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NCERT Exemplar · Q57

Q.State True or False: Given A={0,1,2}A = \{0, 1, 2\}, B={x∈R∣0≤x≤2}B = \{x \in \mathbb{R} \mid 0 \le x \le 2\}. Then A=BA = B.

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The key idea is that AA is a finite set of three specific integers, while BB is an infinite set of all real numbers between 0 and 2 inclusive. Since they contain different elements, A≠BA \neq B, so the statement is False.

The heart of this question is understanding what a set actually is — a collection of distinct objects. Two sets are equal only when they contain exactly the same elements, no more and no less. So the problem reduces to: does every element of AA belong to BB, and does every element of BB belong to AA?

Let’s unpack each set carefully.

  1. Set AA is given explicitly: A={0,1,2}A = \{0, 1, 2\}. This is a finite set with exactly three elements: the integers 0, 1, and 2. Nothing else is in AA.

  2. Set BB is defined by a condition: B={x∈R∣0≤x≤2}B = \{x \in \mathbb{R} \mid 0 \le x \le 2\}. This means BB contains every real number xx that satisfies 0≤x≤20 \le x \le 2. That includes 0, 1, 2 — but also numbers like 0.50.5, 2\sqrt{2}, π−3\pi - 3, 1.9991.999, and infinitely many others. So BB is an infinite set.

  3. Check if A⊆BA \subseteq B: Every element of AA (0, 1, 2) is a real number between 0 and 2, so yes, 0∈B0 \in B, 1∈B1 \in B, 2∈B2 \in B. So A⊆BA \subseteq B is true.

  4. Check if B⊆AB \subseteq A: This is where the statement fails. Take x=0.5x = 0.5. Clearly 0≤0.5≤20 \le 0.5 \le 2, so 0.5∈B0.5 \in B. But 0.50.5 is not one of the three integers in AA. So 0.5∉A0.5 \notin A, meaning B⊈AB \not\subseteq A. …

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