Question

2. Prove that √10
is an irrational
number​

Answer 1

Answer:

Yes! √10 is an irrational number, which can be proved by contradictory method, as follows….

We assume that √10 is a rational number

=> √10 = p/q ( where, p & q belong to the set of integers, q is not equal to 0)

=> Now, let's cancel out all common factors of p/q & represent it in its least form a/b

So, √10 = a/b, here a & b will be co primes. ie, a & b have no common factor other than one.

=> 10 = a² / b²

=> 10 b² = a² ………….(1)

This concludes that, 10 divides a² exactly.

=> 10 divides ‘a’ exactly. ( as a is an integer)

Or, we can say that 10 is a factor of 'a'

which can be stated as…

=> a = 10 * c

=> a² = 100 c² …………(2)

By (1) & (2)

10b² = 100c²

=> b² = 10 c²

This concludes that , 10 divides b² exactly

=> 10 divides b ( as b is an integer)

Or, 10 is a factor of 'b'

This way, we concluded that 10 is a factor of ‘a’ & ‘b’ both. That shows that a & b are not coprimes…

where as we supposed initially, a & b are coprimes.

This contradiction concludes that our initial assumption is wrong.

Hence, √10 should be an irrational number.

[ Hence Proved]

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Answer 2

Refer to the attachment