Question

prove that root5 is irrational

Answer 1

Answer:

Step by step explanation:

Proof - Let us assume on the contrary that $\sqrt{5}$ is a rational number. Then, there exist co-prime positive integers a and b such that

$\sqrt{5}=\frac{a}{b}$ , where a and b are integers and a and b are co prime numbers.

=> $5=\frac{a^{2} }{b^{2} }$

=> $5b^{2}=a^{2}$

=> $b^{2}=\frac{a^{2} }{5}$

=> 5 divides $a^{2}$

=> 5 divides a

Let a=5c

=> $5b^{2}=(5c)^{2}$

=> $5b^{2}=25c$

=> $b^{2}=5c^{2}$

=> $c^{2}=\frac{b^{2} }{5}$

=> 5 divides $b^{2}$

=> 5 divides b

This contradicts the facts that a and b are co-prime number

Since,

Our assumption is wrong

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

Thanks.

Answer 2

Proof : Let assume tha √5 is a rational number in its simplest form p/q.

⇒ √5 = p/q

⇒ 5 = p²/q²

⇒ 5 q² = p²

Therefore 5 | p       [∵ 5 is a factor of p]

Now let m be any natural number which factor of p.

⇒ 5 m = p

⇒ 5²m² = p²

⇒ 25 m² = 5 q²

⇒ 5 m² = q²

Therefore 5 | q      [∵ 5 is factor of q]

This leads to contradiction that 5 is factor of both p/q which we assumed to be in simplest form.

Therefore our assumption was wrong and √5 is an irrational number.

Q.E.D