Question

if p is a prime number, prove that root p is irrational​

Answer 1

Step-by-step explanation:

Let √p=x, which is rational.

We know that p is prime and x≠p {as the condition only satisfies for 1}

Hence, p has only 2 factors, 1 and p itself.

But, p=x*x which contradicts our point that p is prime.

Hence, √p must be irrational.

Hope it helps.

Answer 2

Answer:

Step-by-step explanation:

$Let \sqrt{p}$ be a rational number which implies that:

$\sqrt{p}=\frac{a}{b}$ (where a,b belong to integers) and that a and b have no common factor......eqn(i)

$b\neq 0$

$a^{2} =mb^{2}$$m=\frac{a^{2} }{b^{2} }$

this implies that $a^{2}$ is divisible by m

Let a=mc

on sqaring both the sides,

$a^{2} =m^{2} c^{2} \\mb^{2} =m^{2} c^{2}$ because$(a^{2} =mb^{2} )$

$b^{2} =mc^{2}$ (on reducing m on both sides)

$b^{2}$ is divisible by m

and therefore, b is divisible by m .....eqn(ii)

and from eqn(i) and eqn(ii)...

a and b have a common factor i.e. they are both divisible by m.

This contradicts the assumption that $\frac{a}{b}$ is a rational number as a and b should have no common factor except unity(i).

hence $\sqrt{p}$ is not a rational number and that it is irrational

this is done by the method of contraction

hope it helps you