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=∼A′B′. Every segment is congruent to itself; that is, we always have AB=∼AB.

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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&#039; 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=∼A′B′. Every segment is congruent to itself; that is, we always have AB=∼AB.

sonSnubsreose6v · Accepted Answer

Step 1  Let A, B lies on Horizontal axis.  Lets

scoollato7o · Accepted Answer

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 O⋅P⋅A,OP&amp;lt;OA. Hence, P is inside C. Let B=(95,125). Then O⋅B⋅R and OA=∼OB. This implies that OA&amp;lt;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(R−P) for some real number t such that 0≤t≤1. 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+t−2=0. Since 0≤t≤1, 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

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

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