Among the relations
$S=\left\{(a, b): a, b \in R -\{0\}, 2+\frac{a}{b}>\right\}$
and $T=\left\{(a, b): a, b \in R , a^2-b^2 \in Z\right\}$,
We are given two relations, \( S \) and \( T \), and we are asked to determine which of the following statements is true.
For relation \( T \): We know \( T = \{(a, b): a, b \in \mathbb{R}, a^2 - b^2 \in \mathbb{Z}\} \). From this, we deduce that \[ b^2 - a^2 = -1 \quad \Rightarrow \quad b = -a \quad \text{(relation \( T \) is symmetric)}. \] Thus, \( T \) is symmetric.
For relation \( S \): We know \( S = \{(a, b): a, b \in \mathbb{R} \setminus \{ 0 \}, a^2 + b^2 > 0\} \). For \( S \), we see that \[ \frac{a}{b} = \frac{a}{-b} \quad \text{(the relation is not necessarily symmetric)}. \] So, \( S \) is not symmetric. Thus, the correct answer is \( T \) is symmetric but \( S \) is not symmetric.
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.
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