Question

Prove that √5 is irrational​

Answer 1

Answer:

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Answer 2

ɢɪᴠᴇɴ ᴛʜᴀᴛ,

Prove that √5 is irrational.

Let us assume that √5 is a rational number.

$⟹ \sqrt{5} = \frac{a}{b}$

• [ ∴ a & b are co - primes ]

$⟹ b\sqrt{5} = a$

• [ ∴ squaring on both sides ]

$⟹ {(b \sqrt{5} )}^{2} = {a}^{2} \\ \\ ⟹ {5b}^{2} = {a}^{2} \\ \\ ⟹ {b}^{2} = \frac{ {a}^{2} }{5}$

Hence, 5 divides a² & 5 divides a.

So, a = 5c . Substitute it.

$⟹ {b}^{2} = \frac{ {5c}^{2} }{5} \\ \\ ⟹ {b}^{2} = \frac{25 {c}^{2} }{5} \\ \\ ⟹ {b}^{2} = 5 {c}^{2} \\ \\ ⟹ \frac{ {b}^{2} }{5} = {c}^{2}$

Hence, 5 divides b² & 5 divides b.

Both a & b have a factor of 5. Therefore, a & b are not co - primes.

So, our assumption is wrong.

∴ √5 is an irrational number.

Step-by-step explanation:

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