The set N of natural numbers is:1. unbounded below in R2. bounded abov

1.	The set N of natural numbers is:1. unbounded below in R2. bounded above in R3. unbounded above in R4. bounded above and bounded below in R
Answer» Correct Answer - Option 3 : unbounded above in R Concept: A natural number is a number that occurs commonly and obviously in nature. As such, it is a non-negative number. The set of natural numbers can be denoted by N = {1, 2, 3, 4,....} The set of natural numbers is bounded below and not bounded above in R. We can prove not bounded above in R using contradiction. Proof: Assume by way of contradiction that N is a bounded above. Then, since N is not empty, it follows from the completeness axiom that sup(N) exists. Thus there must be m ∈ N such that sup(N) - 1 < m (sup(N) means supremum of N or least upper bound) ⇒ sup(N) < m + 1 As m ∈ N , also m + 1 ∈ N, Which is a contradiction. ∴ N is not bounded above

The set N of natural numbers is:1. unbounded below in R2. bounded above in R3. unbounded above in R4. bounded above and bounded below in R

Answer» Correct Answer - Option 3 : unbounded above in R

Concept:

A natural number is a number that occurs commonly and obviously in nature. As such, it is a non-negative number. The set of natural numbers can be denoted by

N = {1, 2, 3, 4,....}

The set of natural numbers is bounded below and not bounded above in R.

We can prove not bounded above in R using contradiction.

Proof:

Assume by way of contradiction that N is a bounded above. Then, since N is not empty, it follows from the completeness axiom that sup(N) exists. Thus there must be m ∈ N such that

sup(N) - 1 < m (sup(N) means supremum of N or least upper bound)

⇒ sup(N) < m + 1

As m ∈ N , also m + 1 ∈ N, Which is a contradiction.

∴ N is not bounded above

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