Let S be the set of all real numbers. Then th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) $1$. Reflexivity: For any $a \in S$,we have $1 + a \cdot a = 1 + a^2$. Since $a^2 \ge 0$,$1 + a^2 \ge 1 > 0$. Thus,$(a, a) \in R$ for all $a \in S

Vedclass · Accepted Answer

Reflexive and symmetric but not transitive

Vedclass · Answer

(A) $1$. Reflexivity: For any $a \in S$,we have $1 + a \cdot a = 1 + a^2$. Since $a^2 \ge 0$,$1 + a^2 \ge 1 > 0$. Thus,$(a, a) \in R$ for all $a \in S$. So,$R$ is reflexive.$2$. Symmetry: If $(a, b) \in R$,then $1 + ab > 0$. Since $ab = ba$,it follows that $1 + ba > 0$,which means $(b, a) \in R$. So,$R$ is symmetric.$3$. Transitivity: Consider $a = 1$,$b = -1/2$,and $c = -1$. For $(a, b) = (1, -1/2)$,$1 + (1)(-1/2) = 1 - 0.5 = 0.5 > 0$,so $(1, -1/2) \in R$.For $(b, c) = (-1/2, -1)$,$1 + (-1/2)(-1) = 1 + 0.5 = 1.5 > 0$,so $(-1/2, -1) \in R$.However,for $(a, c) = (1, -1)$,$1 + (1)(-1) = 1 - 1 = 0$,which is not $> 0$. Thus,$(1, -1) 
otin R$. Therefore,$R$ is not transitive.

Let $S$ be the set of all real numbers. Then the relation $R = \{(a, b) : 1 + ab > 0\}$ on $S$ is

Similar Questions

Let $A = \{1, 2, 3, 4\}$ and $R$ be a relation in $A$ defined by $R = \{(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1), (3, 1), (1, 3)\}$. Then $R$ is

On the set of integers $Z$,a relation $S$ is defined as: $S = \{(x, y) \in Z \times Z : |x - y| < 1\}$. Which of the following is true about $S$?

The number of reflexive relations on a set with $4$ elements is equal to

If $R_{1}$ and $R_{2}$ are equivalence relations in a set $A$,show that $R_{1} \cap R_{2}$ is also an equivalence relation.

Let $P$ be the relation defined on the set of all real numbers such that $P = \{(a,b) : \sec^2 a - \tan^2 b = 1\}$. Then $P$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module