Consider the following relations R = {(x, y) | x, y are real numbers a

1.	Consider the following relations R = {(x, y) \| x, y are real numbers and x = wy for some rational number w}; S = \(\bigg\{\bigg(\frac{m}{n},\frac{p}{q}\bigg)\bigg\|\) m, n, p and q are integers such that n, q ≠ 0 and qm = pn}. Then,(a) R is an equivalence relation but S is not an equivalence relation (b) Neither R nor S is an equivalence relation (c) S is an equivalence relation but R is not an equivalence relation (d) R and S are both equivalence relations
Answer» (c) S is an equivalence relation but R is not an equivalence relation R = {(x, y) \| x, y ∈ R, x = wy, w is a rational number} Reflexive: x R x ⇒ x = wx ⇒ w = 1, (a rational number) Hence R is reflexive. Symmetric: x R y \(\not\Rightarrow\) y R x as 0 R 1 ⇒ 0 = (0) . 1 where 0 is a rational number but 1 R 0 ⇒ 1 = (w) 0 which is not true for any rational number. ∴ R is not an equivalence relation S = \(\bigg\{\bigg(\frac{m}{n},\frac{p}{q}\bigg)\bigg\|\) m, n, p, q, ∈ I, n, q, ≠ 0 and qm = pn\(\bigg\}\) Reflexive \(\frac{m}{n}S\frac{m}{n}\) ⇒ mn = nm (True) Symmetric \(\frac{m}{n}S\frac{p}{q}\) ⇒ mq = pn ⇒ pn = mq ⇒ \(\frac{p}{q}S\frac{m}{n}\) (True) Transitive \(\frac{m}{n}S\frac{p}{q}\) and \(\frac{p}{q}S\frac{r}{s}\) ⇒ mq = pn and ps = qr ⇒ mq.ps = pn.qr ⇒ ms = nr ⇒ \(\frac{m}{n}\) = \(\frac{r}{s}\) ⇒ \(\frac{m}{n}\) S \(\frac{r}{s}\) (True) ∴ S is an equivalence relation.

Consider the following relations R = {(x, y) | x, y are real numbers and x = wy for some rational number w}; S = \(\bigg\{\bigg(\frac{m}{n},\frac{p}{q}\bigg)\bigg|\) m, n, p and q are integers such that n, q ≠ 0 and qm = pn}. Then,(a) R is an equivalence relation but S is not an equivalence relation (b) Neither R nor S is an equivalence relation (c) S is an equivalence relation but R is not an equivalence relation (d) R and S are both equivalence relations

Answer»

(c) S is an equivalence relation but R is not an equivalence relation

R = {(x, y) | x, y ∈ R, x = wy, w is a rational number}

Reflexive: x R x ⇒ x = wx ⇒ w = 1, (a rational number)

Hence R is reflexive.

Symmetric: x R y \(\not\Rightarrow\) y R x as

0 R 1 ⇒ 0 = (0) . 1 where 0 is a rational number but

1 R 0 ⇒ 1 = (w) 0 which is not true for any rational number.

∴ R is not an equivalence relation

S = \(\bigg\{\bigg(\frac{m}{n},\frac{p}{q}\bigg)\bigg|\) m, n, p, q, ∈ I, n, q, ≠ 0 and qm = pn\(\bigg\}\)

Reflexive \(\frac{m}{n}S\frac{m}{n}\) ⇒ mn = nm (True)

Symmetric \(\frac{m}{n}S\frac{p}{q}\) ⇒ mq = pn ⇒ pn = mq ⇒ \(\frac{p}{q}S\frac{m}{n}\) (True)

Transitive \(\frac{m}{n}S\frac{p}{q}\) and \(\frac{p}{q}S\frac{r}{s}\)

⇒ mq = pn and ps = qr

⇒ mq.ps = pn.qr ⇒ ms = nr

⇒ \(\frac{m}{n}\) = \(\frac{r}{s}\) ⇒ \(\frac{m}{n}\) S \(\frac{r}{s}\) (True)

∴ S is an equivalence relation.