Question

proove that always exists a rational number between any two rational number

Answer 1

Consider two rational numbers A and B, where B>A. The difference between these number is D = B-A. Hence A+D = B. Note that D is a rational number.

Now divide D by a known irrational number that is >1, such as sqrt(2). Call this E, where: E = D/sqrt(2). E is definitely irrational, and positive, and smaller than D. So A < A+E < A+D. Remember A+D = B. So: A < A+E < B, and A+E is irrational. Hence no matter the values of rational numbers A and B, there is always an irrational number that lies between.


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Answer 2

Let p/q  and  n/m  be two rational numbers.    so  p, q, n, m are real numbers.

Without loss of generality,  let us assume that  p/q < n/m  ie.,  p/q - n/m  < 0
   ie.,  (pm - nq) / (q m )  < 0.    --- (1)

     Let  x = [ p/q  +  n/m ] / 2
         = [ p m + n q ] / [2 q m ]    --- (2)

finding
   p/q  -  x  = p/q  -  (pm+nq) / (2qm)
             = [ 2mp - pm - nq  ] / (2qm)             
             = [ mp - nq ] / (2qm)
             < 0    as per  (1)

  n/m  - x =  n/ m  - (pm+nq)/ (2qm)
               = (2qn - pm - nq ) / (2qm)
               =  (qn - pm) / (2qm)
               > 0  as per  (1)

Hence,    p/q  <  x  <  n/m

Thus there is always a rational number  x  between any two rational numbers.
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if there are two rational numbers  X, Y  (Y > X)  then the rational number

         ( X+Y) / 2  or  X + (Y - X) / k,  or (m X + n Y) / (m+n) 

  lies between given X, Y..    here   k , m , n are positive integers.