To determine if the given relation \( R \) on \( \mathbb{Z} \times \mathbb{Z} \) is reflexive, symmetric, and/or transitive, we analyze each property in the context of the definition of \( R \): \((a, b) R (c, d)\) if and only if \(ad - bc\) is divisible by 5.
Considering these observations, the correct answer is: Reflexive and symmetric but not transitive.
Reflexive : for \((a, b) R (a, b) \)
\( ⇒ ab – ab = 0\) is divisible by 5.
So \((a, b) R(a, b) ∀ a, b ∈ Z \)
∴ R is reflexive Symmetric : For \((a, b) R(c, d) \)
If \(ad – bc\) is divisible by 5.
Then \(bc – ad\) is also divisible by 5.
\(⇒ (c, d) R(a, b) ∀ a, b, c, d ∈ Z \)
∴ R is symmetric Transitive : If \((a, b) R(c, d) \)
\(⇒ ad – bc\) divisible by 5 and \((c, d) R (e, f) \)
\(⇒ cf – de\) divisible by 5
\(ad – bc = 5k_1\) \( k_1\) and \(k_2\) are integers
\(cf – de = 5k_2\)
\(afd – bcf = 5k_1f \)
\(bcf – bde = 5k_2b \)
\(afd – bde = 5(k_1f + k_2b) \)
\(d(af – be) = 5 (k_1f + k_2b) \)
\(⇒ af – be\) is not divisible by 5 for every a, b, c, d, e, f ∈ Z.
\( ∴\) R is not transitive
For e.g., take \(a = 1, b = 2, c = 5, d = 5, e = 2, f = 2\)
The correct option is (B): Reflexive and symmetric but not transitive
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
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There are two ways by which a relation can be represented-
The roster form and set-builder for for a set integers lying between -2 and 3 will be-
I= {-1,0,1,2}
I= {x:x∈I,-2<x<3}