Prove: Given any two rational numbers, there exists a rational number between them.?

Question

The fact that I have to make this so generalized (i.e. with variables) is making me go around in circles. Take a crack at this one:

Prove: Given any two rational numbers, there exists a rational number between them.

I know you start with two rational number defined by a/b and c/d where a,b,c,d are integers. But from there I can't get anything done! Can anyone please help me??

Answer 1

Well, the problem as stated cannot be proven. If a/b = c/d, there is no rational number between them.

But I guess you really mean something like this:
Given a, b, c, d are all integers,
AND that a/b does not equal c/d,
prove that:
there exists a rational number x, such that
x > a/b
AND
x < c/d.

So all you have to do is find some number that's definitely in between a/b and c/d. An obvious candidate is the average of a/b and c/d.

The average is defined as the sum divided by the number of things you're averaging.

The average of 5 and 7 is 6. The average -3 and 1 is -1. So this looks like it should work.

x = (a/b + c/d) / 2
x = (ad/bd + cb/bd) / 2
x = ((ad + cb)/bd) / 2
x = (ad + cb)/2bd

But wait, we were supposed to prove that there was a rational number. Is x a rational number?

ad + cb is an integer, because the multiplication and addition of integers yields an integer.
2bd is an integer, for the same reason.

Therefore x is a rational number. FLAWLESS VICTORY!

So there's our formula.

Let's try this in an example. Let's pick a = 6, b = 7, c = 4, and d = 5.

a/b is 6/7 or 0.857142...
c/d is 4/5 or 0.8

By our formula, a number between them ought to be :

x = (ad + cb)/2bd
x = (6 x 5 + 4 x 7) / 2 x 7 x 5
x = (30+28)/70
x = 58/70 or approximately 0.8285

Tada.

I gave you the answer here, but I hope this helped you understand how such proofs should be undertaken.

Answer 2

Two Rational Numbers

Answer 3

Proof: For any two rational numbers, a and b, there exists a number c that is their average and is also rational and lies between the two values. The sum, difference, product and quotient of any two rational numbers are rational by definition. The sum of the two is therefore rational. Dividing the sum by 2, another rational number, therefore gives a rational result.

Answer 4

Assume that there exist two rational numbers, n and m, that do NOT have any rational numbers between them, where m > n. Then m + n > n+n => (m+n)/2 > (n+n)/2 => (m+n)/2 > n. Similarly, m > (m+n)/2. So, since n < (m+n)/2 < m, (m+n)/2 must be irrational. But if n = a/b and m = c/d, then: (m+n) = (da+bc)/bd and (m+n)/2 = (da+bc)/(2*bd). So (m+n)/2 is clearly irrational and our assumption must be false. Therefore, there are no such numbers m and n.

Answer 5

Take their average! 1/2 ((a/b) + (c/d)) is rational, and between the other two.

Answer 6

so given a/b and c/d, there exists (a/b + c/d) / 2, half way between them.

so for any a, b, c and d, there exists (ad + cb)/2bd