Density of rationals?

Question

Does anyone know how to prove that between any two real numbers, there is a rational number?

Answer 1

Suppose the two real numbers are x and y with x1, so qx+1

Answer 2

This is simple. But the previous answerers have not noticed that you were asking about two REAL numbers, not two RATIONAL numbers.

Let x and y be two real numbers, with x > y.
Let x - y = d, and since x and y are different, d > 0.

Since d > 0, there is some integer n, such that:
1/n < d.

Now consider the set of rational numbers of the form:
p/n, where p is any integer. One of them fits in between x and y.

Specifically, let j/n be the largest of these fractions that is still less than y. Then, since
j/n < y

j/n + 1/n < y +1/n < y + d = y + (x - y) = x
But since j/n + 1/n is > y,
y < j/n + 1/n = (j+1)/n < y + 1/n < x , so

y < (j+1)/n < x

So (j+1)/n is your candidate for a rational number between x and y. There are lots of others, but this one will do!

Answer 3

Whoops, misread the question, sorry.

Let x any y be two distinct real numbers, x>y.

Then x-y>0.

Let N be an integer greater than 1/(x-y).

Now, let P = [2*N*x], where [z] is the largest integer smaller than or equal to z.

Let P' = [2*N*y]. Then P-P'>1, because P = 2*N*x + e where 0<=e<1, and P' = 2*N*y + e' where 0<=e'<1, so:

P-P' = 2*N*(x-y) + (e-e')

Since (e-e')>-1 and N*(x-y)>1, we get that P-P'>1.

So we have that P'+1
Let r = (P'+1)/(2N).

r = (P'+1)/(2N) = (2*N*y+e'+1)/2N = y + (e'+1)/2N > y

r = (P'+1)/(2N) < P/(2N) = (2N*x+e)/(2N) = x+e/(2N)<=x.

So, r is a rational number between x and y.

Answer 4

Rational numbers are real - What's the problem? I think you can show that the average of the two rational numbers is also rational.

Answer 5

huhuhuhu...
abstract algebra....