Table of Contents
Introduction to Mathematical Thinking. Proofs
What is a proof? Fermat’s Last Theorem If n≥3 the equation x^n + y^n = z^n has no solutions for positive whole numbers x, y, z. Goldbach Conjecture Every even number greater than 2 can be expressed as a sum of teo primes. In mathematics, a proof is a logically sound argument that establishes thetruth of the statement being proved. The word argument here is, of course, not the more common everyday use to mean a disagreement between two people. A good proof preemptively counters all counter arguments that a reader might put forward. When professional mathematicians read a proof, they generally do so in a manner reminiscent of a lawyer cross-examining a witness. Constantly probing and looking for flaws. Learning how to prove things forms an important part of college-level mathematics; it takes years to master this skill. What can be achieved in a short period of time is understanding that proving mathematical statements is crucial to mathematicians. We make such a big deal about proofs because they are used for two purposes: establishing truth and communication. A proof must explain why a given statement is true. Proofs written to convince others must be both logically sound and communicatively effective. For complicated groups, the requirement that a mathematician can follow his or her own proof weeks later can also be significant. Some proofs are so deep and complex that only a few experts in the field can understand them. For example, for many centuries most mathematicians believed, or at least held a strong suspicion that x to the n + y to the n = z to the n has no positive whole number solutions for x, y, and z when n is greater than or equal to 3. That was conducted by the French mathematician Pierre de Fermat in the 17th century; however, it wasn’t finally proved until 1994 when British mathematician Andrew Wiles constructed a”long and extremely big proof. Over time it became common knowledge among mathematicians that this was “Fermat’s Last Theorem.” Since it was the last of several mathematical statements Fermat announced that remained unproven for hundreds of years. Most mathematicians, myself included, lack the detailed domain knowledge needed to follow proofs ourselves but it did convince experts in the field—analytic number theory. Since its initial formulation by Pierre de Fermat in1637, Fermat’s Last Theorem has been proven true for all cases but composite numbers with one or more prime factors equal to or greater than 5. Most proofs in mathematics can be read and understood by all professional mathematicians. Though it can take days, weeks, or even months to understand some proofs sufficiently to be convinced by them. Proving a mathematical theorem is much more than gathering evidence in its favor. The Goldbach Conjecture is a famous example of a mathematical statement that has with stood attempts at solution for over 250 years. The conjecture originated when the Swiss mathematician Leonhard Euler suggested that every ev
Optimization intuition: Worked example continued
Right. so this over here is an example. There are no critical points for this graph. We’ve seen lots of functions have no critical points. A critical point is a point on a curve where the tangent line is horizontal, meaning it does not change direction. There are no critical points for this function, and so the maximum is on the boundary of the interval. To find the maximum on the boundary. we check each boundary point individually and determine whether it is one of many local maxima or if it is indeed the global maximum. => No critical points => Max. on boundary. Top edge y=2, x from 0 to 2 f(x.2) = 2x – x^2-2 Choose x to max. Max. of fon top edge = – 1 Diagonal edge y=x x from 0 to 2 f(x, x)=2x-x^2-x=x-x^2 [Graph] Max. of f on diagonal edge = 1/4 Max. off is the max. among three edges. There are three edges. Let’s consider one at a time. First, let’s think about the top edge. The top edge is a horizontal line; y equals 2 and x goes from 0 to 2-that is the edge. Now we want to consider this function on the top edge. So fof x if we’re on the top edge, y equals 2, so we’re thinking about f of (x, 2). And then let’s plug in what that is: 2x minus x squared minus 2, and x goes from 0 to 2. To maximize this function, we must find the x-value that yields the largest output. Such a task can be accomplished through a simple calculus operation. We know that this graph is always negative and so the maximum occurs at -2. So the question was: what does one do with this diagonal edge where y is equal to x? But we’re going to do the edges one at a time, and so first we’re just going to think about the top edge. Conclusion: maximum off on top edge is negative 1. The region R is the red triangle there, and all the level curves of our function f pass through that region. We’ll be looking at the biggest value of f on the top edge, which runs from x equals 0 over here to x equals 2 over there. On that top edge we first go through the negative 2 level curve, but then as we keep going, f increases up to negative 1.5 and then goes back down towards negative 1.5 again before going through it one final time for good. [Graph] Let’s consider the diagonal edge. The top edge is straightforward to draw,and the left edge is also fairly easy. So the function starts at O, then it gets positive, and then it goes back to 0, and then it gets negative. The maximum is going to be somewhere in here–probably about in the middle, which would be over there. But now, how do we compute the thing? Well, we have the diagonal edge. How can we describe this diagonal edge? It’s the line y equals x; x is still going from 0 to 2. And we want to understand the relationship between y and x. We’re thinking about y when x equals 2, and since y equals x, we can substitute 2 for x in the formula for y. That’s 2x minus x squared minus y. But that’s x, so we have that. And then that simplifies to x minus x squared. The function is maximized at the point (x, ×) when x goes from 0 to 2. To find this point, we must first find the derivative of the function and set it equal to zero. In this graph, the x-axis ranges from 0 to 2, while the y-axis ranges from -1 to 1. The quadratic function that best fits this curve is f(x. x) = x2 – ×2. The maximum value of this function occurs at (1/2, 0), which is also the minimum value of the function. The maximum off on the diagonal edge is 1/4. To finish the problem, you take all three edges, find the maximum on each edge, and the biggest of those would be the maximum. We didn’t actually do the left edge, but it turns out not to be the winner because we can see from the picture that the diagonal wins. So the maximum off is 1/4.
en number greater than two can be expressed as the sum of two primes. This property has been verified for many specific even numbers and to date, 2012, it has been verified up to and beyond 1.6 quintillion (16 followed by 18 zeros). Most mathematicians believe it to be true, but it has not yet been proved. All it would take to disprove the conjecture would be to find an even number n for which it could be shown that no two primes sum to n. This conjecture has no known applications or even any significant consequences within mathematics; it is famous solely because it is easy to understand and has resistedall attempts at solution for over 250 years. Proofs have no standard format, but all proofs must be logically sound and establish the truth of some statement. A proof’sclarity is also important because the intended reader must be able to follow your reasoning. Euclid’s proof that there are infinitely many primes is an example of a proof that relies on an unusual insight. Let’s look at it again. Here’s what we did: We showed that if we list the primes in increasing order, then the list can be continued forever. Sowe imagine we’ve listed the primes p1 = 2, p2 = 3, p3 = 5, etc., all the way up to some stage Pn. And we’re sure that we can always add another prime to this list. To do this, we look at this number N which we obtain by multiplying together all theprimes in the list so far and then adding 1. Now this number N consists of the product of all those numbers pi through pi + 1. So it certainly must be bigger than each of those individual numbers pi through pi + 1. If N is prime, one can prove that there is a prime number greater than pN. In this case, the list of prime numbers can be continued by adding N to the end of it. Otherwise, if N is not prime, then there exists a prime number q less than N that divides into it. However, none of the primes p1, p2, etc., listed above would divide evenly into N because if one were to divide N by any of these primes (for example), one would get a remainder of 1. So for each q> pn, we have a prime p dividing N. That is, pn = q × p for some prime q > p. If q does not equal one of these first n primes, it must be later on in the list. In this case,because we showed that there is a prime bigger than pn, we can continue the list by adding this new number to it. The fact that this particular q that divides N is not necessarily the next prime doesn’t matter because showing that there is another prime at all is enough to prove the theorem’s claim. Either way, either if N is prime or if N is not prime, either way there’s another prime to add to the list. It follows that there are infinitely many primes and the theorem is proved. The proof included two creative ideas. The first was to list the primes in increasing order as p1, p2, etc., and to show that the list can be continued forever. The second idea was to construct a number N that guarantees that we can always find another prime number. Many mathematicians would eventually come up with this second idea; it is not an obvious one. This idea is truly genius—one of the greatest ideas in the history of mathematics. THERE ARE INFINITITELY MANY PRIMES 1. List the primes in increasing order as p1, p2, p3…… pn, … show that the list must continue for ever 2. Given the list up to some stage n, p1, p2, …… pn, show there is another prime that can be added to the list. 3. Let N be the number we get when we multiply together all the primes we have listed so far and then add1: N = (p1* p2 * p3*…….* pn ) + 1 Genius! 4. Obviously, N is bigger than all the primes in our list. 5. If N is prime, we know there is a prime bigger than pn, and hence the list can be continued. 6. If N is not prime, then there must be a prime q<N such that й divides N. 7. But none of p1,…… pn divides N, since the division of N by any one of these leaves a remainder of 1. 8. So, q must be bigger than pn. Hence there is a prime bigger than pn, the list can be continued. 9. Either way, there is another prime to add to the list. 10. It follows that there are infinitely many primes. The result is proved. In order to prove that the square root of two is irrational, let us first state the most basic facts about numbers used in this proof. Next, we will begin the actual proof by stating an assumption. We will then proceed with our proof using well-established techniques of mathematical logic and algebra. A proof is a logical argument, and to prove something is true, we must prove that it isn’t false. If you’ve never seen the proof that the square root of 2 is irrational before, this first step might seem pretty mysterious. Why do we begin by assuming what we’re trying to prove is false? The reason this is a great example is that in about six or seven lines we can make it clear while we`re doing something right in the first step. If the square root of 2 were rational, then there would be natural numbers p and q such that p/q equals the square root of 2. A rational number is one that can be expressed as the quotient of two integers. In the case of a positive number it would be two natural numbers. We can always pick those natural numbers, or those integers, to have no common factors. In other words, when we write a rational number as a quotient, we can always cancel out any common factors and express it as a quotient where the two numbers themselves have no common factors. The reason for canceling out common factors is simple: it allows us to rewrite an expression as a sum of its parts. As we’ll see later on, doing so will help us solve equations. Squaring both sides yields 2 = p squared over q squared. Rearranging gives me 2 q squared = p squared; multiply both sides by q squared and the common factors disappear, leaving us with 2 q squared on the left side of my equation and a p squared on the right side. We multiply both sides by q squared again and take out another factor of q squared. This leaves me with a 2p^2; hence p is even because the square of an even number is even and the square of an odd number is odd. Since p is a multiple of 2, then r must be a factor of p; hence p=2r. Substituting this back into the original equation 2q2=p2 gives 4r2=(2r)2, so 4r2=4r2. Cancelling the middle term gives 2q2=4r2 which is an even number since q2 was shown to be even earlier on, and since p was also shown to be even earlier on, it follows that both p and q are even numbers—contradicting our assumption that they had no common factors other than 2. It is impossible to reach an incorrect conclusion through sound reasoning, therefore the only explanation for a false conclusion is that an initial assumption was false. If we reach a false conclusion by sound reasoning then we must have started with a false assumption. Theorem: sqrt2 is irrational. Proof: Assuare, on the contrary, that sqrt2 were ratiared. Then there are natural numbers p, q, with common factors, such that sqrt2 = p/q Squariry: 2=p^2/p^2 Rearrarging: 2q^2=p^2 So p^2 is even. Hence p is even. So p=2r, for some r. 2q2=24)2 = 4r2 Substitutiry for p: 2q^2 = (2r)^2=4r^2 Cancelliry the 2: q^2=2r^2 So q^2 is even. So q is even. So p, q, are both even. But this is impossible, since p, q have no common factors. Hence the original assumption that sqrt2 were rational must be false. Hence sqrt2 must be irrational. QED Thus, the assumption that the square root of 2 is rational must be false. This implies that the square root of 2 is irrational. It is actually not a bad idea to indicate when a proof ends. Mathematicians have different ways of doing this, such as writing a little box at the end or using symbols indicating it is complete. How you express it, how you lay it out on the page is not that critical, provided that you make it clear where each statement ends and another begins. The idea was just to lay it out in a way that can be followed. What makes a proof a proof isn’t the fact that you call it a proof or end with QED; it’s how logically precise each step is and if you can follow its flow. The theorem is a good example to give because it is short, concise, and each step follows from the previous one in a logical sequence. The result of this theorem is significant and changed the course of Greek mathematics. The Greeks had assumed that quotients of integers could be used to measure all lengths in geometric figures; however, this is not true for square roots of 2 or any number whose square is greater than 1. This discovery led mathematicians to consider arithmetic progression as an alternative method for measuring lengths in geometric figures. The Pythagorean proposition is an example of a proof that is both elegant and simple, relying on only simple arithmetic operations. Its significance lies in its ability to demonstrate that square roots are not always integers, which had been previously assumed by other mathematicians. In addition, it proved that quotients of integers were insufficient to measure all lengths in geometric figures. This discovery had a dramatic effect on Greek mathematics and subsequently on the rest of mathematics.
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