Question

prove that root 5 is irrational​

Answer 1

Step-by-step explanation:

Let us assume, on the contrary, that √5 is a Rational Number

We know that, a Rational number is of the form p/q where p and q are integers, and they are coprimes, and q not equal to 0

(When two number have only 1 as it's common factor, they are called coprimes.)

So,

p/q = √5

Squaring both sides we get,

p²/q² = 5

p² = 5q² ----- 1

Thus, 5 is a factor of p²

So, By Theorem

5 is also a factor of p

Let p = 5m

Putting p = 5m in eq.1

(5m)² = 5q²

25m² = 5q²

q² = 25m²/5

q² = 5m²

Thus, 5 is a factor of q²

So, By Theorem

5 is also a factor of q

But we said that p and q are coprime number and have no other common factors, this solution contradicts our statement

This contradiction has risen due to our incorrect assumption statement

Thus, √5 is an irrational number

Hope it helped and you understood it........All the best

Answer 2

$\star\:\:\:\bf\large\underline\blue{Question:-}$

• Prove that $\sqrt{5}$ is irrational.

$\star\:\:\:\bf\large\underline\blue{Proof:-}$

Let $\sqrt{5}$ be rational, say $r$, then

$\sqrt{5} = \frac{p}{q}$, where $p, q$ are integers, $q\neq0$ and $p, q$ have no common factors (except 1).

$\implies 5 = \frac{ {p}^{2} }{ {q}^{2} }$

$\implies {p}^{2} =5 {q}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: ...(i)$

As $5$ divides $5q^{2}$, so $5$ divides $p^{2}$ but $5$ is prime.

$\implies5$ divides $p$.

Let $p=5m$, where $m$ is an integer.

Substituting the value of $p$ in $(i)$, we get,

${(5m)}^{2} = 5 {q}^{2}$

$\implies25 {m}^{2} = 5 {q}^{2}$

$\implies5 {m}^{2} = {q}^{2}$

As, $5$ divides $5m^{2}$, so, $5$ divides $q^{2}$ but $5$ is prime.

$\implies 5$ divides $q$.

Thus, $p$ and $q$ have a common factor $5$. This contradicts that $p$ and $q$ have no common factor (except 1).

Hence, $\sqrt{5}$ is not a rational number. So, we conclude that $\sqrt{5}$ is an irrational number.