Question

Prove that √2 is irrational Hence show that 3√2 is irrational ​

Answer 1

Answer:

Please refer the below explanation

Step-by-step explanation:

Irrationality of a number can be proved by contradiction and assumption method as below,

Let us assume that √2 is a rational number.

We know that a number which can be expressed in the form of p/q where p and q are co-prime ( two or more numbers with HCF as 1) integers, and q is not equal to 0.

Thus ,

√2 = p/q

On squaring,

(√2) ^2 = (p/q)^2

2= p^2/q^2.

2* q^2 = p^2  ---------------------------- Equation 1

This implies that

2 | p^2

There is a theorem that states if a prime number p divides a^2 then p also divides  a

Thus.

2a= p

Substituting this new value into Equation 1

We get,

2*q^2=(2a)^2

2*q^2= 4a^2

q^2= 2a^2

This implies that

q is also divisible by 2.

By now we understand that p and q have common factor as 2. But this contradicting the fact that p and q are co-prime integers. This contradiction is the result of our false assumption that √2 is a rational number.

Therefore, √2 is not a rational number and it is an irrational number.

Similarly to the previous proof ,

let us consider 3√2 to be rational

thus,

3√2= p/q

3 = p/√2q

We know that 3 is rational number but on the Right hand side we have irrational number being multiplied with a rational number( the product will be an irrational). However,

Rational number ≠ Irrational number.

This contradiction has risen due to our wrong assumption that 3√2  is a rational number.

Thus, 3√2 is an irrational number and not a rational number.

Hence proved.