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.
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