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NCERT Exemplar · Q56

Q.State True or False: Let sets RR and TT be defined as R={x∈Z∣x is divisible by 2}R = \{x \in \mathbb{Z} \mid x \text{ is divisible by 2}\}, T={x∈Z∣x is divisible by 6}T = \{x \in \mathbb{Z} \mid x \text{ is divisible by 6}\}. Then T⊂RT \subset R.

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Every multiple of 6 is automatically a multiple of 2, so every element of TT lies in RR, making TT a subset of RR. The statement is True.

Understanding subset relationships through divisibility

When we say T⊂RT \subset R, we mean every element of TT must also be an element of RR. The question boils down to: does divisibility by 6 guarantee divisibility by 2?

The key insight is factorization. Since 6=2×36 = 2 \times 3, any integer divisible by 6 must contain both 2 and 3 as factors. In particular, it must be divisible by 2.

Let me make this precise.

  1. What does TT contain?

    T={x∈Z∣x is divisible by 6}T = \{x \in \mathbb{Z} \mid x \text{ is divisible by 6}\} means T={…,−12,−6,0,6,12,18,…}T = \{\ldots, -12, -6, 0, 6, 12, 18, \ldots\}. Every element can be written as x=6kx = 6k for some integer kk.

  2. What does RR contain?

    R={x∈Z∣x is divisible by 2}R = \{x \in \mathbb{Z} \mid x \text{ is divisible by 2}\} means R={…,−4,−2,0,2,4,6,8,…}R = \{\ldots, -4, -2, 0, 2, 4, 6, 8, \ldots\}. Every element can be written as x=2mx = 2m for some integer mm.

  3. Check if every element of TT belongs to RR.

    Take any x∈Tx \in T. Then x=6kx = 6k for some integer kk. We can rewrite this as:

x=6k=2⋅(3k)x = 6k = 2 \cdot (3k)

Since 3k3k is an integer (call it m=3km = 3k), we have x=2mx = 2m, which means xx is divisible by 2. Therefore x∈Rx \in R.

  1. Conclusion from the logic. …

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