Answer:
Proof given by contradiction.
Step-by-step explanation:
Given that:
![n \geq 2](https://tex.z-dn.net/?f=n%20%5Cgeq%202)
To prove:
is prime if and only if no positive integer > 1 and
divides
.
Solution:
First of all, let
is a composite number i.e. not a prime number such that:
![n =a\times b](https://tex.z-dn.net/?f=n%20%3Da%5Ctimes%20b)
and
and
are prime and
divides
and
also divides
.
Let ![\sqrt n = p](https://tex.z-dn.net/?f=%5Csqrt%20n%20%3D%20p)
or ![n = p\times p](https://tex.z-dn.net/?f=n%20%20%3D%20p%5Ctimes%20p)
1.
:
is prime and is a divisor of
.
2.
:
![n = a\times b = p\times p](https://tex.z-dn.net/?f=n%20%3D%20a%5Ctimes%20b%20%3D%20p%5Ctimes%20p)
We have assumed that ![a > p \Rightarrow b](https://tex.z-dn.net/?f=a%20%3E%20p%20%20%5CRightarrow%20b%20%3Cp)
is a prime number and is a divisor of
.
But we are given that no prime number
divides
but we have proved that
divides
.
So, it is a contradiction to our assumption.
Therefore, our assumption is wrong that
is a composite number.
Hence, proved that
is a prime number.
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