Question

prove that√5 is an irrational number​

Answer 1

Answer:

To prove that √5 is irrational number 

Let us assume that √5 is rational 

Then √5 =  $\frac{a}{b}$

(a and b are co primes, with only 1 common factor and b≠0) 

⇒ √5 =  $\frac{a}{b}$

(cross multiply) 

⇒ a = √5b 

⇒ a² = 5b² -------> α

⇒ 5/a² 

(by theorem if p divides q then p can also divide q²) 

⇒ 5/a ----> 1 

⇒ a = 5c 

(squaring on both sides) 

⇒ a² = 25c² ----> β 

From equations α and β 

⇒ 5b² = 25c²

⇒ b² = 5c² 

⇒ 5/b² 

(again by theorem) 

⇒ 5/b-------> 2 

we know that a and b are co-primes having only 1 common factor but from 1 and 2 we can that it is wrong. 

This contradiction arises because we assumed that √5 is a rational number 

∴ our assumption is wrong 

∴ √5 is irrational number

Answer 2

To prove:-

• $\sf{ \sqrt{5}}$ is an irrational number.

Solution:-

Let us assume that $\sf{ \sqrt{5}}$ is a rational number.

Therefore, it can be expressed in the form $\sf{ \dfrac{p}{q}}$ where p and q are co - prime integers.

$\implies \sf{ \sqrt{5} = \dfrac{p}{q}}$

$\implies \sf{5 = \dfrac{p^2}{q^2}}$ ...[squaring both side]

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

$\implies \sf{5q^2 = p^2}$ ...(1)

$\implies \sf{ p^2 \; is \; a \; multiple \; of \; 5.}$ ...[Euclid division lemma]

$\implies \sf{ p \; is \; also \; a \; multiple \; of \; 5.}$ ...[Fundamental theorem of arithmetic]

$\implies \sf{p = 5m}$

$\implies \sf{p^2 = 25m^2}$ ...(2)

★ From eq. (1) and (2), we get,

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

$\implies \sf{q^2 = 25m^2}$

$\implies \sf{ q^2 \; is \; a \; multiple \; of \; 5.}$ ...[Euclid division lemma]

$\implies \sf{ q \; is \; also \; a \; multiple \; of \; 5.}$ ...[Fundamental theorem of arithmetic]

★ Hence, p and q have a common factor 5.

This become a contradiction that p and q are co prime.

Therefore, $\sf{ \dfrac{p}{q}}$ is not a rational number.

This prove that $\sf{ \sqrt{5}}$ is an irrational number.

$\bold{\underline{\underline{\boxed{\sf{\red{\dag \; Hence \; proved!}}}}}}$

$\rule{200}{2}$