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SubjectsMathematics
रिक्त समुच्चय, सीमित और असीमित समुच्चय, समान समुच्चय
In this Class 10 Mathematics topic from the chapter Sets, students learn how to identify and represent the empty set, which contains no elements, and distinguish finite sets from infinite sets by considering the number of elements they contain. The topic also explains equal sets, where two sets have exactly the same elements regardless of their order. Examples, symbols, and basic comparisons help students apply these ideas accurately.
TOPIC PRACTICE
Up to 7 questions from this page. Select your focus, then start.
Under the standard school convention that ℕ = {1, 2, 3, …}, every natural number n is a factor of 0 because 0 = n × 0. Hence every element of ℕ belongs to A, so A = ℕ. Since the natural numbers are infinite, A is infinite. This distinguishes factors of zero from multiples of zero, so option C is correct.
A multiple of 0 has the form 0 × k, where k is an integer. Regardless of the value of k, 0 × k = 0. Therefore the only integer that can be a multiple of 0 is 0 itself, and A = {0}. It is not ℤ, because nonzero integers cannot be written as 0 times an integer. Hence option B is correct.
The integers from −5 to 5 that are congruent to 1 modulo 3 are −5, −2, 1, and 4. For example, each differs from 1 by a multiple of 3: −6, −3, 0, and 3 respectively. Hence A = {−5, −2, 1, 4} = B. Set equality depends on having exactly the same elements, not on their order or on whether the elements are positive or negative. Therefore option A is correct.
The integers between −6 and 6 that are congruent to 2 modulo 5 can be found from x = 5k + 2. For k = −1, x = −3; for k = 0, x = 2; while k = 1 gives 7, which is outside the interval. Therefore the only eligible elements are −3 and 2, so A = {−3, 2}. Hence option A is correct.
The expression |x - 4| + |x - 9| represents the sum of the distances from x to 4 and 9 on the number line. The distance between 4 and 9 is 5, so for every x in the closed interval [4, 9], the sum is exactly 5. Since an interval contains infinitely many real numbers, A is an infinite set.
A set is unchanged when the order of its elements is rearranged. Therefore, {2, {3}, 1} contains exactly the same three elements as A: 1, 2, and the set {3}; hence it equals A. Notice that 3 and {3} are different objects, so option B is not equal to A. Options C and D also change the nesting or the elements.
If A is a subset of B, every element of A is also in B. Consequently, an element that is outside B cannot be in A; otherwise it would be in B as well. Therefore every element of B′ is an element of A′, which proves B′ ⊆ A′. Taking complements reverses the direction of inclusion. Option A expresses this correctly: the larger set B leaves a smaller complement, while the smaller set A leaves a larger complement.
QUIZ COMPLETE
