Q. 103.6( 43 Votes )

# Prove that &radic

Answer :

Let’s assume that √5 is a rational number.

Hence, √5 can be written in the form a/b [where a and b (b ≠ 0) are co-prime (i.e. no common factor other than 1)]

⸫ √5 = a/b

⇒ √5 b = a

Squaring both sides,

⇒ (√5 b)^{2} = a^{2}

⇒ 5b^{2} = a^{2}

⇒ a^{2}/5 = b^{2}

Hence, 5 divides a^{2}

By theorem, if p is a prime number and p divides a^{2}, then p divides a, where a is a positive number

So, 5 divides a too

Hence, we can say a/5 = c where, c is some integer

So, a = 5c

Now we know that,

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

Putting a = 5c,

⇒ 5b^{2} = (5c)^{2}

⇒ 5b^{2} = 25c^{2}

⇒ b^{2} = 5c^{2}

⸫ b^{2}/5 = c^{2}

Hence, 5 divides b^{2}

By theorem, if p is a prime number and p divides a^{2}, then p divides a, where a is a positive number

So, 5 divides b too

By earlier deductions, 5 divides both a and b

Hence, 5 is a factor of a and b

⸫ a and b are not co-prime.

Hence, the assumption is wrong.

⸫ By contradiction,

⸫ √5 is irrational

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