Question

proof that .1201211122 is irrational?​

Answer 1

Answer:

We know that √2 = 1.201....

So we are required to prove √2 irrational .

Let √2 be rational .

Then √2 = a/b and on squaring both sides we get :

⇒ 2 = a²/b²

⇒ b² = a²/2

Evidently b² is even .

Hence let b = 2 c and then we will have :

4 c² = a²

Then a is also even because a² is even .

But the even numbers become relatively prime and hence there is contradiction .

So √2 is irrational .

Step-by-step explanation:

We proved the statement by contradiction .

We assumed the opposite and we proved that the opposite cannot happen .

This is termed as contradiction .

A rational number is a number that is expressed in p/q form .

p and q are integers .

An irrational number cannot be expressed like that .

Answer 2

ANSWER

As we know that √2 = .1201211122 is an irrational number and it can be easily proved by using assumption method.

Step By Step Explanation

Now let us assume √2 to be a rational number such that it can be expressed in the p/q form.

Now,

√2 = p/q

Now on dividing numerator and denominator by their common factor we get two co-prime numbers, let it be and b.

So, √2 = a/b

(here a and b are co-primes)

In squaring both sides we get =>

2 = (a/b)²

=> 2b² = a²

Hence we can see that 2 is a factor of a² and also a. --------(i)

Now, let a = 2c for some integer c.

So we get =>

2b² = (2c)²

=> 2b² = 4c²

=> b² = 2c²

Hence we can see. that 2 is also an factor of b² and b as well. ------(ii)

From (i) and (ii) we get =>

2 is both a common factor of a and b respectively.

But this is not possible as a and b are co-primes i.e have no common factor other than 1.

Hence our assumption is wrong and √2 =0.1201211122 is irrational