MTH Proofs For Test #1 Fall 2006

Transcription

1 Pat Rossi MTH Proofs For Test #1 Fall 006 Name Instructions: Prove or Disprove the following In the case where the claim is false, provide acounter-example 1 Thesquareofanoddnaturalnumberisodd Proof Let n be an odd natural number Then n can be represented as n =k +1 for some natural number k Observe: n =(k +1) =4k +4k +1= k +k +1=m +1 {z } m ie, n has the form m +1 Hence, n is odd The square of an even natural number is even Proof Let n be an even natural number Then n can be represented as n =k for some natural number k Observe: n =(k) =4k = k =m {z } m ie, n has the form m Hence, n is even 3 is an irrational number Proof Suppose, for the sake of contradiction, that is rational Then there exist

2 integers m and n, with n 6= 0, such that = m n Without loss of generality, we can assume that m and n are relatively prime* = m n = m n n = m m is even m is even k Z such that m =k Thus, n = m =(k) =4k ie, n =4k n =k n is even n is even ie, m and n are both even This contradicts the assumption that m and n are relatively prime Since the assumption that is rational lead us to this contradiction, must be irrational *If m and n are not relatively prime, then let d be the greatest common divisor of m and n There exist relatively prime integers m 1 and n 1 such that m = dm 1 and n = dn 1 Thus we can write = m = dm 1 n dn 1 = m 1 n 1, and is written as the quotient of relatively prime integers 4 The sum or difference of rational numbers is rational Proof Let x, y Q Then m, n, r, s Z with n, s 6= 0such that x = m n and y = r s Observe: x ± y = m n ± r s = ms±nr ns Since integers are closed under addition, subtraction, and multiplication, x±y = ms±nr ns is the quotient of integers, hence rational

3 5 Theproductofrationalnumbersisrational Proof Let x, y Q Then m, n, r, s Z with n, s 6= 0such that x = m n and y = r s Observe: x y = m n r s = mr ns Since integers are closed under multiplication, x y = mr ns hence rational is the quotient of integers, 6 The quotient of rational numbers is rational, provided that the divisor is non-zero Proof Let x, y Q Then m, n, r, s Z with n, r, s 6= 0such that x = m n and y = r s Observe: x = ( m n ) y ( r s) = m s = ms n r nr Since integers are closed under multiplication, x = ms y nr rational is the quotient of integers, hence 7 The sum (or difference) of a rational and an irrational is irrational Proof Let x Q, and y Q c Suppose, for the sake of contradiction, that x + y = z,where z Q Then y = {z} z {z} x y is rational, since it is the difference of rationals Q Q This contradicts the fact that y Q c Since the assumption that z Q leads to a contradiction, it must be the case that z Q c Hence, the sum of a rational x andanirrationaly is irrational z Similarly, the difference of a rational and an irrational is irrational 8 The product or quotient of a rational number and an irrational number is irrational This is false Let y be any irrational number Then 0 y =0is the product of a rational andanirrational,andyetitisrational Similarly, 0 y =0is the quotient of a rational and an irrational, and yet it is rational 3

4 9 The product or quotient of a non-zero rational number and an irrational number is irrational Proof Let x Q, and y Q c Suppose, for the sake of contradiction, that x y = z, where z Q Then y = z x is rational, since it is the quotient of rationals This contradicts the fact that y Q c Since the assumption that z Q leads to a contradiction, it must be the case that z Q c Hence, the product of a non-zero rational x andanirrationaly is irrational z Regarding a quotient, suppose, for the sake of contradiction, that x y = z,where z Q Then y = x z is rational, since it is the quotient of rationals This contradicts the fact that y Q c Since the assumption that z Q leads to a contradiction, it must be the case that z Q c Similarly, since the quotient x is irrational, its reciprocal y also has to be irrational - y x otherwise, it would be the quotient of integers, and hence x would be the quotient of y rationals, which we know to be false 4

5 10 The sum or difference of two irrational numbers may or may not be irrational Proof This true First, we will show that the sum or difference of two irrational numbers can be rational Observe: + =0is the sum of irrationals, and the sum is rational Similarly, =0is the difference of irrationals, and the difference is rational Next, we will show that the sum or difference of two irrational numbers can be irrational Observe: + = is the sum or irrationals To see that this sum is irrational, we will assume the contrary and derive a contradiction If were rational, then would be the quotient of two rational numbers, hence, rational However, = which we know to be irrational Hence, the sum of two irrationals may be irrational Similarly, = is the difference of irrationals, and this difference is irrational (a) Alternate Proof To show that the sum of two irrationals may be rational, consider x = and y = Observe: x, y Q c and yet, x + y = Q To show that the sum of two irrationals may be irrational, consider x = y = Observe: x, y Q c and yet, x + y = Q c 5

6 11 The product or quotient of two irrational numbers may or may not be irrational Proof Toshowthattheproductorquotientoftwoirrationalnumbersmaybe rational, observe that =is the product of irrationals, and this product is rational Similarly, =1is the quotient of irrationals, and this quotient is rational To show that the product or quotient of two irrational numbers may be rational, observe that 1+ is the irrational, as it is the sum of a rational and an irrational Thus, =3+ is the product of irrationals, and this product is irrational Also, 1+ = 1 + = 1 +1 is the quotient of irrationals and this quotient is irrational Alternate Proof: If we accept the fact that the square root of any integer that is not a perfect square is irrational, then we have easier proofs that the product or quotient of irrationals can be irrational: Observe: 3 = 6 is the product of irrationals, and this product is irrational Observe: 6 3 = is the quotient of irrationals, and this quotient is irrational 1 Between any two distinct rational numbers there is another rational number Proof Let x, y Q, with x 6= y Without loss of generality, x<y Define z by z = x+y Observe: x + y is rational, as it is the sum of rationals Thus, z = x+y is rational, as it is the quotient of rationals Furthermore, x = x+x < x+y < y+y = y ie, x =< x+y <y Thus, z = x+y is a rational number that lies between x and y 6

7 13 Between any two real numbers there is a rational number Proof Let x, y R, with x 6= y Without loss of generality, x<y Thus, ε>0 such that y x = ε Select a natural number N such that Nε > 1 Ny Nx = Nε > 1 Since Ny Nx > 1, there exists an integer M such that Nx<M<Ny x< M N <y Since M N is the quotient of integers, it is rational ie, M N is a rational number between x and y 14 Between any two real numbers there is an irrational number Proof: (Slightly incorrect) Let x, y R, with x 6= y Without loss of generality, x<y Thus, ε>0 such that y x = ε Choose an irrational number k such that kε > 1 (Suchanirrationalexists For example, let M be the least integer greater than 1 ε Then k = M + fits the requirements) Thus, ky kx = kε > 1 there exists an integer N such that kx < N < ky x< N <y,where N k k irrational is irrational, since it is the quotient of a rational and an Remark 1 The problem with this proof, is that N maybezerointhatcase, N k irrational is not 7

8 Proof: (Correct) Let x, y R, with x 6= y Without loss of generality, x<y Thus, ε>0 such that y x = ε Choose an irrational number k such that kε > (Suchanirrationalexists For example, let M be the least integer greater than ε Then k = M + fits the requirements) Thus, ky kx = kε > thereexistsanon-zerointegern such that kx<n<ky x< N <y,where N k k andanirrational is irrational, since it is the quotient of a non-zero rational 15 A rational number has either a terminating or a repeating decimal representation (ie, if x Q, then x has a terminating decimal representation, or x has a repeating decimal representation) Proof Suppose that x Q Without loss of generality, we will assume that x>0 (If x =0, clearly it s decimal representation terminates If x<0, then x >0, and our proof will show that x has a terminating or repeating decimal representation, from which it will follow that x will also have a terminating or repeating decimal representation) Since x is a positive rational, it can be represented as the quotient of natural numbers: x = m n To obtain the decimal representation of x, we perform long division of m by n Note that at any given step in the long division process, there are n possible remainders If a remainder of 0 is obtained, the division process is complete, and the decimal representation of x terminates If a remainder of 0 is never obtained, then within n 1 steps, a remainder must repeat Call it r 1 When k appears as a remainder for the second time and is divided by n, it generates the same digit as part of the quotient and it leaves the same remainder as it did the firsttimethatitwasdividedbyn In fact, when r 1 appearsasaremainderfor the second time and is divided by n, it will generate the same finite sequence of digits as part of the quotient and the same finite sequence of remainders This will produce r 1 as a remainder for a third time, and when divided by n will generate the same finite sequence of digits as part of the quotient and the same finite sequence of remainders as it did the first wo times, etc 16 A real number x (with 0 x 1) is rational if and only if it can be written as a terminating, or repeating decimal 8

9 17 Prove or disprove: (a) f : R R given by f (x) =5x +3is onto Proof We must show that y R, x R such that f (x) =y Let y R be given Let x R be given by x = y 3 5 Observe: f (x) =5x +3=5 y =(y 3) + 3= y Thus, given y R, x R (namely x = y 3 5 Hence, f (x) is onto ) such that f (x) =y Scratchwork: We want: x such that f (x) =y 5x +3=y 5x = y 3 x = y 3 5 (b) f : R R given by f (x) =5x +3is one to one Proof Suppose that f (x 1 ) = f (x ) 5x 1 +3 = 5x +3 5x 1 = 5x x 1 = x ie, f (x 1 )=f (x ) x 1 = x Hence, f is one to one (c) f : R R given by f (x) =5x +3is onto Note that since x 0, it follows that 5x +3 3 Therefore our claim is false For example, given y =0, There does not exist an x such that f (x) =0 (d) f : R R given by f (x) =5x +3is one to one This is false Given y =8,x 1 = 1 and x =1are distinct values of x such that f (x 1 )=8=f (x ) Hence, f is not one to one 9

10 (e) f : R R given by f (x) =5x 3 +3is onto Proof We must show that y R, x R such that f (x) =y Let y R be given Let x R be given by x = 1 y ³ Observe: f (x) =5x 3 +3=5 y =5 y =(y 3) + 3= y Thus, given y R, x R (namely x = 1 y 3 3 ) such that f (x) =y 5 Hence, f (x) is onto Scratchwork: We want: x such that f (x) =y 5x 3 +3=y 5x 3 = y 3 x 3 = y 3 5 x = y (f) f : R R given by f (x) =5x 3 +3is one to one Proof Suppose that f (x 1 ) = f (x ) 5x = 5x x 3 1 = 5x 3 x 3 1 = x 3 x 1 = x ie, f (x 1 )=f (x ) x 1 = x Hence, f is one to one 18 The composition of one to one functions is one to one (ie, If f : X Y is one to one, and g : Y Z is one to one, then g f : X Z is one to one) Proof Let the hypothesis be given (ie, Suppose that f : X Y is one to one, and 10

11 g : Y Z is one to one) Suppose also that g (f (x 1 )) = g (f (x )) f (x 1 )=f (x ), since g is one to one x 1 = x, since f is one to one ie, g (f (x 1 )) = g (f (x )) x 1 = x Hence, g f is one to one 19 The composition of onto functions is onto (ie, If f : X Y is onto, and g : Y Z is onto, then g f : X Z is onto) Proof Let the hypothesis be given (ie, Suppose that f : X Y is onto, and g : Y Z is onto) Let z Z Since g is onto, y Y such that g (y) =z Since f is onto, x X such that f (x) =y Observe: g (f (x)) = g (y) =z Thus, given z Z, x X such that (g f)(x) =z Hence, g f is onto 11

12 0 Given f : X Y and g : Y Z, Suppose that g f : X Z isonetooneiseither f or g necessarily one to one? Claim: g is not necessarily one to one Proof Consider f : X Y and g: Y Z as shown below Note that g f : X Z is one to one, as x 1 6= x f (x 1 ) 6= f (x ), but g : Y Z is not one to one f g X Y Z a b 1 3 x y Claim: f must be one to one Proof Let the hypothesis be given (ie, suppose that g f : X Z is one to one) Suppose also, for the sake of deriving a contradiction, that f : X Y is not one to one Then x 1, x X, with x 1 6= x, such that f (x 1 )=f(x ) g (f (x 1 )) = g (f (x )) Thus, x 1, x X, with x 1 6= x, such that g (f (x 1 )) = g (f (x )) g f : X Z is not one to one, contrary to our hypothesis Since the assumption that f is not one to one yields a contradiction, it must be false Hence, f : X Y is one to one Thus, if g f : X Z is one to one, then f : X Y must be onto to one, but g : Y Z is not necessarily one to one 1

13 1 Given f : X Y and g : Y Z, Suppose that g f : X Z is onto Is either f or g necessarily onto? Claim: f is not necessarily onto Proof Consider f : X Y and g: Y Z as shown below Note that g f : X Z is onto, as z Z, x X such that g (f (x)) = z, and yet f : X Y is not onto f g X Y Z a b 1 3 x y Claim: g : Y Z must be onto Proof Let the hypothesis be given (ie, suppose that g f : X Z is onto)let z Z be given Then x X such that g (f (x)) = z y Y (namely y = f (x)), such that g (y) =z ie, Given z Z, y Y such that g (y) =z Hence, g : Y Z must be onto Thus, if g f : X Z is onto, then g : Y Z must be onto, but f : X Yis not necessarily onto 13

14 A function f : X Y has an inverse if and only if it is one to one and onto Proof If f : X Y has an inverse then it is one to one and onto Let the hypothesis be given (ie, suppose that f : X Y has an inverse, f 1 ) Then: f 1 f =1 X, and hence, f 1 f is one to one and onto Also: f f 1 =1 Y, and hence, f f 1 is one to one and onto Since f 1 f is one to one and onto, then by previous exercises (0 and1), f must be one to one, and f 1 must be onto Since f f 1 is one to one and onto, then by previous exercises (0 and1), f 1 must be one to one, and f must be onto Hence, both f and f 1 are one to one and onto If f : X Y is one to one and onto, then it has an inverse Let the hypothesis be given (ie, Suppose that f : X Y is one to one and onto) Note that since f is onto, for any value of y Y, there exists an x X such that f (x) =y Since f is one to one, there is only one x X such that f (x) =y We llcallitx y Thus, we can define f 1 : Y X by f 1 (y) =x y We must now check and make sure that f 1 f =1 X and that f f 1 =1 Y Observe: f 1 f (x y )=f 1 (f (x y )) = f 1 (y) =x y Thus, f 1 f =1 X Also: f f 1 (y) =f (f 1 (y)) = f (x y )=y Hence, f has an inverse Remark Note that if f : X Y that is one to one and onto, then g : Y X that is one to one and onto (eg, f 1 ) Remark 3 If f : S N that is one to one and onto, then g : N S that is one to one and onto also Thus, to show that a set is denumerable, we can show that g : N S that is one to one and onto, or we can show that f : S N that is one to one and onto Either is sufficient 14

15 Remark 4 Since the composition of one to one and onto functions is also one to one and onto, if a set A is known to be denumerable, then any set B that can be put into a one to one correspondence with A is also one to one and onto (Since A is denumerable, one to one one to one f : N onto A Similarly, g : A onto B Thus, (g f) :N B is one to one and onto) The point is this: As an alternate way of showing that a set B is denumerable, we can exhibit a one to one correspondence between B and a set A, where A is known to be denumerable 3 The set of even natural numbers E = {, 4, 6, 8,} is denumerable Proof Observe: N = { 1,, 3, 4, 5, 6, } E = {, 4, 6, 8, 10, 1, } Define f : N E by f (n) =n Clearly from the diagram above, f is one to one and onto Hence, E is denumerable 4 The set of odd natural numbers O = {1, 3, 5, 7,} is denumerable Proof Observe: N = { 1,, 3, 4, 5, 6, } O = { 1, 3, 5, 7, 9, 11, } Define f : N O by f (n) =n 1 Clearly from the diagram above, f is one to one and onto Hence, O is denumerable 5 The set of integers Z = {0, ±1, ± ± 3, ±4,} is denumerable Proof Observe: N = { 1,, 3, 4, 5, 6, 7, } f Z = { 0, 1, 1,,, 3, 3, } Define f : N Z by f (n) = n if n is even n 1 if n is odd Clearly from the diagram above, f is one to one and onto Hence, Z is denumerable 15

16 6 Z nz (and consequently, nz is denumerable) Proof Observe: Z = { 0, 1, 1,,, 3, 3, } nz = { 0, n, n, n, n, 3n, 3n, } Define f : Z nz by f (k) =kn Clearly from the diagram above, f is one to one and onto Hence, Z nz, and therefore, nz is denumerable by Remark 4 on page 15 7 The set of positive rational numbers Q + is denumerable Consider the table of ordered pairs below: (1, 1) (1, ) (1, 3) (1, 4) (1, 5) % % (, 1) (, ) (, 3) (, 4) (, 5) % % (3, 1) (3, ) (3, 3) (3, 4) (3, 5) % (4, 1) (4, ) (4, 3) (4, 4) (4, 5) % (5, 1) (5, ) (5, 3) (5, 4) (5, 5) If we consider the ordered pair (i, j) in the i th row and j th column to represent the quotient of integers i, then every positive rational number appears in the table at least j once (eg, the rational number m appears in the n mth row and n th column) Furthermore, the arrows in the table induce an exhaustive ordering of the positive rational numbers as follows: 1, 1,, 3, 1 3, 1 4, 3, 3, 4, 5, 1 5, (Note that we have discarded repititions of rationals if they occur eg, we have discarded (, ) becauseitisequaivalentto(1, 1) which is already on our list) Note also that since the positive rationals are ordered, they are in a one to one correspondence with the natural numbers Hence, the positive rational numbers are denumerable 16

17 8 The set of negative rational numbers Q is denumerable Proof The function f : Q + Q given by f m n = m n onto For if f (x 1 )=f (x ), is clearly one to one and Then x 1 = x x 1 = x, thus f is one to one Also, given y Q, we can choose x Q +, given by x = y This yields f (x) = x = ( y) =y Thus, f is onto 9 The union of a denumerable set and a finite set is denumerable (you can assume that the two sets are disjoint) Proof Let A = {a 1,a,,a k } and B = {b 1,b,b 3, } Then A is finite and B is denumerable Observe: N = { 1,, 3,, k, k+1, k+, k+3, } f (A B) = { a 1, a, a 3,, a k b 1, b, b 3, } Define f : N (A B) by f (n) = a n b n k if n k if n>k Clearly from the diagram above, f is one to one and onto Hence, (A B) is denumerable 30 The union of two (disjoint) denumerable sets is denumerable Proof Let A = {a 1,a,a 3,} and B = {b 1,b,b 3,} Observe: N = { 1,, 3, 4, 5, 6, } f (A B) = { a 1, b 1, a, b, a 3, b 3, } Define f : N (A B) by f (n) = a n+1 b n if n is odd if n is even Clearly from the diagram above, f is one to one and onto Hence, (A B) is denumerable 17

18 31 The union of finitely many (disjoint) denumerable sets is denumerable (ie, if A 1,A,,A n are denumerable, then n i=1a i is denumerable) Proof Suppose that A 1,A,,A n are denumerable Then we can name their elementsasfollows: A 1 = {a 11,a 1,a 13,} A = {a 1,a,a 3,} A n = {a n1,a n,a n3,} Consider: N = { 1,, n, n +1, n +, n, n +1, n +, 3n, n i=1a i = { a 11, a 1,, a n1, a 1, a, a n, a 13, a 3,, a n3, The function f : n i=1a i N given by f (a ij )=(j 1) n + i as shown above, is clearly one to one and onto Hence, n i=1a i is denumerable Alternate Proof (By induction on n) (Step 1) Show that our proposition is true for n =1 1 i=1a i = A 1, which is denumerable, by hypothesis (Step ) Assume that k i=1a i is denumerable, and show that k+1 i=1 A i is denumerable Observe: k+1 i=1 A i = k i=1a i Ak+1, which is denumerable, since it is the union of two denumerable sets Hence, n i=1a i is denumerable for all n N 18

19 3 The union of denumerably many denumerable sets is denumerable (ie, if A 1,A,,A n,are denumerable, then i=1a i is countable) (Again, you can assume that the sets are disjoint) Proof Let sets A 1,A,,A n, be denumerable and given by: A 1 = {a 11,a 1,a 13,} A = {a 1,a,a 3,} A 3 = {a 31,a 3,a 33,} A n = {a n1,a n,a n3,} (Note that a ij is the j th element of the i th set) Consider the table of elements from i=1a i listed below: a 11 a 1 a 13 a 14 a 15 % % a 1 a a 3 a 4 a 5 % % a 31 a 3 a 33 a 34 a 35 % a 41 a 4 a 43 a 44 a 45 % a 51 a 5 a 53 a 54 a 55 The table contains every element of i=1a i For example, the j th element of set A i is given by a ij This element is found in the i th row and j th column of the table Furthermore, the arrows in the table induce an exhaustive ordering of the elements of i=1a i as follows: a 11,a 1,a 1,a 31,a,a 13,a 14,a 3,a 3,a 41,a 51,a 4,a 33,a 4,a 15, Note also that since the entire set of elements of i=1a i is ordered, they are in a one to one correspondence with the natural numbers Hence, the union of denumerably many denumerable sets is denumerable 33 The set of rational numbers is denumerable (countable) Proof Q + {0} is the union of a denumerable set and a finite set, hence it is denumerable The entire set of rationals can be expressed as Q =(Q + {0}) Q, which is the union of two denumerable sets, hence denumerable 19

20 34 The real numbers 05 and are equal (ie, 05 =04999 ) Suppose that x =04999 Observe: 10x =4999 Hence: 9x =10x x =(4999 ) (04999 )=45 ie, 9x =45 Hence, x =05 But x =04999 also Hence 05 = Remark: The pervious proof hinges upon the supposition that we know how to add and subtract non-terminating decimals and that when we do, things work out" just as we think they should 35 Alternate Proof Suppose, for the sake of deriving a contradiction, that 05 6= Then 05 > and consequently, ε >0 such that = ε By the Axiom of Archimedes, n N such that n> log (ε) Observe: Since > {z }, n decimal places It follows that ε = < {z } =10 n < 10 log(ε) = ε n decimal places Thus, ε<ε,a contradiction Since the assumption that 05 6= leads to a contradiction, the assumption must be false Hence, 05 6= The set of real numbers in the interval [0, 1] is uncountable (non-denumerable) Proof (By contradiction) Suppose, for the sake of deriving a contradiction, that the set of real numbers in the interval [0, 1] is denumerable Then there exists an exhaustive ordering of the set of real numers in the interval [0, 1] {x 1,x,x 3,,x n,} 0

21 Note that this ordering contains ALL of the real numbers in the interval [0, 1] Consider the decimal expansions of these numbers: x 1 =0x 11 x 1 x 13 x =0x 1 x x 3 x 3 =0x 31 x 3 x 33 x n =0x n1 x n x n3 x nn Observe: Here, x ij is the j th digit past the decimal point in the decimal expansion of the i th real number x i Also: If x i can be written in terminating and non-terminating form (eg, 05 can be written as ), then we choose the non-terminating form (The number 0 will be represented as 0000 ) Define y [0, 1] as follows: y =0y 1 y y 3 y n expansion of y where y i is the i th digitpastthedecimalpointinthedecimal For n =1,, 3, define the digit y n as follows: 5 if x nn 6=5 y n = 1 if x nn =5 Observe: y [0, 1] and yet y 6= x n for any n N The reason for this is that, by construction of y, the n th digit of y is different from the n th digit of x n (ie, y n 6= x nn ) for all n N Hence, y 6= x n n N This contradicts our assumption that our list contains ALL of the real numbers in the interval [0, 1] Since the assumption that the set of real numbers in the interval [0, 1] is denumerable led to this contradiction, the assumption must be false Hence, the numbers in the interval [0, 1] is non-denumerable (uncountable) 1

22 37 The set of real numbers in the interval (0, 1) has the same cardinality as the set of real numbers in the interval [0, 1] (ie, f :[0, 1] (01) that is one to one and onto) Proof Let {q 1,q,q 3,,q n,} be an ordering of the rational numbers in the interval (0, 1) (Such an ordering exists, since the rationals in the interval (0, 1) are denumerable) Define f :[0, 1] (01) by x if x Q c f (x) = q 1 if x =0 q if x =1 q n+ if x = q n for n N Thefunctionisshowngraphically,below R [0,1] = Irrationals in (0, 1) 1, 0, q 1, q, q 3, ÿ, q n, ÿ f i irrat (0,1) R (0,1) = Irrationals in (0, 1) q 1, q, q 3, q 4, q 5, ÿ, q n+, ÿ Clearly, f :[0, 1] (01) is one to one and onto Corollary The set of real numbers in the interval (0, 1) is non-denumerable (incountable) Remark: By assuming that Q [(0,1) (the set of rational numbers in the interval (0, 1)) is denumerable, we have assumed the (intuitively) obvious fact that the subset of a denumerable set is denumerable (or finite) Time permitting, we may or may not prove this fact

23 38 The entire set of real numbers R is uncountable (non-denumerable) Proof Since the set of real numbers in the interval (0, 1) is non-denumerable, it suffices to exhibt a function f :(0, 1) R that is one to one and onto Define f :(0, 1) R by f (x) =cot(πx) y x -10 f (x) =cot(πx) Clearl, from the graph of f (x), we see that f (x) is one to one and onto 39 The entire set of real numbers R is uncountable (non-denumerable) Proof (Alternate Proof) Since the set of real numbers in the interval (0, 1) is non-denumerable, it suffices to exhibt a function f :(0, 1) R that is one to one and onto 1 1 for 0 <x< 1 x Define f :(0, 1) R by f (x) = 1 1 for 1 x<1 x Observe: f (x) 0 1 x<1 Similarly, f (x) < 0 0 <x< 1 3

24 f is one to one Suppose that f (x 1 )=f (x ) 0 1 x 1 1= 1 x 1 1 x 1 = 1 x x = x 1 x =x 1 x = x 1 ie, (x )=f (x ) 0 x 1 = x Similarly, if f (x 1 )=f (x ) < 0, then 1 1 x 1 =1 1 x 1 x 1 = 1 x x =x 1 x = x 1 ie, f (x )=f (x ) < 0 x 1 = x Hence, f is one to one f is onto Suppose that y 0 Let x 1, 1 be given by x = 1 1 (y+1) Observe: f (x) = 1 1 = 1 x (1 (y+1)) 1 = 1 1 ( (y+1)) 1 = (y+1) 1 = (y +1) 1=y ie, f (x) =y Suppose that y<0 Let x 0, 1 be given by x = 1 (1 y) Observe: f (x) =1 1 ie, f (x) =y =1 1 x 1 (1 y) Thus, given y R, x R such that y = f (x) Hence, f is onto =1 1 =1 (1 y) =y 1 (1 y) 4

25 Scratch Work If y 0, then we want x such that y = f (x) y = 1 x 1 y +1= 1 x x = 1 y+1 1 x = 1 (y+1) 1 x = 1 (y+1) x 1= 1 (y+1) x =1 1 (y+1) If y 0, then we want x such that y = f (x) y =1 1 x y 1= 1 x x = 1 y 1 x = 1 1 y x = 1 (1 y) 40 The set of irrational numbers Q c is uncountable Proof (By contradiction) Suppose, for the sake of contradiction, that Q c is countable (denumerable) Then R = Q Q c is the union of two denumerable sets, hence, denumerable This contradicts the fact that R is uncountable (non-denumerable) Since the assumption that Q c is countable (denumerable) leads to a contradiction, Q c must be uncountable (non-denumerable) 5