Question

a R b if $1+ab>0$ . what type of relation it is.

Answer 1

Hint: First observe the relation given in question. Find all the properties related to the relation. Try to substitute values or variables to check their properties. Try to use the definition of Reflexive, transitive, symmetries, Antisymmetry. First and the foremost thing to do is use the definition of relation and say that the given relation satisfies all the conditions needed for it to be a relation. Next see the range on which relation is defined. While checking properties, substitute the elements only which satisfy the range given in the question.

Complete step-by-step answer:

Relation:- the relations in math are also called a binary relation over set x, y is a subset of the Cartesian product of the sets x, y, that is the relation contains the elements of the set of $x\times y$ .
$x\times y$, is a set of ordered pairs consisting of element x in X and y in Y. It encodes the information, (x, y) will be relation set if and only if x is related to the elementary in the relation properly again.
Reflexive Relation: A relation R is said to be reflexive, if a is an element in the range. If the pair (a, a) belongs to relation then relation is reflexive. In other words, if a is related to itself by given relation then relation is reflexive.
Symmetric Relation: A relation R is said to symmetric, if a, b are elements in the range. If the pair (a, b) belongs to relation the (b, a) must belong to relation. In other words if a is related to b, then b must also be related to a for a relation to be symmetric.
Transitive Relation: If (a, b) belongs to relation and (b, c) also belongs to the relation, then (a, c) must belong to relation to make it transitive.
Antisymmetric Relation: If (a, b) and (b, a) are given belong to relation, then a must be equal to b to make the relation as anti symmetric then it is an equivalence relation.
Given relation in the question is given by (items of a, b):
a R b if $1+ab>0$.
Case-1: Checking whether R is reflexive relation.
So, if $a=b$ .
By substituting $a=b=k$ in the relation, we get it as $1+{{a}^{2}}$ .
By basic knowledge we know ${{a}^{2}}>0$ So, $1+{{a}^{2}}>0$ .
So, relation is reflexive.
Case-2: checking whether R is symmetric.
If a, b belong to R. Then $1+ab>0$ .
Now, by substituting b, a we get $1+ba$ . We know that $ab=ba$ . So, $1+ba>0$ .
So, the relation is symmetric.
Case-3: Checking whether Relation is transitive.
So, we have $1+ab>0,1+bc>0$ as (a, b); (b, c) belongs to the relation.
By these we cannot say (a, c) belongs. Because we disprove it by example.
Take $a=-6$ $b=-1$ $c=\dfrac{1}{3}$
$\begin{align}
  & a.b\Rightarrow 1+ab=1+67>0 \\
 & b,c\Rightarrow 1-\dfrac{1}{3}=\dfrac{2}{3}>0 \\
\end{align}$
Now, if we substitute a,c $1+ac=1-2=-1$ .
So, it is not transitive.
Relation is Reflexive, symmetric.

Note: Be careful while getting transitive relation as many students confuse and think that the relation is transitive but it is not. While checking symmetric take $ab=ba$ condition only if a, b are numbers. Here they are numbers. So, you can take it. But in case of matrices you must not take it.