Saturday, February 23, 2013

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 + 25 is irrational.
       Let take that 3 + 25 is a rational number.
      So we can write this number as
                                    3 + 25           = a/b
      Here a and b are two co prime number and b is not equal to 0
      Subtract 3 both sides we get
                                    25                 = 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 + 25 is a irrational number

3. Prove that the following are irrationals:
      (i) 1/2 (ii) 75 (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 75 is a rational number.
            So we can write this number as
                                    75                 = 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

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