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*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.*

*hope helped u dear frnd @ :-)*

*thank Q*### This Is a Certified Answer

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.

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.

===============

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.