Assuming it exists, then this is the order-preserving isomorphism from integer to rational number. First of all, we know that this mapping is a one-to-one mapping.

T={f(x) | x is an integer}

If t is a set of rational numbers q, then each rational number must have an original image.

Then we might as well set up.

The integer 0 under f looks like rational number a,

The integer 1 under f looks like rational number b,

Then (a+b)/2 is a rational number, but there is no original graph, because there is no integer between 0 and 1.