A basis for the right quantum algebra and the “1 = q” by Foata D., Han G.-N.

By Foata D., Han G.-N.

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16 Robson, 2007, p. 104. 15 Egypt and Mesopotamia 27 can be translated into the modern equation 11x 2 + 7x = 6 14 . The scribe multiplied by 11 to turn the equation into a quadratic in 11x, namely, (11x)2 + 7(11x) = 68 34 . He or she then solved 11x = 7 2 2 1 1 3 7 √ 7 + 68 − = 81 − = 9 − 3 = 5 . ’’ Nevertheless, because the problem was probably manufactured to give a simple answer, it is easy to see that the unknown side x is equal to 12 . This idea of “scaling’’ also enabled the scribes to solve quadratic-type equations that did not directly involve squares but that, instead, involved figures of different shapes.

107. Egypt and Mesopotamia 29 of the actual sides of the square into rectangles of length 1. Taking 14 of the entire sum means considering only the shaded gnomon which is one-fourth of the original figure. When we add a square of side 1 to that figure, we get a square the side of which we can then find. Subtracting 1 from the side gives half the original side of the square. Other problems on BM 13901 treat various situations involving squares and sides, with each of the solution procedures having a geometric interpretation.

14: To construct a square equal to a given rectilinear figure. We can think of this latter construction as an algebraic problem because we are asked to find an unknown side of a square satisfying certain conditions. Since Euclid had already shown that a rectangle could be constructed equal to any rectilinear figure, this problem translates into the modern equation x 2 = cd, where c, d are the lengths of the sides of the rectangle constructed. Again, an analysis may prove helpful. Assume that c = d (for otherwise the problem would be solved).

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