
The smallest natural number is
[a] 0
[b] 1
[c] 2
[d] None of these
Answer
598.5k+ views
Hint: Recall the definition of natural number. List them and try finding which is the smallest natural number from the list. Finally claim that number is the smallest natural number.
Complete step-by-step answer:
Natural Number: Number 1,2,3,… are called natural numbers. They are also known as counting numbers as these numbers are used for counting.
As can be seen from the definition 1 is the smallest natural number.
Claim: 1 is the smallest natural number.
Proof: Suppose not
Let there exists another natural number $a$ such that $a<1$ and a is a natural number.
Also since natural numbers have at least 1 as the difference between them
Hence we have $1-a\ge 1$
Since every natural number is >0, we have
$a>0$
Hence we have $1\ge 1+a>1$
Hence 1>1, which is a contradiction. Hence 1 is the smallest natural number.
Hence option [b] is correct.
Note: [1] The number 1 is called the greatest lower bound of natural numbers. A number m is said to be the greatest lower bound of a set A if \[\forall a\in A,m\le a\] and $\forall c$ such that $c\le a\forall a\in A$ we have $m\ge c$.
[2] Real numbers have greatest lower bound property,i.e. if a set A which is a subset of R has a lower bound, then A has greatest lower bound also. The natural numbers have a lower bound 0. Hence according to this property they have the greatest lower bound also which is true since 1 is the greatest lower bound of natural numbers.
Complete step-by-step answer:
Natural Number: Number 1,2,3,… are called natural numbers. They are also known as counting numbers as these numbers are used for counting.
As can be seen from the definition 1 is the smallest natural number.
Claim: 1 is the smallest natural number.
Proof: Suppose not
Let there exists another natural number $a$ such that $a<1$ and a is a natural number.
Also since natural numbers have at least 1 as the difference between them
Hence we have $1-a\ge 1$
Since every natural number is >0, we have
$a>0$
Hence we have $1\ge 1+a>1$
Hence 1>1, which is a contradiction. Hence 1 is the smallest natural number.
Hence option [b] is correct.
Note: [1] The number 1 is called the greatest lower bound of natural numbers. A number m is said to be the greatest lower bound of a set A if \[\forall a\in A,m\le a\] and $\forall c$ such that $c\le a\forall a\in A$ we have $m\ge c$.
[2] Real numbers have greatest lower bound property,i.e. if a set A which is a subset of R has a lower bound, then A has greatest lower bound also. The natural numbers have a lower bound 0. Hence according to this property they have the greatest lower bound also which is true since 1 is the greatest lower bound of natural numbers.
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