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Math Discovery Examples

Math Discovery Algebra (Space)

Space* A blank sheet is blank. We may or may not refer to that blankness. We may give it a name: identity, zero, one, empty set. The blankness is that origin point, that average, that center which is often unsaid but we may want to note as the natural, clever reference point, as in the case of the swimmer's hat that floated downstream (pg.64) Next, we can expand around the center by balancing positive and negative, numerator and denominator. We thereby introduce parity (Z2), odd or even, affirm or reject, where to reject rejection is to affirm. Next, we can expand terms as polynomials, as with "and" and "or", and thus create equations that construct and relate roots. Finally, we can consider a vector space in which any point can serve as the center for a basis. We thereby construct external "space". It is fully fledged in that it can define vectors, thus model time with arrows that have beginnings they come from and ends they go to.10

Center* A blank sheet is blank. We may or may not refer to that blankness. We may give it a name: identity, zero, one, empty set. The blankness is that origin point, that average, that center which is often unsaid but we may want to note as the natural, clever reference point, as in the case of the swimmer's hat that floated downstream (pg.64) (Blank sheet, what is so central that it is often left unsaid, origin of a coordinate system, natural or clever point of view, symmetrize an equation, average principle, choice of notation, convenient notation)65

• Change your point of view* Changing your point of view is typically choosing the origin for a coordinate system. Changing the point of view is just another manifestation of peripheral vision. Sometimes a problem is hard only because we choose the "wrong" point of view. Spending a few minutes searching for the "natural" point of view can pay big dividends. pg. 64 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.1522
  • The swimmer's hat* A person dives from a bridge into a river and swims upstream through the water for 1 hour at constant speed. She then turns around and swims downstream through the water at the same rate of speed. As the swimmer passes under the bridge, a bystander tells her that her hat fell into the river as she originally dived. The swimmer continues downstream at the same rate of speed, catching up with the hat at another bridge exactly 1 mile downstream from the first one. What is the speed of the current in miles per hour? ... what if we look at things from the hat's point of view? pg. 64 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.1523
• What is essential? And what is peripheral? What are they asking for? And what are they not asking?
• Simple form* Given any diophantine equation [an equation whose variables assume only integer values] ... Is the problem in "simple" form? Always make sure that you have divided out all common factors, or assume the variables share no common factors, etc. The purpose of math is to construct a simplification. pg.264 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc2220
  • Axiom of the empty set* Wikipedia: There is a set such that no set is a member of it.1162
  • The decimal point* Dividing a number by 10, we are taught to move the decimal point to the left, but actually, it is the number that moves to the right. The decimal point is a fixed reference point. Similary, multiplying a number by 10, it may seem that the decimal point moves to the right, but actually, the number moves to the left.807
  • Zero is just a place holder* Note that zero is just a place holder and so we don't have to multiply or add by zero in the algorithms. Zero is a fiction (as a numerical value outside of the system) and those operations are placeholder operations (meaningful only with regard to the particular system). Gospel Math. 1849
• Considering the simplest point in an arbitrary coordinate system* In the following problem, the "north pole" is the simplest point to consider, yet the coordinate system was arbitrary, and so the conclusion is valid for all points. We need some "notation". Let us assume that there is a universal coordinate system, such as longitude and latitude, so that we can refer to the "same" location on any planet. For example, if the planets were little balls floating in a room, the location "north pole" would mean the point on a planet which was closest to the ceiling. Given such a universal coordinate system, what can we say about a planet P which has a private point at location x? Without loss of generality, let x be at the "north pole". Clearly, the centers of all the other planets must lie on the south side of P's "equatorial" plane. But that renders the north poles of these planets public ... we have shown pretty easily that If location x is private on one planet, it is public on all the other planets. pg. 64 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.1521
• Convenient notation* Think about convenient notation. pg.29, The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.1426
• Elegant solution* Algebra is commonly taught as a series of computational techniques. ... algebra is also an aesthetic subject. Sometimes one has to slog through messy thickets of algebraic expressions to solve a problem. But these unfortunate occasions are pretty rare. A good problem solver takes a more confident approach to algebraic problems. The wishful thinking strategy teaches her to look for an elegant solution. Cultivate this mindset: employ a light, almost delicate touch, keeping watch for opportunities that avoid ugly manipulations in favor of elegant, often symmetrical patterns. pg.162, The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.2168
• Invariant with respect to permutation of some of the roots of a polynomial* The substitution u:=x + 1/x, which helped solve x**4+x**3+x**2+x+1=0 worked because u is invariant with respect to a permutation of some of the roots. This idea is the germ of ... Galois theory. pg. 103, The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.1650
• Invariants* An invariant, as the name suggests, is merely some aspect of a problem - usually a numerical quantity - that does not change, even if many other properties do change. pg. 102, The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.1643
  • Euler's formula* Given a polyhedron without holes, the numbers of vertices, edges and faces satisfy V - e + f = 2, which is an invariant. See pg. 103, The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.1645
• Mean Value Theorem* If f(x) is continuous on [a,b] and differentiable on (a,b), then there is a point u in (a,b) at which f'(u) = (f(b) - f(a))/(b-a). ... the proof is just one sentence: Tilt the picture for Rolle's theorem! The mean value theorem connects a "global" property of a function (its values at the endpoints a and b) with a "local" property (the value of its derivative at a specific point) and is thus a deeper and more useful fact than is apparent at first glance. ... Suppose f is differentiable on all real numbers and there is a constant k < 1 such that |f'(x)|<=k for all real x. Show that f has a fixed point. ... Since the derivative is at most k in absolute value, and since k < 1, the graph of y=f(x) to the right of the y-axis will be trapped within the dotted line "cone" and will eventually "catch up" with the graph of y=x. The mean value theorem lets us prove this is a satisfying way. Suppose that for all x>=0, we have f(x) does not equal to x. Then by the Intermediate Value Theorem we must have f(x)> x. Pick b>0 (think large). By the mean value theorem, there is a u in (0,b) such that f'(u) = (f(b)-f(0))/(b-0) ... Since f(b) > b, we have ... f'(u) > 1 - v/b Since b can be arbitrarily large, we can arrange things so that f'(u) becomes arbitrarily close to 1. But this contradicts |f'(u)| <=k < 1 ... The satisfying thing about this argument was the role that the mean value theorem played in guaranteeing exactly the right derivative values to get the desired contradiction. pg.297-298 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc2243
• Motel room paradox* Three women check into a motel room which advertises a rate of $27 per night. They each give $10 to the porter, and ask her to bring back 3 dollar bills. The porter returns to the desk, where she learns that the room is actually only $25 per night. She gives $25 to the motel desk clerk, returns to the room, and gives the guests back each one dollar, deciding not to tell them about the actual rate. Thus the porter has pocketed $2, while each guest spent 10-1 = $9, a total of 2 + 3 x 9 = $29. What happened to the other dollar? ... The actual "invariant" here is not $30, but $27, the amount that the guests spend, and this will always equal the amount that the porter took ($2) plus the amount that went to the desk ($25). pg. 22, 102, The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.1647
• Notation* Adding notation is typically referencing everything with regard to an origin, a reference point. We need some "notation". Let us assume that there is a universal coordinate system, such as longitude and latitude, so that we can refer to the "same" location on any planet. For example, if the planets were little balls floating in a room, the location "north pole" would mean the point on a planet which was closest to the ceiling. Given such a universal coordinate system, what can we say about a planet P which has a private point at location x? pg. 64 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.1520
• {$x^4=(x-1)^4$}. Solve by working with the center {$y=x-\frac{1}{2}$}. Can also solve by subtracting so that all of the terms are on one side. Then factor {$a^2-b^2$}.
• The center of a center is the -1-simplex. This arises in reduced homology.
• Zero is a center.
• A kernel of a group homomorphism is a center.
• The identity element (the do-nothing action) in a group is the center.
• The zero vector space is the center, the beginning and the end, for a finite exact sequence.
• A category may have a zero object, such as the zero vector space in the category of vector spaces, or the group with one element in the category of groups.
• 维基百科: Origin

Balance* We can expand around the center by balancing positive and negative, numerator and denominator. We thereby introduce parity (Z2), odd or even, affirm or reject, where to reject rejection is to affirm. (Parity, Z2: affirm-reject, multiplication by one, addition of zero, union with empty set, expansion around center) 66

• Add zero creatively* Many [factoring] problems involve combinations of these formulas, along with basic strategies (for example, wishful thinking), awareness of symmetry, and the value add zero creatively tool. Zeitz gives the example of factoring x**4 + 4, thinking wishfully that it was the difference of two squares, and making more perfect squares appear by adding 0 = 4x**2 - 4x**2. pg. 163, The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc. 1378
  • Completing the square by adding zero* x**2 + a*x = x**2 + a*x + a**2/4 - a**2/4 = (x + a/2)**2 - (a/2)**2 One way to discover this completing-the-square formula is to add zero creatively, as above. pg. 163, The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc. 1380
• Appeal to Physical Intuition* Let x1, x2, ... xn be positive real numbers with product P and sum S. Prove that the largest value of P is attained when all the xi are equal. ... Imagine the n positive numbers as "physical" points on the number line, each with unit weight. The balancing point (center of mass) of these weights is located at the arithmetic mean value A = S/n. Notice that if we move the points around in such a way that they continue to balance at A, that is equivalent to saying that their sum stays constant. Our strategy, inspired by the symmetry-product principle, is to consider situation where the xi are not all equal and show that we can make them "more equal" and increase their product without changing their sum. If the points are not all clustering at A, then at least one will be to the left of A (call it L) and another will be to the right of A. Of these two points, move the one which is closest to A right up to A, and move the other so that the balancing point of the two points hasn't changed. ... This proof is called "algorithmic" ... pg.195-196 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.2192
• Easy invariants* Be on the lookout for "easy" invariants. Check to see if you can rearrange your problem to get simple numbers such as zero or one. pg.106, The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc. 1742
• Homogeneous coloration* Parity is invoked by a coloration which is homogeneous or not. Is it possible to tile a 66 x 62 rectangle with 12 x 1 rectangles? ... Color the squares of the 66 x 62 rectangle with 12 colors in a cyclic "diagonal" pattern... This coloring has the nice property that any 12 x 1 rectangle in the tiling consists of 12 differently colored squares ... each color occurs in the same number of squares. We will call such a coloration "homogeneous". ... We can break it up into 4 sub-rectangles ... the entire large rectangle is not homogeneous... pg.111, The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc. 1746
• Make an expression uglier* Sometimes you may want to make an expression uglier because it then yields more information. ... Which is bigger 1998/1999 or 1999/2000? ... here is an argument that uses the define a function tool: Let f(x) = x/(x+1) ... How does this function grow? We have f(x) = x/(x+1) = 1/(1 + 1/x) and now it is easy to check that as x>0 increases, the 1/x term decreases, causing f(x) to increase ... "If the denominator increases, the fraction decreases, and vice versa." ... f(x) is monotonically increasing for positive x. ... Notice how we actually made an expression uglier... it is much easier to analyze the behavior of the function. pg.165, 191-192 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.2171
• Modular arithmetic* Modular arithmetic arguments are typically parity arguments (divisible by N - not divisible by N). Let N be a 4-digit number with decimal representation abcd. Then n = 10**3 a + 10**2 b + 10 c + d. ... 10**k = 1**k = 1 (mod 9) ... n = 10**3 a + 10**2 b + 10 c + d = 1 a + 1 b + 1 c + d (mod 9) ... The important thing is to be aware of the possibility that an invariant may be a quantity modulo m for a properly chosen m. pg.110, The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc. 1744
• Multiply cleverly by one* The sister to the add zero creatively tool is the multiply cleverly by one. pg. 163, The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc. 1379
• Overcount and rectify* In explaining combinations and thus the coefficients of the binomial theorem... To count the number of ways a joint event occurs, multiply together the number of choices for each sub-event. To rectify uniform overcounting, divide by the overcounting factor. Such overcounting occurs, for example, when some of the objects are indistinguishable but we label them to make them easier to count, then take off the labels. In general, the number of ways you can select a subset of r distinct elements from a set of n distinct elements, where the order of selection doesn't matter, is P(n,r)/r! = (n r). This is called a combination. If the order does matter, then the number of ways is P(n,r) and it is called a permutation. Compare overcounting with the use of "negative numbers". pg.207-208 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc2200
• Parity* Integers are divided into two parity classes: even and odd. The even integers are divisible by 2, the odds are not. Note that zero is even. ... The parity of a sum of a set of integers is odd if and only if the number of odd elements is odd. The parity of a product of a set of integers is odd if and only if there are no even elements in the set. ... knowledge of parity is sometimes all that is needed, especially if parity is involved in the statement of the problem. ... Whenever a problem involves integers, ask yourself if there are any parity restrictions. Experiment with different values than the given if necessary. ... Parity works amazingly well, but it is rather crude. After all, we are reducing the infinite universe of integers into a tiny world inhabited by just two entries, "even" and "odd". pg. 104, 106, 110 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.1511
  • 127 people in a tennis tournament* If 127 people play in a singles tennis tournament, prove that at the end of the tournament, the number of people who have played an odd number of games is even. ... each game has exactly two people playing it... the sum counts every game that has been played exactly twice! ... the sum above is even, and is a sum of an odd number (127) of elements. If an odd number of them were odd, the sum would not be even... pg. 102, The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.1644
  • Parity problem: Dominos on a chessboard* Remove the two diagonally opposite corner squares of a chessboard. Is it possible to tile this shape with thirty-one 2 x 1 "dominos"? ... At first, it seems like a geometric/combinatorial problem with many cases and subcases. But it is really just a question about counting colors. The two corners that were removed wre both (without loss of generality) white, so the shape we are interested in contains 32 black and 30 white squares. Yet any domino, once it is placed, will occupy exactly one black and one white square. The 31 dominos thus require 31 black and 31 white squares, so tiling is impossible. pg. 60 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.1509
• Parity of a sum or product* Let a1, a2, ..., aN represent an arbitrary arrangement of the numbers 1, 2, 3,...N. Prove that, if N is odd, the product (a1-1)(a2-2)...(aN-N) is an even number. ... The crux move: consider the sum (a1-1) + (a2-2) + ... + (aN-N) pg. 105 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.1743
• Simplicity* You have been taught that "simplification" is to combine things in "like terms". This sometimes simplifies an expression, but the good problem solver has a more focused, task-oriented approach, motivated by the wishful thinking strategy. Avoid mindless combinations unless this makes your expressions simpler. Always move in the direction of greater simplicity and/or symmetry and/or beauty (the three are often synonymous). pg.165, The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.2170
• Adversary. Terrence Tao.

Polynomials* We can expand terms as polynomials, as with "and" and "or", and thus create equations that construct and relate roots. (Or, And, method of undetermined coefficients, expansion, construction)67

• Addition means combine like units, list different units* We can practice this principle by considering "units", but also numbers used as units ("tens", "thousands", "millions", "sevenths", "percentages"), imaginary units ("zillions"), any numbers ("twos", (-1)), unknown numbers ("X"), squares of unknown numbers ("X**2"). We can mix units such as 2x + 3y -5x + 4y and likewise. Gospel Math. 1847
• Designing algorithms with index cards* I have the student write on an index card, "I'm thinking of a number". Then they create subsequent index cards, each with their own instruction: "Double the number", "Add 5 to the number", "Halve the number", "Square the number" and so on. Then I have them input a number and get the output and write down all of the intermediate answers. They do this for several numbers. Then we can consider an abstract input X. We can discuss if the algorithm can be simplified, and can it be reversed. Basic Math.1955

• And* Simple multiplication. If there are a varieties of soup and b varieties of salad, then there are ab possible ways to order a meal of soup and salad. pg.205 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.2198
• Combinatorial tactics* But mostly, combinatorics is a tactical game. You have already learned the fundamental tactics of multiplication, division, addition, permutations, and combinations. pg.212 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc2206
• Complement* Each selection of 10 winners from a group of 17 is simultaneously a selection of 7 losers from this group. ... The Symmetry Identity ... (n r) = (n n-r) ... The combinatorial argument shows why it is true, while algebra merely shows us how it is true. pg.209 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc2202
• Complete the square* 2185
• Counting the Complement* A particular application of [changing your point of view] in counting problems is If the thing you wish to count is confusing, try looking at its complement instead. How many n-bit strings contain at least 1 zero? ... an easier approach is to first note that there are 2**n possible n-bit strings, and then count how many of them contain no zeros. A complement is possible when there is a closed system, which is what polynomials define. pg.225 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc2209
• Division Algorithm* Let f(x) and g(x) be polynomials in K[x], where K is one of Z, Q, R, C or Zn. Then f(x) = Q(x)g(x) + R(x) where Q(x), R(x) are elements of K[x] and the degree of R(x) is less than the degree of g(x). ... For example ... by doing "long division" we get ... The important thing is that the quotient Q(x) is also in Z[x], i.e., also has integer coefficents. pg.180 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.2184
• Eliminate radicals* Assume that you can make good progress. What will the result look like? If this expression is to simplify, we will probably be able to eliminate radicals. If we multiply any two terms, we can use the difference of two squares formula and get expressions which contain only one radical. pg.166 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.2173
• Extracting squares* The tactic of extracting squares includes many tools in addition to completing the square. ... It is easy enough to see how this works, but why is another matter. For now, hindsight will work: remember that many useful squares lurk about and come to light when you manipulate "cross-terms" appropriately (making them cancel out or making them survive as you see fit). pg.164-165, The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.2169
• Factor Theorem* A reinterpretation of this theorem from a problem solver's perspective is To know the zeros of a polynomial is to know the polynomial. In other words, if you don't know the zeros of the polynomial under consideration, either expend some effort to find them, or shift your focus to a new polynomial whose zeros are apparent. pg.182 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.2176
• Infinitely many primes* There are infinitely many primes ... We start by assuming that there are only finitely many primes p1, p2, ... , pN. Now (the ingenious crux move!) consider the number Q = (p1p2...pN) + 1. Either it is prime, or it is composite. ... We construct a value p1p2...pN that satisfies all conditions exactly and then adjust it ever so slightly +1 so that it satisfies none of them. In other words, we create a model p1p2...pN that satisfies all and then we force that model to break down. We design this value with regards to the closed system as a whole. pg.244 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc2215
• Modulo n filtering* Another essential tactic [in analyzing diophantine equations] is to "filter" the problem modulo n for a suitably chosen n. This tactic often helps to show that no solutions are possible, or that all solutions must satisfy a certain form. (Use of the division algorithm is closely related to the factor tactic.) ... the complete theory of Pythagorean triples: Find all solutions to x**2 + y**2 = z**2 ... we can assume that our solution is not just primitive, but relatively prime in pairs ... Next, a little parity analysis; ie., let's look at things modulo 2. Always begin with parity. You never know what you will discover. ... exactly one of x and y is even... pg.265-267 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc2226
• Number theoretic functions* Of the infinitely many functions with domain N, we will single out a few that are especially interesting. Most of these functions are multiplicative ... f(ab) = f(a)f(b) whenever a and b are relatively prime ... Define sigma_r(n)= sum over divisors d of n of d**r ... Let F(n) = sum over divisors d of n of f(d). If f is multiplicative, then F will be multiplicative as well. Define phi(n) to be the number of positive integers less than or equal to n which are relatively prime to n. ... We can use the principle of inclusion-exclusion to evaluate phi(n) ... We define [the Mobius function] mu(n) = 1 if n=1, =0 if p**2 divides n for some prime p, = (-1)**r if n = p1p2...pr, each p a distinct prime. This is a rather bizarre definition, but it turns out that the Mobius function very conveniently "encodes" PIE. ... show that sum over divisors d of n of mu(d) = 1 if n=1; =0 if n>1. The values of mu(n) alternate sign depending on the parity of the number of prime factors of n. This is what makes the Mobius function related to PIE. ... phi(n) = sum over divisors d of n of mu(d)(n/d). pg.257-261 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc2221
• Or* Simple addition. If there are a varieties of soup and b varieties of salad, then there are a+b possible ways to order a meal of soup or salad (but not both soup and salad). pg.205 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.2197
• Substitution* Let a, b, c be positive real numbers such that abc=1. Prove that 1/(a**3(b+c)) + 1/(b**3(c+a)) + 1/(c**3(a+b)) >= 3/2. ... What is the worst thing about this problem? It is an inequality involving fairly ugly fractions. Wishful thinking tells us that it would be nicer if the fractions either were less ugly or did not exist at all. ... There is a pretty obvious substitution - but only obvious if you have the idea of substitution in the forefront of your consciousness. The substitution is x=1/a, y=1/b, z=1/c, which transforms the original inequality (use the fact that xyz=1) into x**2/(y+x) + y**2/(z+x) + z**2/(x+y) >= 3/2. This inequality is still not that easy to deal with, but the denominators are much less complicated, and the problem has been reduced in complexity. pg.170 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.2178
• Symmetric functions of zeroes* we can get a series of expressions for the coefficients of a polynomial in terms of its zeros x**4 + a3x**3 + a2x**2 + a1x + a0 = (x-p)(x-q)(x-r)(x-s) ... Equating like terms, we have a3 = - (sum of all zeros), a2 = +(sum of all products of two different zeros), a1 = -(sum of all products of three different zeros), a0 = +(product of the zeros), where it is understood that "different" here has a purely symbolic meaning; i.e. we multiply only zeros with different labels, such as p and q, even if their numerical values are the same. ... we see, the pattern, and can write the formulas in general ... These formulas are very important, and should be committed to memory ... note the role that the power of -1 plays. pg.184-185 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.2188
• The Fundamental Theorem of Algebra* The Factor Theorem above tells us that x-a is a factor of polynomial P(x) if a is a zero. But how do we know if a polynomial even has a zero? The Fundamental Theorem of Algebra guarantees this: Every polonomial in C[x} has at least one complex zero. pg.182 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc.2187
• Viewing a problem modulo m* Viewing a problem modulo m for a suitably chosen m is a wonderful simplification tactic because it reduces the infinite universe of integers to the finite world of Zm. You have encountered this idea before with parity ... which is just the case m=2 ... Often (but not always) we turn to prime values of m, since primes are simpler, more "fundamental" objects which are generally easier to understand. In general, When beginning a number theory investigation, assume that the variables are prime or at least relatively prime. Often the general case follows from the prime case, with just a few "technical" details. ... Fermat's Last Theorem. Let n>=3. Prove that the equation x**n + y**n = z**n has no non-zero integer solutions. ... we shall point out two simplifications ... Without loss of generality, n is prime ... we may assume that ... x, y, z have no common factor (other than 1). ... One reason that primes are so convenient is that unique "multiplicative inverses" exist. Viewing modulo m emphasizes the closed nature of a system. pg.252-253 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc2219
• x**2 + y**2 = 13* This is a diophantine equation, so we should try factoring. "But it doesn't factor", you say. Sure it does! We can write (x+yi)(x-yi) = 13, where i is of course equal to the square root of -1, The only problem, and it is a huge one, is that i is not an integer. But let's stay loose, bend the rules a bit, and make the problem easier. It is true that the square root of -1 is not an integer, but what if we looked at the problem in Z13? Notice that 5**2 = 25 == -1 (mod 13). In other words, i "makes sense" modulo 13. pg.274-275 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc2227
• (x-a) x is unknown (conscious question); a is known (unconscious answer); and the difference x-a gives the discrepancy. So a polynomial can be thought of as constructed as the product of the discrepancies - or it can be constructed as coefficients for powers. Note that EVERY polynomial with complex coefficients can be constructed as a product of such discrepancies. So it's a type of completeness. What makes that true?

Vector space* We can consider a vector space in which any point can serve as the center for a basis. (Superposition, linear combination, duality) 68

• Indicator function* We shall now present a proof of the complement function of PIE, using the "binary" language of indicator functions. Recall that the indicator function of A is denoted by 1A and is a function with domain U (where U is a "universal set" containing A) and range {0,1} defined by 1A(x) = 0 if x not in A, =1 if x in A, for each x in U. ... 1A(x)1B(x) = 1 A-intersect-B(x). 1 - 1A(x) = 1 complement-of-A(x). ... the product of two indicator functions is the indicator function of the intersection of two sets and the indicator function of a set's complement is just one subtracted from the indicator function of that set. ... Define the function g(x) = (1 - 1A1(x))(1 - 1A2(x)) (1 - 1A3(x)) ... N0 = sum over x in U of g(x). pg.230-31 The Art and Craft of Problem Solving, Paul Zeitz, 1999, John Wiley & Sons, Inc2213
• Linear combination* ...the result was attained in two steps. First, we were lucky enough to spot a particularly accessible case, a special situation, and gave a solution well adapted, but restricted, to this special situation ... Then, by combining particular cases to which the restricted solution is applicable, we obtained the full, unrestricted solution, applicable to the general case... The first step deals with a particular case which is not only especially accessible, but also especially useful; we can appropriately call it a leading particular case: it leads the way to the general solution. The second step combines particular cases by a specific algebraic operation. ... n particular solutions, after being multiplied by given constants, are added to form the general solution. ... we add and subtract equations dealing with the special situation to obtain the general proof. Let us call the algebraic operation employed ... linear combination or superposition. We may use the terms introduced to outline our pattern: Starting from a leading special situation we attain the general solution by superposition of particular cases. "Mathematical Discovery: On Understanding, Learning and Teaching Problem Solving" by George Polya, 1962, John Wiley & Sons.2252
• Hatcher: "‘n-simplex’ will really mean ‘ n simplex with an ordering of its vertices’." This makes it possible to define an orientation.