Answer:
Proof given by contradiction.
Step-by-step explanation:
Given that:
[tex]n \geq 2[/tex]
To prove:
[tex]n[/tex] is prime if and only if no positive integer > 1 and [tex]\leq[/tex] [tex]\sqrt n[/tex] divides [tex]n[/tex].
Solution:
First of all, let [tex]n[/tex] is a composite number i.e. not a prime number such that:
[tex]n =a\times b[/tex]
and [tex]a[/tex] and [tex]b[/tex] are prime and [tex]a[/tex] divides [tex]n[/tex] and [tex]b[/tex] also divides [tex]n[/tex].
Let [tex]\sqrt n = p[/tex]
or [tex]n = p\times p[/tex]
1. [tex]\underline{a < p}[/tex]:
[tex]a[/tex] is prime and is a divisor of [tex]n[/tex].
2. [tex]\underline{a>p}[/tex]:
[tex]n = a\times b = p\times p[/tex]
We have assumed that [tex]a > p \Rightarrow b <p[/tex]
[tex]b[/tex] is a prime number and is a divisor of [tex]n[/tex].
But we are given that no prime number [tex]\leq \sqrt n[/tex] divides [tex]n[/tex] but we have proved that [tex]b < \sqrt n[/tex] divides [tex]n[/tex].
So, it is a contradiction to our assumption.
Therefore, our assumption is wrong that [tex]n[/tex] is a composite number.
Hence, proved that [tex]n[/tex] is a prime number.
