Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Selected Solutions to Homework 5 1. Complete the worksheet handed out in class on Wednesday. Be sure to write up the construction steps and proofs that they create the desired information carefully. Comment: For the most part the solutions on this were good. A couple of brief comments are warranted however. Be sure to at know how the constructions work. In particular, don’t just say copy the angle, be sure that you explain how to do so. When you are teaching, this is part of what you will need to do. 2. In class last week√ we defined the geometric mean of two numbers a and b to be equal to ab. Show that if f (x) = ln(x), then the geometric mean of the numbers a and b is the value c such that f (c) is the average of the values of f (a) and f (b). Would this still be true if we used the common logarithm (base 10) in place of the natural logarithm? Explain why or why not. √ Solution: Let √ √ c = ab. Then f (a) = ln(a), f (b) = ln(b), and f (c) = f ( ab) = ln( ab). Now by the power rule for natural logs, √ 1 f (c) = ln( ab) = ln (ab) 2 1 = ln(ab) 2 1 = (ln(a) + ln(b)), by the product rule for logs 2 1 = (f (a) + f (b)) 2 Thus f (c) is the average of f (a) and f (b). If you replace the natural logarithm with any logarithm (whether the logarithm base 10 or any other) then because the only properties of the natural logarithm we used in the proof hold for all logarithms, the result would be true for any logarithm. 3. Suppose you have invested $50 in an account with continually compounding interest. The banker tells you that according to his table, after 5 years you will have exactly $125. After leaving the bank, you 1 realize that you needed to know how much money you would have after 2 and a half years. How much money would you have if you withdraw it then? (assume no penalties) Explain. Solution: One method for solving this is to use the general formula for the principal of an investment after n years at a given interest rate P (n) = P0 ern where r is the interest rate. So that 125 = 50e5r . Then we have r = 51 ln(5/2). Using this we can solve for P (2.5) to get the result. A second method would be to avoid finding the interest rate would be to simply use that P (2.5) = 50e2.5r = 50(e5r )1/2 . We now bring the 50 inside the square root to give P (2.5) = (50 · 50e5r )1/2 = (50 · 125)1/2 . Thus the value P (2.5) is the geometric mean of 50 and 125. A third method would be to look at the underlying principle behind interest calculations. Namely, the key fact about interest is that it always accrues by the same power over equal amounts of time. That is, if you have P0 dollars at the start, and after time t you have P0 · a dollars, then at time 2t you have P0 · a2 . Consequently, using t = 2.5, and the information given in the problem q the amount of money after 2 , and the amount 50a = 5 = 2t years is 125 = 50a . Hence a = 125 50 q √ 50 125/50 = 125 · 50 or the geometric mean of 50 and 125. This is a general notion of the geometric mean. When a problem involves exponentiation or the property that a constant ration occurs over a constant time, then the inverse function corresponds to a logarithm and you have that the geometric mean arises as the half-way point. 4. Suppose you are standing at the origin, and at every lattice point (point in the plane with integer coordinates) there is a fencepost (of 0 thickness). Is there any direction that you can look so that there is not a fencepost in front of you? That is, if you look infinitely far, must you see a fencepost in your direct line of sight? Explain. 2 Solution: Suppose you see a fencepost at the lattice point (a, b), where a and b are integers. Since you are standing at the point (0, 0), you are looking along a line of slope b/a. Thus, you are looking along a line of rational slope. Hence, if you look along a line with irrational slope, then you will√never see a fence post. Consequently, if you look along the line y = 2 · x, you will never see a fencepost. (In fact, if you look in a random direction, the probability that you will see a fencepost is 0 because the rational numbers are insignificant next to the irrational numbers.) 5. Prove that if a line has two points with rational coordinates (all four of which are different), then any point (x, y) on the line with one coordinate rational must have the other coordinate rational. Solution: Let the points (a, b) and (c, d) be two points on the given line. Since these points both have rational coordinates, the slope of the d−b line, m = c−a is a rational number (not equal to 0 as b 6= d and finite since c 6= a). Now suppose (x, y) is another point on the line with x or y−b y rational. Then x−a = m, so that y = m(x − a) + b. Consequently, if x is rational, since m, a, and b are all rational and the rational numbers are closed under multiplication, subtraction, and addition, it follows that y is also rational. A similar argument shows that if y is a rational number then x is a rational number too. 6. A number field is a non-empty set of real numbers that is closed under addition, multiplication, subtraction, and division by non-zero elements. In particular, 0 and 1 must be in the field. Prove that any number field must contain all of the rational numbers. That is, show that if F is a number field, then for every pair of integers a and b with b 6= 0, a/b ∈ F . Solution: Since F is a number field, 1 ∈ F . Let a be an integers. Since |a| = 1 + 1 + . . . + 1 (a 1s) and F is closed under addition, |a| ∈ F . If a = |a|, then a ∈ F . Otherwise, a = 0 − |a| ∈ F (since F is closed under subtraction) and again a ∈ F . Thus every integer is in F . Let q be a rational number. Then there exist integers a and b with b 6= 0 such that q = a/b. Since F is closed under division by non-zero elements, it 3 follows that q ∈ F . As q was an arbitrary rational number, it follows that F contains all the rational numbers. QED 7. Show that if you can construct segments of length a and b, and you have a pair of coordinate axes given, then you can construct the point on the Cartesian plane with coordinates (a, b). (This is not hard, but requires a little explanation.) Suppose a and b are constructible lengths. Then given a pair of coordinate axes, to construct the point (a, b), do the following. First, mark off the length a along the x-axis with your compass point at the origin. This is then the point (a, 0). Next construct a line perpendicular to the x axis at (a, 0) by constructing a circle centered at (a, 0) intersecting the x-axis at points (0, 0) and (2a, 0). Using a compass length longer than a, then draw circles centered at (0, 0) and (2a, 0) of equal radius. These circles intersect in two points. The line connecting these two points is the perpendicular to the x-axis at (a, 0). Now mark off the length b on this perpendicular line with the compass point at (a, 0). The point at this intersection has coordinates (a, b). 8. The three explanations of why a negative times a negative is a positive. Comments: Each of the three explanations has their own best setting. They also all have their own background. The first two explanations begin with the notion that we have a choice of defining what a negative times a negative should be. This is certainly one perspective. People chose to manipulate the symbols in this way because it has, in the first explanation, a geometric meaning, and in the second explanation, a natural extension of a pattern with the natural numbers. Neither explanation corresponds to a model of why a negative multiplied by a negative should be a positive, and one of the difficulties with the explanation is that no good elementary models exist (at least that I know of). The third explanation is inherently different from the first two because it starts with the assumption that we want the distributive property for multiplication over addition to hold. The third explanation is then a proof that if we require the distributive law to hold true then we have no other choice but to define a negative times a negative to be a positive. 4 In terms of prior knowledge, the first explanation requires an understanding of the number line (which may or may not be the case with the students that you will teach), an understanding of function at some elementary level, and finally an understanding that a positive times a negative is a negative. The last is at least easy to convince people of. It also requires a willingness to believe that multiplication shouldn’t collapse the number line (unless multiplying by 0). The second explanation requires less prior knowledge. It requires some experience with patterning and arithmetic series patterns (though not under that name). It also requires knowing that a positive times a negative is a negative. The third explanation requires the most background, which is probably an underlying reason why most of you chose not to use it. In particular, a student would need some familiarity with proof, some understanding of the distributive law on a more abstract level (even though I avoided using −a), and a mathematical patience, by which I mean a willingness to work through the problem without seeing how the answer will arise. This last is an important willingness to cultivate in students, because as you move to a higher and higher level of mathematics you will need to have more and more patience to wait for the punchline of a proof. 5