is irrational

Notation

               (a, b)

               a | b

               Q

               Q’

               Z

 

        H.C.F. of a and b

        a divides b , e.g. 2 | 6

        set of rational numbers

        set of irrational numbers

        set of integers

 

 

Here are different proofs, hope you enjoy reading them.

 

(1) 

     We therefore get  a2 = 2b2.

 

         For the square of an integer, the unit place can be 0, 1, 4, 5, 6 or 9.

     Therefore  the unit place of 2b2  can only be  0, 2 or 8.

     As a2 = 2b2, the unit place should also be the same.

     Hence, the unit place of  a2, 2b2 can only be 0.

     We then deduce that  10  is a common factor of a and b,

     contradicting to  (a, b) = 1.

 

(2)   

      b > 1, otherwise we get  .

      There exists prime factor  p > 1 (can be b itself) such that  p | b.

                Since

                               

       

      In any of the case  p | a, contradicting to (a, b) = 1.

 

(3)   

     Let  p > 1 be a prime factor of b.

     Now,  a2 = 2b2  Þ  b2 = a2 – b2 = (a + b)( a – b)

     then    p | (a + b)   or    p | (a - b)

     In any one of the cases,   p | a.

     But  p | (a, b) contradicting to  (a, b) = 1.

 

(4) 

    We therefore get  a2 = 2b2.

       Since 2b2  is even, hence a2  is even.

    Then, we get  a  is also even.   Let  a = 2c, where  c Î Z.

    Put it in a2 = 2b2, we get  b2 = 2c2.

    With the same reasoning, we get  b is also even.

    Therefore 2 is a factor of (a, b), contradicting to  (a, b) = 1.

 

(5) 

    We therefore get  a2 = 2b2.

    Since  b > 1, then  a > 1.

    By the Fundamental theorem of arithmetic (unique factorization in prime factors),

        ,  

    Substitute in  a2 = 2b2,

    L.H.S. has even number of prime factors and

    R.H.S. has odd number of prime factors (including the number 2)

    Contradiction.

 

(6)  ,

        Let a be the smallest numerator so that this holds.

   

        a2 = 2b2, a2 – ab = 2b2 - ab, 

    a ( a – b ) = b ( 2b – a )

   

    Since  a > b (prove this yourselves)

    \ 2b – a < a , contradicting with a is the smallest numerator.

 

(7) , 

    ( a + 1 ) ( a – 1 ) = 1

   

    Since  is an infinite fraction, it cannot be a rational number.

 

(8)  is a root of the equation  x2 – 2 = 0

   

    From Theory of equation,  b | 1  (the coefficient of )

    But then   , which is not an integer.

 

(9)        Suppose

                       

    Then    and    is an integer.

    So b = 1 and , which is  an integer.

    Contradiction.

 

(10)       Suppose

                       

   

     Therefore

                       

                           where  

     Since    ,  b< a < 2b

             = a – ( 2b – a) = 2 ( a – b ) > 0

                      = b – ( a – b ) = 2b – a > 0

       therefore  .

    Similarly,  if we construct  

    We can therefore get two infinite sequences  with integral numbers

        

    and  ,

    But there are only finite number of integers between a and b.

    Contradiction.