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  • Let <img alt="R=\{ \left ( x,y \right ) : x,y \epsilon N\ and \ x^{2}-4xy+3y^{2}=0 \}," src="/latex-image/?R%3D%5C%7B%20%5Cleft%20%28%20x%2Cy%20%5Cright%20%29%20%3A%20x%2Cy%20%5Cepsilon%20N%5C

Let R=\{ \left ( x,y \right ) : x,y \epsilon N\ and \ x^{2}-4xy+3y^{2}=0 \}, where N  is the set of all natural number. Then the relation R is:

Option: 1

reflexive but neither symmetric nor transitive


Option: 2

symmetric and transitive


Option: 3

reflexive and symmetric


Option: 4

reflexive and transitive


Answers (1)

best_answer

\begin{array}{l} \mathrm{R}=\left\{(x, y): x, y \in \mathrm{N} \text { and } x^{2}-4 x y+3 y^{2}=0\right\} \\ \text { Now, } x^{2}-4 x y+3 y^{2}=0 \\ \Rightarrow(x-y)(x-3 y)=0 \\ \therefore x=y \text { or } x=3 y \\ \therefore \quad \mathrm{R}=\{(1,1),(3,1),(2,2),(6,2),(3,3), (9,3), \ldots \ldots\} \end{array}

Since (1,1), (2,2), (3,3),......  are present in the relation, therefore R is reflexive.

Since (3,1) is an element of R but (1,3) is not the element of R therefore R is not symmetric.

\\\text{Here } (9,3)\in \mathrm{R}\text{ and } (3,1)\in \mathrm{R}\,\,but\,\,(9,1) \,\,not \,\,in\,\,\in \mathrm{R}\\

Hence, not transitive

Posted by

Devendra Khairwa

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