Question

prove that underroot 2 is irrational number​

Answer 1

Step-by-step explanation:

because it is not form in p/q in which p is not equal to q

Answer 2

Assume to reach the contradiction that √2 is a rational number. Then √2 can be written as a fraction, because it is known that every rational number can be written as a fraction.

Let,

$\longrightarrow\sf{\sqrt2=\dfrac{p}{q}}$

where p and q are two integers which are assumed to be coprime integers, i.e., they've no common factors except 1, and $\sf{q\neq0.}$

Then,

$\longrightarrow\sf{2=\left(\dfrac{p}{q}\right)^2}$

$\longrightarrow\sf{2=\dfrac{p^2}{q^2}}$

$\longrightarrow\sf{p^2=2q^2}$

This implies p² is exactly divisible by 2, so is p, since it is an integer.

Let p = 2m for some integer m. Then,

$\longrightarrow\sf{(2m)^2=2q^2}$

$\longrightarrow\sf{4m^2=2q^2}$

$\longrightarrow\sf{q^2=2m^2}$

This implies q² is exactly divisible by 2, so is q, since it is an integer.

Now we got both p and q as multiples of 2, which contradicts our earlier assumption that both p and q are coprime integers.

Thus our assumption is contradicted and proved that √2 is an irrational number.