Mathematics

Example : Prove that (i) √2 is irrational number and (ii) √3 is irrational number. Similarly √5, √7, √11…... are irrational numbers.

Solution:
(i) Let us assume, to the contrary, that is √2 rational. So, we can find an integers named r and s (≠ 0) such that √2 = r/s.
Suppose r and s not having a common factor other than 1. Then, we divide by the common factor to get √2 = a/b, where a and b are coprime. So, b√2 = a.
Squaring on both sides and rearranging, we get 2b² = a². Therefore, 2 divides a². Now, by Theorem it following that 2 divides a.
So, we can write a = 2c for some integer c.
Substituting for a, we get 2b² = 4c², that is, b² = 2c².
This means that 2 divides b², and so 2 divides b (again using Theorem with p = 2).
Therefore, a and b have at least 2 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that is rationalnumber. So, we conclude that √2 is an irrational number.

(ii) Let us assume, to contrary, that is √3 rational. That is, we can find integers a and b (≠ 0) such that √3 = a/b.
Suppose a and b not having a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime. So, b√3 = a.
Squaring on both sides, and rearranging, we get 3b² = a².
Therefore, a² is divisible by 3, and by Theorem, it follows that a is also divisible by 3.
So, we can write a = 3c for some integer c.
Substituting for a, we get 3b² = 9c², that is, b² = 3c².
This means that b² is divisible by 3, and so b is also divisible by 3 (using Theorem with p = 3).
Therefore, a and b have at least 3 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that √3 is rational.
So, we conclude that √3 is irrational number.

Example: Prove that 7 - √3 is an irrational number.

Solution :

Method I : Let 7 - √3 is an rational number.
(p, q are integers, q ≠ 0)

√3 =

Here p, q are integers. Therefore, is also integer.
LHS = √3 is also integer but this is contradiction that √3 is irrational so our assumption is wrong that 7 - √3 is rational number.
7 - √3 is irrational number and hence proved.

Method II : Let 7 - √3 is a rational number. We know sum or difference of two rationals is also rational number. Therefore, 7 - (7 - √3) = √3 = A rational Number. But this is contradiction that √3 is irrational.
Therefore, √3 is irrational number and hence proved.

Example : Prove that √5/3 and 2√7 are irrationals

Solution:

(i) Let √5/3 be a rational number. Multiply it by 3, we get = √3
(Product of two rationals is also rational) but this is contradiction that √5 is irrational number.
Therefore, √5/3 is irrational number and proved.

(ii) Let 2√7 is rational number. Multiply it by 1/2, we get
(Division of two rational no. is also rational number)
Therefore, √7 is rational number but this is contradiction that √7 is irrational number.
2√7 is irrational number and it has proved.