Let Q^{2} be the rational plane of all ordered pairs (

Tara Alvarado 2021-12-13 Answered
Let Q2 be the rational plane of all ordered pairs (x,y) of rational numbers with the usual interpretations of the undefined geometric terms used in analytical geometry. Show that axiom C-1 and the elementary continuity principle fail in Q2. (Hint: the segment from (0,0) to (1,1) can not be laid off on the x axis from the origin.
Axiom C-1: If A, B are two points on a line a, and if A' is a point upon the same or another line a′ , then, upon a given side of A′ on the straight line a′ , we can always find a point B′ so that the segment AB is congruent to the segment A′B′ . We indicate this relation by writing AB=AB. Every segment is congruent to itself; that is, we always have AB=AB.
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sonSnubsreose6v
Answered 2021-12-14 Author has 21 answers
Step 1
Let A, B lies on Horizontal axis.
Lets

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scoollato7o
Answered 2021-12-15 Author has 26 answers

Suppose that the Elementary Continuity Principle holds for this interpretation. Let O=(0,0),A=(3,0),P=(1,0) and R=(3,4) in Q2
Let C be the circle in Q2 with center O and radius OA. Note that C consists of all points (x, y) in Q2 such that x2+y2=9.
Since OPA,OP<OA. Hence, P is inside C. Let B=(95,125). Then OBR and OA=OB. This implies that OA<OR and, hence, that R is outside C. Hence, by the Elementary Continuity Principle, there exists a point Q in Q2 such that Q lies on P R and C.
Since Q lies on PR,Q=P+t(RP) for some real number t such that 0t1. Thus, P=(1,0)+t(2,4)=(1+2t,4t). Since P is in Q2, we conclude that t is a rational number. Since P lies on C, OP=OA. Hence, (1+2t)2+(4t)2=9. In other words, 5t2+t2=0. Since 0t1, it follows that t=1+4110 and, hence, 10t+1=41. Since t is a rational number, it follows that 10t+1 is a rational number. In other words, 41 is a rational number. On the other hand, since 41 is not a perfect square, 41 is not a rational number. This is a contradiction. Hence, the Elementary Continuity Principle fails to hold for this interpretation

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