REAL NUMBERS (CLASS X - CBSE) - Exercise 1.3

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Exercise 1.3

1.  Prove that Ö5 is irrational.

Solution : Let us assume, to the contrary, that Ö5 is rational.

So, we can find integers a and b (≠ 0)  such that Ö5  =   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 Ö5 = a

Squaring on both sides and rearranging we get,

                   5b² = a²    -----  (1)

So, 5 divides a²

Now, by theorem 1.3, it follows that 5 divides a.

So, we can write  a = 5c  for some integer c

Substituting for a in equation(1) we get,

     5b² = (5c)² = 25c²

  i.e., b² = 5c²

This means that 5 divides b² and so 5 divides b (again using theorem 1.3 with p=5)

Therefore a and b have at least 5 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 Ö5 is rational.

So we conclude that Ö5 is irrational.

2. Prove that  3 + 2Ö5  is irrational

Solution : Let us assume that 3 + 2Ö5 is rational.

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

      3 + 2Ö5   =     a       

                            b

Rearranging this equation we get,

          2Ö5 =    a   - 3   

                       b           

          Ö5  =   a-3b

                      2b

Since a and b are integers, we get  a-3b  is rational and so Ö5 is rational

                                                     2b

This contradicts the fact that Ö5 is irrational.

This contradiction has arisen because of our incorrect assumption that 3 + 2Ö5 is rational.

So, we conclude that 3 + 2Ö5 is irrational.

3. Prove that the following are irrationals :

(i)   1

     Ö2

Solution : Let us assume that  1  is rational.

                                             Ö2       

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

        1    =     a       

      Ö2           b

Rearranging this equation we get,

          Ö2 =   b  

                     a               

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

                                                    a

This contradicts the fact that Ö2 is irrational.

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

Ö2 

So, we conclude that   1   is irrational.

                                 Ö2

(ii) 7Ö5

Solution : Let us assume that  7Ö5 is rational.

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

       7Ö5   =     a       

                       b

Rearranging this equation we get,

          Ö5 =   a  

                    7b              

Since a and b are integers, we get   a  is rational and so Ö5 is rational

                                                    7b

This contradicts the fact that Ö5 is irrational.

This contradiction has arisen because of our incorrect assumption that   7Ö5 is rational.

So, we conclude that  7Ö5 is irrational.

(iii) 6 + Ö2

Solution : Let us assume that 6 + Ö2 is rational.

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

      6 + Ö2   =     a       

                          b

Rearranging this equation we get,

          Ö2 =    a   - 6   

                     b             

          Ö2  =   a-6b

                       b

Since a and b are integers, we get  a-6b  is rational and so Ö2 is rational

                                                      b

This contradicts the fact that Ö2 is irrational.

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

So, we conclude that 6 + Ö2 is irrational.

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