"Let R1 and R2 be two relation defined as follows : R1 = {(a, b) $\in$ R2 : a2 + b2 $\in$ Q} and R2 = {(a, b) $\in$ R2 : a2 + b2 $\notin$ Q}, where Q is the set of all rational numbers. Then :"

Question

Accepted Answer

"For R₁ :

Let a = 1 + $\sqrt 2$, b = 1 $-$ $\sqrt 2$, c = ${8^{{1 \over 4}}}$

aR₁b : a² + b² = 6 $\in$ Q

bR₁c : b² + c² = 3 $-$ 2$\sqrt 2$ + 2$\sqrt 2$ = 3 $\in$ Q

aR₁c : a² + c² = 3 + 2$\sqrt 2$ + 2$\sqrt 2$ $\notin$ Q

$\therefore$ R₁ is not transitive.

For R₂ :

Let a = 1 + $\sqrt 2$, b = $\sqrt 2$, c = 1 $-$ $\sqrt 2$

aR₂b : a² + b² = 5 + 2$\sqrt 2$ $\notin$ Q

bR₂c : b² + c² = 5 $-$ 2$\sqrt 2$ $\notin$ Q

aR₂c : a² + c² = 3 + 2$\sqrt 2$ + 3 $-$ 2$\sqrt 2$ = 6 $\in$ Q

$\therefore$ R₂ is not transitive."

Let R₁ and R₂ be two relation defined as follows :
R₁ = {(a, b) $\in$ R² : a² + b² $\in$ Q} and
R₂ = {(a, b) $\in$ R² : a² + b² $\notin$ Q},
where Q is the set of all rational numbers. Then :

Solution

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Let R1 and R2 be two relation defined as follows : R1 = {(a, b) $\in$ R2 : a2 + b2 $\in$ Q} and R2 = {(a, b) $\in$ R2 : a2 + b2 $\notin$ Q}, where Q is the set of all rational numbers. Then :

Solution

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More Sets, Relations and Functions questions

Drill 25 more like these. Every day. Free.

Let R₁ and R₂ be two relation defined as follows :
R₁ = {(a, b) $\in$ R² : a² + b² $\in$ Q} and
R₂ = {(a, b) $\in$ R² : a² + b² $\notin$ Q},
where Q is the set of all rational numbers. Then :