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By Sharipov R.A.
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Sample text
The codirectedness −→ −→ −→ −→ conditions AB ⇈ CD and AC ⇈ BD follow from the disposition of points A ≺ B ≺ C ≺ D. From [AB] ∼ = [CD] and from the obvious relationship [BC] ∼ = [BC], applying the first item of the axiom A15, we derive [AC] ∼ = [BD]. Conversely, from [AC] ∼ = [BD] and § 4. SLIPPING VECTORS. ADDITION OF VECTORS . . 61 [BC] ∼ = [BC], upon applying the item (2) of the axiom A15, we get [AB] ∼ = [CD]. The theorem is proved. 2. 1. Proof. Assume that pAB and pCD are the mappings of congruent translation on a straight line a and assume that pAB = pCD = p.
Then we choose and fix some point A lying on the plane α, but not lying on the line a. Let’s draw the line AO and apply the axiom A10 to the points A and O on this line. As a result we find a point B on the line AO such that the point O lies in the interior of the segment [AB]. The points A and B belong to the set α \ a. They are not equivalent since the segment [AB] intersects the line a at the point O. Hence, the equivalence classes Cl(A) and Cl(B) are distinct. Let’s prove that an arbitrary point X of the set α \ a belongs to one of these classes.
E. a + b = b + a; (2) it is associative, i. e. (a + b) + c = a + (b + c); (3) there is a vector 0, such that 0 + a = a + 0 = a for an arbitrary vector a; (4) for any vector a there is an opposite vector a′ such that a + a′ = a′ + a = 0. 3. 7) since the property of associativity is peculiar to the composition of any mappings. 7) and from the definition of the zero vector p0 = id. The rest is to prove the fourth property. −→ Let AB be a geometric realization for a slipping vector a. Let’s denote by a′ the slipping vector whose geometric realization is the −→ vector BA.