Let S be the set of all rational numbers of the form mn- where m - Z and n-1-2-3- Prove that - on S defined by a - b-a b is not a binary operation-

S=a=mn #160;: #160;m #8712;Z, #160;n #8712;1, #160;2, #160;3Let #160;a=13, #160;b=53 #8712;Sa #160;* #160;b #160;=ab #160; #160; ...

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Standard XII

Mathematics

De Morgan's Law

Let S be the ...

Question

Let S be the set of all rational numbers of the form $\frac{m}{n}$ , where m ∈ Z and n = 1, 2, 3. Prove that * on S defined by a * b = ab is not a binary operation.

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Solution

$S = \{a = \frac{m}{n} : m \in Z, n \in \{1, 2, 3\}\} Let a = \frac{1}{3}, b = \frac{5}{3} \in S a * b = a b = \frac{1}{3} \times \frac{5}{3} = \frac{5}{9} \notin S [∵ 9 \notin \{1, 2, 3\}] Therefore, ∃ a, b ∈ S, such that a * b ∉ S$

Thus, * is not a binary operation.

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Let S be the set of all rational numbers of the form mn, where m ∈ Z and n = 1, 2, 3. Prove that * on S defined by a * b = ab is not a binary operation.

S=a=mn : m∈Z, n∈1, 2, 3Let a=13, b=53∈Sa * b =ab =13×53 =59∉S ∵ 9 ∉1, 2, 3 Therefore, ∃ a, b ∈ S, such that a * b ∉ S Thus, * is not a binary operation.

Let S be the set of all rational numbers of the form $\frac{m}{n}$ , where m ∈ Z and n = 1, 2, 3. Prove that * on S defined by a * b = ab is not a binary operation.

$S = \{a = \frac{m}{n} : m \in Z, n \in \{1, 2, 3\}\} Let a = \frac{1}{3}, b = \frac{5}{3} \in S a * b = a b = \frac{1}{3} \times \frac{5}{3} = \frac{5}{9} \notin S [∵ 9 \notin \{1, 2, 3\}] Therefore, ∃ a, b ∈ S, such that a * b ∉ S$

Thus, * is not a binary operation.