Solutions to Some Examination Problems MATH 300 Monday 25 April 2016

Transcription

1 Name: s to Some Examination Problems MATH 300 Monday 25 April 2016 Do each of the following. (a) Let [0, 1] denote the unit interval that is the set of all real numbers r that satisfy the contraints that 0 r and r 1 and let R denote the set of all real numbers. Invent a function f : [0, 1] R so that f is both one-to-one and maps [0, 1] onto R. Verfiy that your function has the desired properties. Conclude that [0, 1] and R has the same cardinality. (b) Prove that the set of all triples of natural numbers is countable. Still to come. Consider the following sets of points on the X Y plane. C = {(x, y) x, y R and x 2 + y 2 = 1}, D = {(x, y) x, y R and x 2 + y 2 1}, and D o = {(x, y) x, y R and x 2 + y 2 < 1}. C is the unit circle, D is the unit disk, and D o is the open unit disk that is, the unit disk with the circle C removed. Prove that the unit disk D and the open unit disk D o have the same number of elements. [ You might consider first a radius connecting the center of the disk with a point of the circle. In this case, the situation is like proving [0, 1] and [0, 1) have the same number of elements. ] Still to come. Do each of the following. (a) Let [0, 1] denote the unit interval that is the set of all real numbers r that satisfy the contraints that 0 r and r 1 and let C denote the unit circle that is the set {(x, y) x 2 +y 2 = 1}. Invent a function f : [0, 1] C so that f is both one-to-one and maps [0, 1] onto C. Verfiy that your function has the desired properties. Conclude that [0, 1] and C has the same cardinality.

2 (b) Prove that the set of all polynomial of degree at most 2, with natural number coefficients, is countable. Still to come. Prove that the graph of the sine function, namely {(x, sin x) x is a real number}, can be put into one-to-one correspondence with the unit interval, namely [0, 1]. We do this in three stages. Stage 0 Here we show that the since curve can be put into one-to-one correspondence with the real line R. Here is the one-to-one correspondence p: p(x) = (x, sin x) Observe that each real number x corresponds to a point on the sine curve. This mapping evidently hits every point on the curve, so it maps R onto the sine curve. It is one-to-one since the input can be recovered from the output: the input is just the first coordinate of the output. Stage 1 Here we show that R can be put into one-to-one correspondence with the open interval ( π 2, π 2 ). Just observe that tan: ( π 2, π 2 ) R has the desired properties. Stage 2 In this stage we show that the closed unit interval [0, 1] can be put into one-to-one correspondence with the open interval ( π, ) ( π 2 2. To start, observe that if we define g so that g(x) = π 2 x 1 2), the g will be a one-to-one correspondence from [0, 1] onto [ π, ] π 2 2. The function g just shifts everything down by a half and then scales it up by stretching it a bit. But we want to get to the open interval, so we need a one-to-one map h from [ π, ] ( π 2 2 onto π, ) π 2 2. The only difference between these two intervals in that the domain has both of its endpoints will the target interval has neither. It would be nice if we could just use the identity function. But it doesn t quite work. We have to modify it. Here is how 1 If x = π 2 1 If x = 1 for some positive integer n n+1 n h(x) = 1 If x = π 2 1 If x = n+1 1 for some positve integer n n x In all other cases. So putting all there one-to-one correspondence together we get the map p tan g h gives us a one-to-one correspondence from the closed unit interval [0, 1] onto the sine curve. 2

3 Prove that any two circles on the Euclidean plane, regardlesss of their centers and their radii, have the same number of points. In a circle, draw a radius that is horizontal. Then for any point P on the circle, draw a line segment connecting P to the center of the circle. Let θ be the angle, read counterclockwise, that this segment makes with the horizontal radius. The association of P with θ is a one-to-one correspondence between the points on the circle and the real numbers in [0, 2π) = {r 0 r < 2π}. So the circle and [0, 1) have the same cardinality. Since the size and position of the circle we considered did not come into play in our reasoning, we see that all circles have the same cardinality. Problem 5 Let P be the set of all ordered pairs (a, b) where a and b are real numbers. Let us define a two-place relation on P by (a, b) (c, d) if and only if a 2 c 2 = d 2 b 2 where (a, b) and (c, d) belong to P. Prove that is an equivalence relation on P. Draw a diagram on the X Y plane of the equivalence class that contains the point (0, 1). We must check that is reflexives, symmetric, and transitive. Reflexivity Let (a, b) be an arbitrary member of P. We need to show that (a, b) (a, b). But observe (a, b) (a, b) if and only if a 2 a 2 = b 2 b 2. But the right side of this works out to the true statement 0 = 0 regardless of how a and b are chosen. So is reflexive. Symmetry Suppose that (a, b) (c, d). We must show that (c, d) (a, b). Just observe the following logical equivalences: (a, b) (c, d) a 2 c 2 = d 2 b 2 (a 2 c 2 ) = (d 2 b 2 ) c 2 a 2 = b 2 d 2 (c, d) (a, b) Transitivity Here we assume (a, b) (c, d) and (c, d) (e, f) and we must demonstrate that (a, b) (e, f). 3

4 From the two equivalences we have assumed, using the definition of we can pull off the following: a 2 c 2 = d 2 b 2 c 2 e 2 = f 2 d 2 add those equations a 2 e 2 = f 2 b 2 (a, b) (e, f) The Diagram The equivalence class containing the poitn (0, 1) is, by definition, the set {(x, y) (x, y) (0, 1)}. But observe that (x, y) (0, 1) x = 1 2 y 2 x 2 + y 2 = 1 This last is just the equation describing the unit circle. So this equivalence class is just the unit circle. Problem 5 Let P be the set of all ordered pairs (a, b) where a can be any integer and b can be any positive integer. That is P = {(a, b) a, b Z and b > 0}. Let us define a two-place relation on P by (a, b) (c, d) if and only if ad = bc where (a, b) and (c, d) belong to P. Prove that is an equivalence relation on P. We must check that is reflexives, symmetric, and transitive. Reflexivity Let (a, b) be an arbitrary member of P. We need to show that (a, b) (a, b). But observe (a, b) (a, b) if and only if ab = ba. Since multiplication of integers is commutative, we see that is reflexive. 4

5 Symmetry Suppose that (a, b) (c, d). We must show that (c, d) (a, b). Just observe the following logical equivalences: (a, b) (c, d) ad = bc bc = ad cd = da (c, d) (a, b) Transitivity Here we assume (a, b) (c, d) and (c, d) (e, f) and we must demonstrate that (a, b) (e, f). From the two equivalences we have assumed, using the definition of we can pull off the following: ad = bc cf = de multiply cleverly adf = bcf bcf = bde by transitivity of equality adf = bde cancel the nonzero d af = be (a, b) (e, f) Problem 5 Let P be the set of all ordered pairs (a, b) where a and b are real numbers. Let us define a two-place relation on P by where (a, b) and (c, d) belong to P. (a, b) (c, d) if and only if b d = 2(a c) Prove that is an equivalence relation on P. Draw a diagram on the X Y plane of the equivalence class that contains the point (0, 1). We must check that is reflexives, symmetric, and transitive. Reflexivity Let (a, b) be an arbitrary member of P. We need to show that (a, b) (a, b). But observe (a, b) (a, b) if and only if b b = 2(a a) 5

6 and on the right we see that this is just a way to write 0 = 0 no matter what values a and b have. So our relation is reflexive. Symmetry Suppose that (a, b) (c, d). We must show that (c, d) (a, b). Just observe the following logical equivalences: (a, b) (c, d) b d = 2(a c) (b d) = 2(a c) d b = 2(c a) (c, d) (a, b) Transitivity Here we assume (a, b) (c, d) and (c, d) (e, f) and we must demonstrate that (a, b) (e, f). From the two equivalences we have assumed, using the definition of we can pull off the following: b d = 2(a c) = 2a 2c d f = 2(c e) = 2c 2e add the equations b f = 2a 2f = 2(a e) (a, e) (e, f) The Diagram The equivalence class of the point (0, 1) is, by definition, the set {(x, y) (x, y) (0, 1)}. But just observe that (x, y) (0, 1) means y 1 = 2(x 0). That is, y = 2x + 1. The graph of this is a straight line of slope 2 with Y -intercept at 1. Problem 5 Let P be the set of all ordered pairs (a, b) where a and b are real numbers. Let us define a two-place relation on P by (a, b) (c, d) if and only if a 2 c 2 = b d where (a, b) and (c, d) belong to P. Prove that is an equivalence relation on P. Draw a diagram on the X Y plane of the equivalence class that contains the point (0, 0). We must check that is reflexives, symmetric, and transitive. Reflexivity Let (a, b) be an arbitrary member of P. We need to show that (a, b) (a, b). But observe (a, b) (a, b) means a 2 a 2 = b b. 6

7 The stuff on the right is just a way of saying 0 = 0, so it is true for all choices of a and b. So is reflexive. Symmetry Suppose that (a, b) (c, d). We must show that (c, d) (a, b). Just observe the following logical equivalences: (a, b) (c, d) a 2 c 2 = b d (a 2 c 2 ) = (b d) c 2 a 2 = d b (c, d) (a, b) Transitivity Here we assume (a, b) (c, d) and (c, d) (e, f) and we must demonstrate that (a, b) (e, f). From the two equivalences we have assumed, using the definition of we can pull off the following: a 2 c 2 = b d c 2 e 2 = d f a 2 e 2 = b f by adding the two equations (a, b) (e, f) The Diagram The equivalence class containing the point (0, 0 is, by definition {(x, y) (x, y) (0, 0)}. But observe (x, y) (0, 0) x = y 0 x 2 = y The graph of x 2 = y is just that familiar parabola with is vertex at the origin. Problem 6 Let A and B be nonempty sets and suppose that f is a function that maps A onto B. Prove that these is a function g that maps B into A so that f g is the identity function on B. We must devise a function g mapping B into A so that f g is the identity function on B. So consider an arbitrary element b of B. Since f maps A onto B, there must be at least one a in A so that f(a) = b. Now there might be a lot of such elements in A. But we only need one of them. Let us pick one and call it g(b). We do this for every element b of B. In the way, each b will be assigned a single value (since we only made one choice), and that value is in A. So g is indeed a function from B into A. Moreover, we carefully pick g(b) so that f(g(b)) = b. That means f g is the identity function of B. 7

8 Problem 6 Let A and B be nonempty sets and suppose that f is a function that maps A into B and that g is a function that maps B into A. Assume further that f g is the identity map on B. Prove that f must map A onto B and that g must be one-to-one, Let us first see that g is a one-to-one function. To this end, let b and b be elements of B so that g(b) = g(b ). We need to see that b = b. But observe g(b) = g(b ) f(g(b)) = f(g(b ) by applying f to both sides b = b since f g is the identity function on B Now let us prove that f maps A onto B. To this end, let b be an arbitrary element of B. We have to com up with some a A so that f(a) = b. But recall f(g(b)) = b since f g is the identity function on B. So we can let our desired a be g(b). Problem 6 Let A and B be nonempty sets and suppose that f is a one-to-one function that maps A into B. Prove that there is a function g that maps B into A so that g f is the identity function on A. Given the one-to-one function f we must invent the desired function g. Now f is a certain set of ordered pairs. So the members of f have the form (a, b) where a A and b B. Of course, only certain ordered pairs actually belong to f. As a first attempt, we make g 0 = {(b, a) a A and b B and f(a) = b}. Now we have to answer three questions: Is g 0 really a function? Is g 0 f the identity function on A? Is the domain of g 0 really B? To answer the first question we have to see that (b, a) g 0 and (b, a ) g 0 implies a = a. The way we defined g 0 as a set of ordered pairs reduces the task above to showing that (a, b) f and (a, b) f implies a = a. But this is just the assertion that f is one-to-one. So this part is okay. g 0 is a function. To see whether g 0 f is the identity function on A, we need to show g 0 (f(a)) = a for any a A. So let a be an arbitrary element of A. Let b = f(a). So by the definition of g 0 we get (b, a) g 0. Or as we customarily write g 0 (b) = a. This gives us g 0 (f(a)) = a, as desired. 8

9 Last, we want that the domain of g 0 is B. But is might not be. The domain of g 0 is just the set of all possible outputs of f. Because f might not map A onto B, we cannot conclude the B is the domain of g 0. But we can fix this. Since A is not empty, let us pick an element of A, call it d. Then define g as follows { g 0 (b) if f(a) = b for some a A g(b) = d otherwise So what we have done is extend g 0 by saying what value to give those extra elements of B. Extending g 0 to g still gives us a function and we will get g f to be the identity on A, since none of those extra elements of B come into play in evaluating g(f(a)). Problem 6 Let A and B be nonempty sets and suppose that f is a function that maps A into B and that g is a function that maps B into A. Assume further that g f is the identity function on A. Prove that f must be one-to-one and that g must map B onto A. Let us first prove that f is one-to-one. To this end, let a and a be arbitrary elements of A so that f(a) = f(a ). We need to prove that a = a. But observe f(a) = f(a ) g(f(a)) = g(f(a )) by applying g to both sides. a = a since g f is the identity function on A. Now let us prove that g maps B onto A. To this end, let a be an arbitrary element of A. We have to come up with some b B so that g(b) = a. But recall g(f(a)) = a since g f is the identity function on A. So we can let our desired b = f(a). 9