E-320: Teaching Math with a Historical Perspective Oliver Knill, 2012Lecture 5: WorksheetsWe look at the quadratic and the cubic equation and then at puzzles like the 15 puzzle o r Rubiktype puzzles.The quadratic equationThe solution of the quadratic equation x2+bx+c = 0 isone of the major achievements of early algebra. It relieson the method of completion of the square and isdue to the Persian mathematician Al Khwarizmi.Artistrendering of Al Khwarizmi for anadvertisement.The completion of the square is the idea to add b2/4on both sides of the equation and move the constant tothe right. Like this x2+ bx + b2/4 becomes a square(x + b/2)2. Geometrically, one has added a square to aregion to get a square. From (x + b/2)2= −c + b2/4 wecan solve x and get the famous formula for the solutionof the quadratic equationx =sb24− c −b2.Since o ne can take bot h the positive and the negativesquare root, there are two solutions.xb2xb21) Write down the solution formula for the equation ax2+ bx + c = 0.2) If x1, x2are the two solutions to x2+bx+c = 0, then the sum of the two solutions is x1+x2= −b.3) If x1, x2are the two solutions of x2+ bx + c, then the pr oduct of the solutions is x1x2= c.4) What are the solutions to is x4− 4x2+ 3 = 0?5) Find the solutions to x6− 4x4+ 3x2= 0.The cubic equation1) Lets look at t he cubic equation x3− 7x + 6. Can you figure out the roots?2) Verify that if a, b, c are solutions to a cubic equation satisfy a + b + c = 0 if and only if it isdepressed: x3+ px + q = 0. Hint: Write (x − a)(x − b)(x − c).Lecture 5: Symmetry groupsWe look at a ll the rotational symmetries of a square and realize it as a group. Then, we do thesame for all rotational and reflection symmetries of a rectangle.The rotation symmetries of a squareGiven a square in the plane centered at the origin. We can rotate the square by 90, 180 or 270degrees and get the same shape. Given two such rotations, we can perform one a fter the otherand get an other rotation. All the rotations leaving the square invariant form a group: one can”add” these operations and get a new operation.+ turn 0 turn 90 turn 1 80 turn 2 70turn 0 turn 0 turn 90 turn 1 80 turn 2 70turn 9 0 turn 90 turn 180 turn 270 turn 0turn 1 80 turn 1 80 turn 270 turn 0 turn 90turn 2 70 turn 2 70 turn 0 turn 90 turn 1 80We can write the multiplication table in a more compact way by writing 1 for the turn 90degrees, 2 for the turn 180 degrees and 3 for the turn 270 degrees:+ 0 1 2 30 0 1 2 31 1 2 3 02 2 3 4 13 3 0 0 2If we look at all rotations and reflections which leave the square inva riant, there are 8 groupelements. Beside the four rotations, we have 2 reflections at the diagonals and 2 reflections at themain axes.1) What do you obtain if you reflect at the x axes, then reflect at the dia gonal x=y?2) What do you obtain if you reflect first at the diagonal x=y and the reflect at the x axes?3) We have seen a bove that the rotations which leave t he square invariant, form a group whichsatisfies xy = yx. It is commutative. Does the group of rota tions and reflections form a symmetrygroup which is commutative?The symmetries of a rectangleAssume now we have a rectangle which is not a square. We look at all possible symmetries of thisobject. They include the identity, the refection at the axes as well as the reflection at the center(which is a turn 180 degrees).4) What do you obtain if you compose a reflection at the x-axes with the reflection at the y-axes?5) Is the symmetry group of a rectangle commutative?The rotational symmetry group as well as the full rotation-reflection symmetry group can beintroduced for any geometrical object. Like triangles, cubes, octahedrons or polyhedra for tilingsin the plane. Understanding the symmetries of an object produces a link between algebra andgeometry.The 15 puzzleThe 15 puzzle was invented by Noyes Palmer Chapmanin 1874. Chapman was a postmaster fr om Canastota in NewYork. From there the puzzle moved over to Syracuse, Watch-hill, Hartfort and was first seriously sold in Boston. SamLoyd offered a 1 000 dollar prize for the solution of t he case,when two pieces are switched. Since the number of trans-positions plus the distance of the empty space to the 15’thposition is always an even number, it can not be solved if itis odd initially.1) Argue that the puzzle has less or equal than 16! group elements.The floppyThe floppy cube was designed by Katsuhiko Okamoto. With192 possible positions it is much less complex than the Rubikcube. We will learn how to solve it in class.The Rubik’s cubeThe Rubik’s cube is quite a la rge puzzle. 3) Argue that therubic cube has less than 8! · 12! · 38· 212group