Let R be a relation on N defined by x + 2y = | vedclass

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

Question

(C) The relation $R$ is defined on the set of natural numbers $N$ by the equation $x + 2y = 8$.We need to find pairs $(x, y)$ such that $x, y \in N$ a

Vedclass · Accepted Answer

$\{2, 4, 6\}$

Vedclass · Answer

(C) The relation $R$ is defined on the set of natural numbers $N$ by the equation $x + 2y = 8$.We need to find pairs $(x, y)$ such that $x, y \in N$ and $x + 2y = 8$.If $y = 1$,then $x + 2(1) = 8 \implies x = 6$.If $y = 2$,then $x + 2(2) = 8 \implies x = 4$.If $y = 3$,then $x + 2(3) = 8 \implies x = 2$.If $y = 4$,then $x + 2(4) = 8 \implies x = 0$. Since $0 
otin N$,this is not a valid pair.For any $y > 3$,$x$ will be negative,which is not in $N$.Thus,the relation $R = \{(2, 3), (4, 2), (6, 1)\}$.The domain of $R$ is the set of all first elements of the ordered pairs in $R$.Therefore,the domain is $\{2, 4, 6\}$.

Let $R$ be a relation on $N$ defined by $x + 2y = 8$. The domain of $R$ is

Similar Questions

In the context of a power set containing a subset,the relation 'is a subset of' $(\subseteq)$ is:

$A$ relation $R$ is defined from $\{2, 3, 4, 5\}$ to $\{3, 6, 7, 10\}$ by $xRy \iff x$ is relatively prime to $y$. Then the domain of $R$ is

The solution set of $x \equiv 3 \pmod{7}$,where $x \in \mathbb{Z}$,is:

Given $A = \{1, 2, 3, 4, 5\}$ and $B = \{1, 4, 5\}$. If $R$ is a relation from $A$ to $B$ such that $(x, y) \in R$ with $x > y$,then the range of $R$ is:

$A = \{1, 2, 3, 5\}$ and $B = \{4, 6, 9\}$. Define a relation $R$ from $A$ to $B$ by $R = \{(x, y) : \text{the difference between } x \text{ and } y \text{ is odd}; x \in A, y \in B\}$. Write $R$ in roster form.

Vedclass Test Series

Exam Paper Generator

Online Exam Module