Prove directly that the map given by is a field isomorphism.
Note that elements of can be written uniquely in the form where , so that this rule indeed defines a function.
Now , so that preserves addition.
Similarly, we have , so that preserves multiplication.
Thus is a ring homomorphism, and hence a field homomorphism.
Note that if , then we have , and so if we have , where and are rational numbers- a contradiction since is not rational. So , and thus , and we have . So is injective.
Finally, is surjective since we have for all and .
So is a field isomorphism.