Geometric Identity |
Geometric intervals that project each into the other are identical. |
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Because a number is not a geometric element, and a geometric element is not a number,Numerical equality and geometric identity are distinct and disjoint states.By extension, inequality and inidentity are also distinct and disjoint states. We explore how we may establish whether or not
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We will need the following facts:For a geometric interval to exist at all,
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In fact, it might altogether be better to speak of interels rather than of intervals, because inter-vals are between numerical values, but we want that which falls between geometric elements, not between numbers - so, inter-els! |
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Projective Comparison of Intervals on Coplanar Lines |
“Perspectivity” and “Projectivity” |
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![]() A profoundly important special case, illustrated below, arises if point BR1 degenerates into point BS0, on line s, so that intervals kR0 and kS0 “abut” – that is to say, the intervals neither separate nor overlap. In the instance above, the right end of the left interval would become ‘one’ with the left end of the right interval—which should remind us that, while elements do not have ends, intervals do. In this case, these ends are points, so these intervals are interpunctual intervals. There are also interlinear intervals, and interplanar intervals. ![]() This, together with iteration, can be used to construct a chain of identical intervals on a line, which might serve as a sort of projective gradation, or scale — but one must take great care not to construe from it that the intervals have size, least of all equal size. |
Following from the above, two points, say A0 and A1, in a line, say u, form two intervals, say i0 and i1, on that line (fig. 1).
Let A0 lie to the left of A1 from the observer's vantage. Then, interval i0, ‘between’ A0 and A1, lies to the right of A0, and the second interval, i1, ‘between’ A0 and A1, lies to the left of A0, continuing leftwards to the right of A1 (because lines have no ends). Similarly, let points B0 and B1, on a second line, v, For the intervals on line u, to be geometrically identical That is, A0 on u must project into B0 on v, via line m0, say, and vice versa, and A1 on u must project into B1 on v, via line m1, say, and vice versa. This immediately requires that The foregoing is a perspectivity, which is a range of points centrally projected directly from one line to another, and clearly, from the above, its lines and points are necessarily coplanar. We see that whenever we have one perspectivity, we must have two. Here, we have one in point P, and another in point Q. They come in pairs. Next, consider two lines, r and s Now, let both R and S project points A0 & A1 in line r, into points BR0 & BR1, and BS0 & BS1, in line s, which forms two pairs of intervals, kS0 & kS1, and kR0 & kR1, in line s. Since intervals kRn & kSn are “perspectivities” of a single common interval, in, they are each geometrically identical to that common interval, and are, therefore, identical to each other. A double projective comparison, such as this, of two intervals by indirection via a third, is a projectivity. |
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“Lincoln” (Daniel Day-Lewis) on |
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Projective Comparison of Skew (i.e., non-coplanar) Intervals |
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If two lines, say m and n, are skew to each other, they have no condition of incidence, with the result that intervals on these lines cannot be compared by single-stage, central projection—but they can be compared by two-stage, central projection to, and from, a third interval on an intermediate line, say r. That is to say, by a projectivity. This works because, while skew lines cannot be incident, the planes (here, α and β) in or on which they lie must be – in a line.And this line then serves as our
intermediate (here, r) for the two perspectivities carried by the two
incident planes. Note that line m on plane α meets line r in point G, distinct from point H, in which line n on plane β meets line r. This confirms that lines m and n are indeed skew to each other. Corresponding intervals i[0,1] and k[0,1] are thus geometrically identical, as they are in perspective with the same intervals, q[0,1], on line r. |
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A Note on Absolute Size and Equality, with reference to Figure 4 If the intervals i[0,1] and k[0,1] (fig. 4) are to have absolute size and to be ‘equal’ to each other in the Euclidean sense, ![]() However,
since lines m and n
are by definition skew, only one of points G and H can so lie, We conclude that it is always possible to compare skew intervals projectively for geometric identity, but never possible to compare skew intervals projectively for numerical equality. The often-twinned assumptions that |
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We have considered identity of intervals made by point-pairs on lines. (type a, above) We would need also to consider identity of intervals made by line-pairs on points, (type b) and to consider identity of intervals made by plane-pairs on lines. (type c) | (
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** Perhaps surprisingly, a line in a plane has no ‘points in common’ with that plane (intuition might suggest that they have all the points of the line in common). But lines and planes are not made of points, so there are no points available here, from either the line or the plane, to be “in common”. |