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Extra resources for An easier solution of a Diophantine problem about triangles, in which those lines from the vertices which bisect the opposite sides may be expressed rationally

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Of y, this is a semiordered set. If, on the contrary, x < y denotes that x is a child of y, there is no semiordered set owing to the absence of transitivity. ) of the natural numbers and also in the set of the real numbers, then the usual relation < denotes an ordering relation. In these sets the sign < retains its former meaning. If, on the contrary, we wish to define another ordering or semiordering relation in one of these sets, then we denote it by the sign -<. This applies to other analogous cases.

It still remains to prove that if S is an infinite set, then it is possible to map it one-to-one onto a subset of it. According to Theorem 9 we take a countably infinite subset A of S and put B = S - A. 3) it is possible to map A one-to-one onto a proper subset of itself. Let us define an extension of this mapping such that every element of B is its own image. Thus we obtain a one-to-one mapping of S onto a proper subset. This completes the proof of Theorem 8. It follows from Theorem 8 that every (non-empty) finite set S is equivalent to only one set of the form (1, .

As these attempts have been unsuccessful, mathematicians have now accepted the axiom of choice. § 16. Transfinite Induction By means of (complete) induction we are able to prove only such propositions as can be suitably split up into countably many component propositions. The following is more general. THEOREM 20 (theorem of transfinite induction). Let a proposition A, correspond to each element v of a well-ordered set I. If the truth of A follows from the assumption of the truth of all propositions Ax (x < v) for each v, then all the propositions A, (v E I) are true.

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### An easier solution of a Diophantine problem about triangles, in which those lines from the vertices which bisect the opposite sides may be expressed rationally by Euler L.

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