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

By Foata D., Han G.-N.

Read or Download A basis for the right quantum algebra and the “1 = q” principle PDF

Similar algebra books

An Introduction to Homological Algebra (2nd Edition) (Universitext)

With a wealth of examples in addition to ample purposes to Algebra, it is a must-read paintings: a in actual fact written, easy-to-follow consultant to Homological Algebra. the writer presents a remedy of Homological Algebra which methods the topic when it comes to its origins in algebraic topology. during this fresh version the textual content has been absolutely up to date and revised all through and new fabric on sheaves and abelian different types has been further.

Commutative Algebra, Vol. 2

This moment quantity of our treatise on commutative algebra offers mostly with 3 uncomplicated issues, which transcend the roughly classical fabric of quantity I and are customarily of a extra complicated nature and a newer classic. those subject matters are: (a) valuation idea; (b) thought of polynomial and gear sequence earrings (including generalizations to graded jewelry and modules); (c) neighborhood algebra.

Additional info for A basis for the right quantum algebra and the “1 = q” principle

Sample text

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).