"Thus history reveals that the solutions to certain problems required the invention of new numbers. For example, trying to explain the meaning of sqrt(-1) or to solve the equation x^2 = -1 led to the creation of imaginary numbers."
This is the fourth and final page of the section "Number Worlds." As this introduction implies, the number world we read about here is the world of complex numbers.
Here are some excerpts from this page:
"And imaginary numbers led mathematicians to expand the world of numbers to include all real and imaginary numbers and more! With their introduction, all numbers invented thus far could be classified as complex.
"The real number line shows the organization of all real numbers and their respective distances and sizes relative to zero. The imaginary numbers use an imaginary number line. Combining the real and imaginary number lines, we get a way to picture the complex numbers on what is called the complex number plane. Every point in the plane they form is associated with only one complex number -- no other complex number has that location. (4, 0) means 4 + 0i which equals the number 4. The question now is -- Are there any numbers which do not appear on this plane? The chapter The Magic of Numbers has some examples."
The chapter Pappas mentions in this last sentence is Chapter 4 of her book. We read this chapter much earlier in the year, during most of April, so you can read my April posts for examples of numbers that aren't complex. (The two major examples mentioned in the April posts are the quaternions and Cantor's infinite numbers.)
There are two pictures on this page. One of them goes with yesterday's page -- the list of natural numbers is given, with one of them saying:
"Even though we're infinite, not one of us can be found to solve this problem. ? + 5 = 3."
The other picture demonstrates how to find the number -3 1/2 + 2i on the complex plane. First there is a real line with =3 1/2 labeled, and then an imaginary line with 2i labeled. These lines combine to form a complex plane with -3 1/2 + 2i labeled.
By the way, Pappas never finishes the story of how negative -- and ultimately complex -- numbers were finally granted full numberhood. She mentions Cardano among those who used negative numbers to solve equations -- and in fact he's one of the first to use imaginary numbers as well. (I briefly mention this in an old post from nearly three years ago.) But Cardano used negative and imaginary values only in the hopes that they would cancel out and lead to positive real solutions. In fact, when he first gave the Cubic Formula (the analog of the Quadratic Formula for polynomial equations of degree three), if an equation has three positive real roots, the Cubic Formula will always produce imaginary terms that cancel out to produce the real solutions. So Cardano used imaginary numbers only to get to the positive real solutions that he wanted.
To see why negative and imaginary numbers were finally admitted, we actually consider what Pappas says about mathematical worlds -- they have definitions, axioms (or postulates) and theorems. Both negative and imaginary numbers can be made to satisfy postulates -- indeed, of the "Postulates from Arithmetic and Algebra" from Lesson 1-7 of the U of Chicago text real numbers satisfy all of them, and complex numbers satisfy all except the "Postulates of Inequality and Operations." And so these postulates were the main reason for accepting the numberhood of negative and imaginary numbers.
Lesson 7-7 of the U of Chicago text is called "Sufficient Conditions for Parallelograms." (In the modern Third Edition of the text, this appears as Lesson 7-8.)
This is what I wrote two years ago about this lesson:
Lesson 7-7 of the U of Chicago text is on sufficient conditions for parallelograms. This is how we prove that a figure is a parallelogram. Dr. Franklin Mason would call these the "parallelogram tests," or, as he often abbreviates it, the "pgram tests."
The sufficient conditions for a parallelogram are that if a quadrilateral has opposite sides congruent, or if it has opposite angles congruent, or if its diagonals bisect each other, or if it has one pair of sides both congruent and parallel, then the figure is a parallelogram.
Think about it. Suppose we are given a quadrilateral with opposite sides congruent. It's easy to use SSS to prove that it's a parallelogram. The two pairs of given sides already give us SS, and for the third S, we draw in a diagonal and note that it's congruent to itself. And so we use the traditional SSS proof to prove this theorem.
In general, if we are given that a figure is symmetrical -- because it's an isosceles triangle or a kite or any figure that has its own Symmetry Theorem -- then we just use that theorem. But if we're given things like congruent sides or angles, then these are probably the S or A parts of one of the traditional Congruence Theorems SSS, SAS, or ASA and so we should use those.
Therefore this section is an excellent demonstration of the power of the SSS, SAS, and ASA Congruence Theorems, even in a Common Core class where the focus is on transformations.
Parallelograms don't have reflection symmetry. On the other hand, they do have rotational symmetry, and here is an example of a proof where rotational symmetry comes in handy:
-- Prove that if two perpendicular lines intersect at the center of a square, then they divide that square into four congruent regions.
Let's call the square ABCD. A few cases are easy to prove. Suppose the two perpendicular lines are the diagonals of the square AC and BD, which meet at the center O. These lines are perpendicular because the diagonals of any kite are perpendicular, and a square is a kite. These lines intersect at the center O because the diagonals of any rectangle are congruent and bisect each other, and a square is, of course, a rectangle. Then this is enough to prove that the triangles AOB, BOC, COD, and DOA are congruent by SAS.
Another easy case is if the two lines are the perpendicular bisectors of the four sides. Each of the four regions has three right angles (one an angle of the square, and two more right angles by the definition of perpendicular) and hence four right angles (since the sum of the angles of a quadrilateral is 360), and hence each is a rectangle. And as each one's length and height is half that of the original square (by the definition of bisector), hence each is a square. So there are four congruent squares, each with a side length half that of the original square.
But the tough case is when the two lines are neither diagonals nor perpendicular bisectors. The four points where the two lines intersect AB, BC, CD, and DA respectively, let's call these points E, F, G, and H respectively. These quadrilaterals -- AEOH, BFOE. CGOF, and DHOG -- are not special quadrilaterals like kites or trapezoids, yet we must prove them congruent.
Here's my claim -- a 90-degree rotation about O maps AEOH to BFOE. The proof -- we should have proved earlier this week (our extra discussion for Section 7-6) that a rotation maps regular polygons to themselves, so ABCD is mapped to BCDA. Clearly O, the center of rotation, is mapped to itself, so it remains to show where points such as E are mapped. Since EG and FH are perpendicular, rotating line EG must give us line FH, so E'is clearly on line FH. Since E is on AB and AB maps to BC, E' must lie somewhere on line BC. And BC and FH intersect at F, so E' is F.
Similarly, F' is G, G' is H, and H' is E.Therefore the rotation maps AEOH to BFOE -- and since there is an isometry mapping AEOH to BFOE, they are congruent. Similarly both are congruent to CGOF and DHOG. QED
This question was inspired by one I saw many years ago on the Theoni Pappas calendar. [2017 update: Yes, the Pappas calendar I'll finally be reading in three weeks!]
No comments:
Post a Comment