Question

prove that underroot 5 is an irrational number​

Answer 1

Assume to reach the contradiction that √5 is a rational number.

The assumption implies that √5 can be written as a fraction, because it is known that every rational numbers can be written as a fraction.

Let,

$\longrightarrow\sf{\sqrt5=\dfrac{p}{q},\ \ q\neq0}$

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

We see that,

$\longrightarrow\sf{5=\left(\dfrac{p}{q}\right)^2}$

$\longrightarrow\sf{5=\dfrac{p^2}{q^2}}$

$\longrightarrow\sf{p^2=5q^2}$

This implies p² is exactly divisible by 5, so is p, since it is an integer.

Let p = 5m for some integer m. Then,

$\longrightarrow\sf{(5m)^2=5q^2}$

$\longrightarrow\sf{25m^2=5q^2}$

$\longrightarrow\sf{q^2=5m^2}$

This implies q² is exactly divisible by 5, so is q, since it is an integer.

Now we got both p and q as multiples of 5, which contradicts our earlier assumption that both p and q are coprime integers.

Thus our overall assumptions are contradicted and thus proved that √5 is an irrational number.