REAL NUMBERS (CLASS X - CBSE) Pre - exercise 1.3 examples

Theorem 1.3 :  Let p be a prime number. If p divides a², then p divides a, where a is a positive integer.

Proof : Let prime factorization of a be as follows,

          a = p₁, p₂, . . . , p_n, where p₁, p₂, . . . , p_n are primes, not necessarily distinct.

          \a² = (p₁, p₂, . . . , p_n ) (p₁, p₂, . . . , p_{n)  =} p₁², p₂², . . . , p_n²

Now, we are given that p divides a². Therefore from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a². However, using the uniqueness part of  the Fundamental Theorem of Arithmetic, we realize that the only prime factors of a² are p₁, p₂, . . . , p_n. So, p is one of p₁, p₂, . . . , p_n.

Now, since a = p₁, p₂, . . . , p_n, p divides a.

Theorem 1.4 :  Ö2 is irrational

Proof : Let us assume, to the contrary, that Ö2 is rational.

So, we can find integers r and s (≠ 0)  such that Ö2  =   r                                                                                                                                 s

Suppose r and s have a common factor other than 1. Then we divide by the common factor  to get Ö2 =   a   where  a and b are co-prime.

                               b 

So,              b Ö2 = a

Squaring on both sides and rearranging we get,

                   2b² = a²    -----  (1)

So, 2 divides a²

Now, by theorem 1.3, it follows that 2 divides a.

So, we can write  a = 2c  for some integer c

Substituting for a in equation(1) we get,

     2b² = (2c)² = 4c²

This means that 2 divides b² and so 2 divides b (again using theorem 1.3 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 Ö2 is rational.

So we conclude that Ö2 is irrational.

Example - 9 : Prove that Ö3 is irrational

Solution : Let us assume, to the contrary, that Ö3 is rational.

So, we can find integers a and b (≠ 0)  such that Ö3  =   a                                                                                                                                 b

Suppose a and b have a common factor other than 1. Then we divide by the common factor  and  assume that  a and b are co-prime.

                            

So,              b Ö3 = a

Squaring on both sides and rearranging we get,

                   3b² = a²    -----  (1)

So, 3 divides a²

Now, by theorem 1.3, it follows that 3 divides a.

So, we can write  a = 3c  for some integer c

Substituting for a in equation(1) we get,

     3b² = (3c)² = 9c²

  i.e., b² = 3c²

This means that 3 divides b² and so 3 divides b (again using theorem 1.3 with p=3)

Therefore a and b have at least 3 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 Ö3 is rational.

So we conclude that Ö3 is irrational.

Example - 10 : Show that 5 - Ö3 is irrational.

Solution : Let us assume that 5 - Ö3 is rational.

We can find a co-prime a and b (b≠ 0) such that

      5 - Ö3   =     a       

                         b

Rearranging this equation we get,

          Ö3 =  5  -  a   =   5b - a

                           b           b       

Since a and b are integers, we get  5b - a  is rational and so Ö3 is rational

                                                       b

This contradicts the fact that Ö3 is irrational.

This contradiction has arisen because of our incorrect assumption that 5 - Ö3 is rational.

So, we conclude that 5 - Ö3 is irrational.

Example - 11 : Show that 3Ö2 is irrational.

Solution : Let us assume that 3Ö2 is rational.

We can find a co-prime a and b (b≠ 0) such that

      3Ö2   =     a       

                      b

Rearranging this equation we get,

          Ö2 =   a  

                    3b              

Since a and b are integers, we get   a  is rational and so Ö2 is rational

                                                    3b

This contradicts the fact that Ö2 is irrational.

This contradiction has arisen because of our incorrect assumption that  3Ö2 is rational.

So, we conclude that 3Ö2 is irrational.