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Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

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.

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

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)

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.