ R
So, if (x, y) ∈ R, and then (y, x) ∉ R ∀ x, y ∈ N
R is not symmeteric.
Check for Transitivity:
∀ x, y, z ∈ N
If (x, y) ∈ R and (y, z) ∈ R
Then, (x, z) ∈ R
We have,
4x + y = 10
⇒ y = 10 − 4x
Where x, y ∈ N
So, put x = 1
⇒ y = 10 - 4(1)
⇒ y = 10 - 4
⇒ y = 6
Put x = 2
⇒ y = 10 - 4(2)
⇒ y = 10 - 8
⇒ y = 2
We can't take y > 2, because if we put y = 3
⇒ y = 10 - 4(3)
⇒ y = 10 - 12
⇒ y = -2
But, y ≠ -2 as y ∈ N
so, only ordered pairs possible are
R = {(1, 6), (2, 2)}
This relation R can never be transitive.
Because if (a, b) ∈ R, then (b, c) ∉ R.
R is not reflexive.
Hence, the relation is neither reflexive nor symmetric nor transitive.
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