- #1
Bashyboy
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- 5
Homework Statement
Here the formulation of Dedekind's axioms that I am using:Suppose that line ℓ is partitioned by the two nonempty sets ##M_0## and ##M_1## (i.e., ##\ell = M_0 \cup M_1##) such that every point between two points of ##M_i## is is also in ##M_i##, for ##i = 0,1##. Then there exists a point ##X \in \ell## such that one of the partitions is equal a ray of ##\ell## with origin point ##X## and the other partition is equal to the complement.
Homework Equations
The Attempt at a Solution
By way of contradiction, suppose that the rational plane satisfies Dedekind's axiom (DA). Consider the ##x##-axis ##\ell = \{(a,0)~|~ a \in \mathbb{Q}\}## in the rational plane ##\mathbb{Q}^2##, and the sets ##M_0 = \{(a,0)~|~ a \in \mathbb{Q}^+, a^2 > 2 \}## and ##M_1 = \ell / M_0##. Clearly these two sets partition the line ##\ell##. Therefore, by DA we can conclude that ##\exists! X \in \ell## such that ##X## is the origin point of ##M_0## (or ##M_1##, as both cases should be equivalent), meaning that ##X \in M_0 \implies X = (x,0)##, where ##x > 0## and ##x^2 > 2##.
Here is where I am struck; I cannot find explicitly find the contradiction. It seems that I could show one of two things, although I may be wrong: (1) ##x## is forced to be ##\sqrt{2}##, which would be contradiction, or (2) ##M_0## is not a ray after all.
To show that ##M_0## is not a ray after all, it seems as though I would have to get an inequality of the form ## \sqrt{2} < y < x~~ \forall y \in \mathbb{Q}##, because none of the ##y##'s would be contained in ##M_0## nor in ##M_1##; however, I am being a bonehead and cannot see clearly.
Would someone mind giving me a hint?