I'm a bit late to see this comment, but it was disappointing that none of the replies so far show why this is not true. Take any irrational numbers x and y, an

I'm a bit late to see this comment, but it was disappointing that none of the replies so far show why this is not true. Take any irrational numbers x and y, and suppose for convenience that x < y. Let d = y-x i.e. the length of the interval between x and y. Let b be an integer so large that 1/b is less than d. Consider fractions of the form a/b. As a increases, a/b will eventually (strictly) exceed the value of x. Suppose we fix a to be the smallest such value, i.e. (a-1)/b < x < a/b. Adding d to both sides we get: a/b - 1/b + d < x + d Using the fact that 1/b < d and x + d = y, we get a/b < y. Therefore x < a/b < y.