X mathematics solutions chapter 1 exercise 1.3

EXERCISE 1.3

1.    Prove that √5 is irrational.

                             Let take √5 as rational number

                        If   a and b are two co prime number and b is not equal to 0.

                                     We can write √5      = a/b

                        Multiply by b both side we get

                                                            b√5     = a

                        To remove root, Squaring on both sides, we get

                                                            5b^2    = a^2    ……………(1) 

      Therefore, 5 divides a^2 and according to theorem of rational number, for any prime number p which is divides a^2 then it will divide a also.

                                    That means 5 will divide a. So we can write

                                                            a          = 5c

                        and plug the value of  a in equation (1) we get

                                                            5b^2    = (5c)^2

                                                            5b^2    = 25c^2

                        Divide  by 25 we get

                                                            b^2/5   = c^2

            again using same theorem we get that b will divide by 5

            and we have already get that a is divide  by 5

            but  a and b are co prime number. so it is contradicting .

                                    Hence √5  is a non rational number    

2.    Prove that 3 + 2√5 is irrational.

       Let take that 3 + 2√5 is a rational number.

      So we can write this number as

                                    3 + 2√5           = a/b

      Here a and b are two co prime number and b is not equal to 0

      Subtract 3 both sides we get

                                    2√5                 = a/b – 3

                                    2√5                 = (a-3b)/b

      Now divide by 2 we get

                                    √5                    = (a-3b)/2b

      Here a and b are integer so (a-3b)/2b is a rational number so √5 should be a rational number But √5 is a irrational number so it contradict the fact   

                                    Hence result is 3 + 2√5 is a irrational number

3. Prove that the following are irrationals:

      (i) 1/√2 (ii) 7√5 (iii) 6 + √2

     

      (i)   Let take that 1/√2  is a rational number.

            So we can write this number as

                                    1/√2                = a/b

            Here a and b are two co prime number and b is not equal to 0

            Multiply by √2 both sides we get

                                    1                      = (a√2)/b

            Now multiply by b

                                    b                      = a√2

            divide by a we get

                                    b/a                   = √2

      Here a and b are integer so b/a is a rational number so √2 should be a rational number But √2 is a irrational number so it is contradict  

                                                            Hence result is 1/√2 is a irrational number

(ii)        Let take that 7√5 is a rational number.

            So we can write this number as

                                    7√5                 = a/b

            Here a and b are two co prime number and b is not equal to 0

            Divide by 7 we get

                                    √5)                  =a/(7b)

      Here a and b are integer so a/7b is a rational number so √5 should be a rational number but √5 is a irrational number so it is contradict  

                                                            Hence result is 7√5 is a irrational number.

(iii) Let take that  6 + √2 is a rational number.

      So we can write this number as

                                     6 + √2            = a/b

      Here a and b are two co prime number and b is not equal to 0

      Subtract 6 both side we get

                                    √2                    = a/b – 6

                                    √2                    = (a-6b)/b

Here a and b are integer so (a-6b)/b is a rational number so √2 should be a rational number But √2 is a irrational number so it is contradict  

                                                            Hence result is 6 + √2 is a irrational number