If $R$ be a relation defined as $aRb iff |a - b| > 0$, then the relation is
- reflexive
- symmetric
- transitive
- symmetric and transitive
The Correct Option is D
Solution and Explanation
Reflexivity - Let a be an arbitrary element.
Then,
$\left|a-a\right|=0 > 0 \Rightarrow a R a$
This, R is not reflexive on R.
Symmetry - Let a and b be two distinct elements, then (a,b) $\in$R
$\Rightarrow\left|a-b\right|>0\Rightarrow\quad\left|b-a\right|>0$
$\quad\quad\quad \quad \quad \quad \left(\because \left|a-b\right|=\left|b-a\right|\right)$
$\Rightarrow \left(b,a\right)\in R$
Thus, $\left(a, b\right) \in R \Rightarrow \left(b, a\right) \in R.$ So, R is symmetric.
Transitivity - Let $\left(a, b\right) \in R$ and $\left(b, c\right) \in R.$
Then$\quad |a - b| > 0$ and $|b - c| > 0$
$\Rightarrow \quad\quad |a - c| > 0 \Rightarrow \left(a, c\right) \in R$
So, R is transitive.
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Concepts Used:
Functions
A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. Let A & B be any two non-empty sets, mapping from A to B will be a function only when every element in set A has one end only one image in set B.
Kinds of Functions
The different types of functions are -
One to One Function: When elements of set A have a separate component of set B, we can determine that it is a one-to-one function. Besides, you can also call it injective.
Many to One Function: As the name suggests, here more than two elements in set A are mapped with one element in set B.
Moreover, if it happens that all the elements in set B have pre-images in set A, it is called an onto function or surjective function.
Also, if a function is both one-to-one and onto function, it is known as a bijective. This means, that all the elements of A are mapped with separate elements in B, and A holds a pre-image of elements of B.
Read More: Relations and Functions