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Every consistent first-order axiomatic system has models of every cardinality. Given this fact, if your modified axioms has ${\mathbb{Q}}^3$ as a model, then ${\mathbb{Q}}^3$ will not be a unique model. However it still makes sense to ask if ${\mathbb{Q}}^3$ is a unique model.
In Hilbert's original writings, he came up with a concrete countable model to his axiomatization of Euclidean geometry. This model is the smallest Pythaogrean field containing ${\mathbb{Q}}$, i.e. $a^2+b^2$ \sqrt{a^2+b^2}$is in the field if$a$and$b$are. If you allow adding irrational points, this countable model is an instance of a discrete model. 2 deleted 1 characters in body Every consistent first-order axiomatic system has models of every cardinality. Given this fact, if your modified axioms has${\mathbb{Q}}^3$as a model, then${\mathbb{Q}}^3$will not be a unique model. However it still makes sense to ask if${\mathbb{Q}}^3$is a unique model. So long as your modification uses only countably many symbols, then there will always be a countable models. Countable models are "discrete" iin in a sense. In Hilbert's original writings, he came up with a concrete countable model to his axiomatization of Euclidean geometry. This model is the smallest Pythaogrean field containing${\mathbb{Q}}$, i.e.$a^2+b^2$is in the field if$a$and$b$are. If you allow adding irrational points, this countable model is an instance of a discrete model. 1 Every consistent first-order axiomatic system has models of every cardinality. Given this fact, if your modified axioms has${\mathbb{Q}}^3$as a model, then${\mathbb{Q}}^3$will not be a unique model. However it still makes sense to ask if${\mathbb{Q}}^3$is a unique model. So long as your modification uses only countably many symbols, then there will always be a countable models. Countable models are "discrete" iin a sense. In Hilbert's original writings, he came up with a concrete countable model to his axiomatization of Euclidean geometry. This model is the smallest Pythaogrean field containing${\mathbb{Q}}$, i.e.$a^2+b^2$is in the field if$a$and$b\$ are. If you allow adding irrational points, this countable model is an instance of a discrete model.