Course Notes on Real Numbers and Functions PDF

Contents 1 The real numbers 3 1.1 Equations and inequations in R........................ 5 1.2 The absolute value function.......................... 10 1.3 Supremum and infimum...

Contents 1 The real numbers 3 1.1 Equations and inequations in R........................ 5 1.2 The absolute value function.......................... 10 1.3 Supremum and infimum............................ 15 1.4 Archimedean principle............................. 24 1.5 The integer part function........................... 30 1.6 Density.................................... 35 1.7 The principle of mathematical induction................... 39 1.7.1 Simple mathematical induction reasoning............... 39 1.7.2 Strong mathematical induction reasoning............... 44 2 Real sequences 46 2.1 Definitions and notations........................... 46 2.2 Subsequences................................. 48 2.3 Supremum and infimum............................ 49 2.4 Monotone sequences.............................. 51 2.5 Limit of a real sequence............................ 53 2.6 Convergent sequences, divergent sequences.................. 63 2.7 Equivalent sequences and negligible sequences................ 74 2.7.1 Equivalent sequences.......................... 74 2.7.2 Negligible sequences.......................... 79 2.8 Adjacent sequences.............................. 81 2.9 Cauchy sequences............................... 84 3 Continuity of a function 101 3.1 Introduction.................................. 101 3.1.1 Parity and symmetry of a function.................. 102 3.1.2 Periodic functions........................... 105 3.1.3 Bounded functions........................... 106 3.1.4 Monotone functions.......................... 106 3.2 Limit of a function.............................. 108 3.2.1 Neighborhood............................. 109 3.2.2 Limit as x tends to x0......................... 110 3.2.3 Limit from the left and from the right as x tends to x0........ 114 3.2.4 Limit as x tends to +∞ or −∞................... 116 3.2.5 Operations on the limits........................ 119 1  c Dr. Ali Ayad & Dr. Ali Fares 2 3.2.6 Asymptotic behavior of a function................... 124 3.2.7 Equivalent functions and negligible functions............. 127 3.3 Continuity of a function............................ 132 3.4 Sequence defined by a function........................ 135 3.5 Theorems on continuous functions...................... 139 3.6 Some basic functions and their inverses.................... 151 3.6.1 The nth power function and the nth root function........... 151 3.6.2 The rational power function...................... 154 3.6.3 The logarithm of base a and the exponential of base a........ 155 4 Differentiability of a function 158 4.1 Definitions and properties........................... 158 4.2 Theorems on differentiable functions..................... 177 4.3 The inverse functions of the trigonometric functions............. 189 4.3.1 The arc sine function......................... 189 4.3.2 The arc cosine function........................ 191 4.3.3 The arc tangent function....................... 194 4.4 The hyperbolic functions and their inverses.................. 200 4.4.1 The hyperbolic argument sine function................ 205 4.4.2 The hyperbolic argument cosine function............... 207 4.4.3 The hyperbolic argument tangent function.............. 210 4.5 L’Hôpital’s rule................................ 220 5 Finite expansions 226 5.1 Taylor’s formula................................ 226 5.2 Finite expansions............................... 231 5.2.1 Definitions and properties....................... 231 5.2.2 Finite expansions of some basic functions............... 236 5.2.3 Operations on finite expansions.................... 239 5.2.4 Finite expansions in a neighborhood of infinity............ 249 5.2.5 Generalized finite expansions..................... 251 5.2.6 Applications.............................. 253 Chapter 1 The real numbers Recall the following classical sets: N = {0, 1, 2, 3,... , n,... } is the set of natural numbers. N∗ = N − {0} = {1, 2, 3,... , n,... } is the set of (strictly) positive natural numbers. Z = {... , −2, −1, 0, 1, 2,... } is the set of integers.   a ∗ Q= , such that a ∈ Z, b ∈ N is the set of rational numbers. b A number which is not rational is called an irrational number. R is the set of all rational and irrational numbers (called real numbers). Exercise 1 a 1) Show that every rational number can be written in the form b with a ∈ Z and b ∈ N∗ , a and b are relatively prime. 2) a) i) Show that if a ∈ Z such that a2 is even, then a is even. √ ii) Show that 2 is an irrational number. b) i) Show that if a ∈ Z such that a2 is divisible by 3, then a is divisible by 3. √ ii) Show that 3 is an irrational number. √ √ 3) Prove that 2 + 3 is an irrational number. ln 2 4) Prove that ln 3 is an irrational number. Solution 1) Let r ∈ Q, then there exist u ∈ Z and v ∈ N∗ such that r = uv. Let d be the positive greatest common divisor of u and v. Put a = ud ∈ Z and b = v d ∈ N∗ , then a and b are relatively prime and u a u = vd = = r. b d v 3 c Dr. Ali Ayad & Dr. Ali Fares 4 2) a) i) Let a ∈ Z such that a2 is even. If a is odd, then there exists k ∈ Z such that a = 2k + 1, so a2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. Hence a2 odd, which is impossible. Hence a is even. √ √ a ii) Suppose that 2 ∈ Q, then there exist a ∈ Z and b ∈ N∗ such that 2 = b a2 with a and b are relatively prime. Then 2 = b2 , and therefore a2 = 2b2. So a2 is even and therefore a is even by the part (i), so there exists m ∈ Z such that a = 2m, hence a2 = 4m2 , so 4m2 = 2b2 , hence b2 = 2m2 , so b2 is even and therefore b is even by the part (i). Hence 2 is a common divisor √ of a and √b, which is impossible since a and b are relatively prime. Thus 2 ∈ / Q, i.e., 2 is irrational. b) i) Let a ∈ Z such that a2 is divisible by 3. Let r be the remainder of the division of a by 3, then there exists q ∈ Z such that a = 3q + r, with r ∈ {0, 1, 2}. Then a2 = (3q + r)2 = 9q 2 + 6qr + r 2 = 3(3q 2 + 2qr) + r 2. If r = 1, then a2 = 3(3q 2 + 2q) + 1, so 1 is the remainder upon division of a2 by 3, this contradicts the hypothesis. If r = 2, then a2 = 3(3q 2 + 4q + 1) + 1, so 1 is the remainder upon division of a2 by 3, this contradicts the hypothesis. Hence r = 0, and therefore a is divisible by 3. √ √ ii) Suppose that 3 ∈ Q, then there exist a ∈ Z and b ∈ N∗ such that 3 = ab 2 with a and b are relatively prime. Then 3 = ab2 , hence a2 = 3b2. So a2 is divisible by 3 and therefore a is divisible by 3 by the part (i), so there exists m ∈ Z such that a = 3m, hence a2 = 9m2 , so 9m2 = 3b2 , therefore b2 = 3m2 , so b2 is divisible by 3 and therefore b is divisible by 3 by the part (i). Hence 3 is a common divisor √ of a and √ b, which is impossible since a and b are relatively prime. Thus 3 ∈ / Q, i.e., 3 is irrational. √ √ √ √ √ √ 3) Let α = 2 + 3, then α − 2 = 3, so (α − 2)2 = 3, i.e., α2 − 2α 2 + 2 = 3. Hence √ α2 − 1 2=. 2α √ If α ∈ Q, then α2 ∈ Q, and therefore 2 ∈ Q, which is impossible. Hence α ∈ / Q, i.e., α is an irrational number. 4) Suppose that ln 2 ln 3 ∈ Q, then there exist a ∈ Z and b ∈ N∗ such that ln ln 2 3 = ab with a and b are relatively prime. So b ln 2 = a ln 3, and therefore ln(2 ) = ln(3a ). Hence b 2b = 3a , which is impossible since 2b is even and 3a is odd. Thus ln 2 ln 3 is an irrational number. 2  c Dr. Ali Ayad & Dr. Ali Fares 5 1.1 Equations and inequations in R The set R, equipped with the binary relation ≤, satisfies the following properties: Reflexivity: x ≤ x for all x ∈ R. Antisymmetric: for any x, y ∈ R, if x ≤ y and y ≤ x, then x = y. Transitivity: for any x, y, z ∈ R, if x ≤ y and y ≤ z, then x ≤ z. The relation ≤ is a total order, i.e., for any x, y ∈ R, x ≤ y or y ≥ x. For any x, y ∈ R, x < y ⇔ (x ≤ y and x 6= y). For any x, y ∈ R, x < y ⇒ x ≤ y. The converse is not necessary true. For any x, y, z ∈ R, if x ≤ y and y < z, then x < z. Similarly, if x < y and y ≤ z, then x < z. For any x, y ∈ R, x < y or x = y or y < x. For any x, y ∈ R, we have: x ≤ y ⇔ x + a ≤ y + a. For any x, y ∈ R, we have: – If a > 0, then x ≤ y ⇔ ax ≤ ay. – If a < 0, then x ≤ y ⇔ ax ≥ ay. For any x, y ∈ R, we have: xy = 0 ⇔ (x = 0 or y = 0). xy 6= 0 ⇔ (x 6= 0 and y 6= 0). Let a, b, c ∈ R such that a 6= 0. Let’s solve, in R, the equation (∗) ax2 + bx + c = 0. Put ∆ = b2 − 4ac, and call it the discriminant of the equation (∗). – If ∆ = 0, then the equation (∗) admits a double real root b x1 = x2 = −. 2a – If ∆ > 0, then the equation (∗) admits two distinct real roots, which are: √ √ −b − ∆ −b + ∆ x1 = and x2 =. 2a 2a  c Dr. Ali Ayad & Dr. Ali Fares 6 – If ∆ < 0, then the equation (∗) has no real roots. Let a, b, c ∈ R such that a 6= 0. c – If a + b + c = 0, then the real roots of the equation (∗) are x1 = 1 and x2 = a. – If a − b + c = 0, then the real roots of the equation (∗) are x1 = −1 and x2 = − ac. Let u, v ∈ R. If S = u + v and P = uv, then u and v are the roots of the equation x2 − Sx + P = 0. Let a, b, c ∈ R such that a 6= 0. The sign of the trinomial ax2 + bx + c depends on the signs of a and the discriminant ∆ = b2 − 4ac. – Suppose that ∆ = 0. Then the sign of ax2 + bx + c is the same as the sign of a b except for x = − 2a. – Suppose that ∆ > 0. If a > 0, then ax2 + bx + c > 0  if (x < x1 or x > x2 ), ax2 + bx + c < 0 if x1 < x < x 2. If a < 0, then ax2 + bx + c > 0  if x1 < x < x2 , ax2 + bx + c < 0 if (x < x1 or x > x2 ). These two cases can be written in the following way: the sign of ax2 + bx + c is the same as the sign a outside roots, and it is the opposite sign of a inside roots. x x1 x2 sg(ax2 + bx + c) sg(a) 0 -sg(a) 0 sg(a) where sg(y) is the sign of the real number y. – Suppose that ∆ < 0, then the sign of ax2 + bx + c is the same as the sign of a. Example 1.1.1 Let’s solve, in R, the equation x2 + 4x − 5 = 0. Since a + b + c = 0, then the solutions of this equation are x1 = 1 and x2 = −5. Let’s solve, in R, the equation x3 + x2 − 2x = 0. We have: x3 + x2 − 2x = 0 ⇔ x(x2 + x − 2) = 0 ⇔ x(x − 1)(x + 2) = 0 ⇔ (x = 0 or x = 1 or x = −2). So the set of solutions of the equation x3 + x2 − 2x = 0 is S = {−2, 0, 1}.  c Dr. Ali Ayad & Dr. Ali Fares 7 Let’s solve, in R, the inequation x2 + x − 2 ≤ 0. Since ∆ = 9 > 0 and a = 1 > 0, then x2 + x − 2 ≤ 0 ⇔ −2 ≤ x ≤ 1. Exercise 2 Show that the following proprerties are satisfied: √ √ 1) For any x, y ∈ R+ , if x < y, then x < y. 2) For any x, y ∈ R, 2xy ≤ x2 + y 2. √ 3) For any a, b ∈ R+ , ab ≤ a+b 2. 4) For any x, y, z ∈ R, xy + xz + yz ≤ x2 + y 2 + z 2. 5) For any x, y, z, t ∈ R, we have: h i x ≤ y and z ≤ t ⇒ x + z ≤ y + t. 6) For any x, y, z, t ∈ R, we have: h i 0 ≤ x ≤ y and 0 ≤ z ≤ t ⇒ 0 ≤ xz ≤ yt. h i 0 ≤ x < y and 0 ≤ z < t ⇒ 0 ≤ xz < yt. 7) For any x, y ∈ R, we have: [0 ≤ x ≤ y] ⇒ x2 ≤ y 2 and [0 ≤ x < y] ⇒ x2 < y 2. 8) For any x, y ∈ R, we have:     1 1 1 1 [0 < x < y] ⇒ < and [x < y < 0] ⇒ <. y x y x 9) For any x, y ∈ R, we have:   1 1 [x > 0 and y > 0] ⇒ (x + y) + ≥ 4. x y Solution 1) Let x, y ∈ R+ such that x < y. Then, multiplying by the conjugate, we obtain: √ √ √ √ √ √ ( y − x)( y + x) y−x y− x= √ √ = √ √ >0 y+ x y+ x √ √ √ √ since y − x > 0 and y+ x > 0. So x< y. c Dr. Ali Ayad & Dr. Ali Fares 8 2) Let x, y ∈ R. Since 0 ≤ (x − y)2 = x2 + y 2 − 2xy, then 2xy ≤ x2 + y 2. √ √ 3) Let a, b ∈ R+. Put x = √ a and y = b, then 2 a and y 2 = b. By the part (2), √ x = a+b 2 2 √ 2xy ≤ x + y , so 2 a b ≤ a + b. Hence ab ≤ 2. 4) Let x, y, z ∈ R. We have: 2(xy + xz + yz) = 2xy + 2xz + 2yz ≤ (x2 + y 2 ) + (x2 + z 2 ) + (y 2 + z 2 ) (by the part (2)) 2 2 2 = 2(x + y + z ). So xy + xz + yz ≤ x2 + y 2 + z 2. 5) Let x, y, z, t ∈ R such that x ≤ y and z ≤ t, then y − x ≥ 0 and t − z ≥ 0. As (y + t) − (x + z) = (y − x) + (t − z) ≥ 0, then x + z ≤ y + t. 6) Let x, y, z, t ∈ R such that x ≥ 0 and z ≥ 0, then xz ≥ 0. Suppose that 0 ≤ x ≤ y and 0 ≤ z ≤ t. To show that xz ≤ yt, we distinguish the following two methods: 1st method: By multiplying the inequality x ≤ y by z ≥ 0, we obtain xz ≤ yz. By multiplying the inequality z ≤ t by y ≥ 0, we obtain yz ≤ yt. So xz ≤ yt. 2nd method: Since t ≥ 0, x ≥ 0, y − x ≥ 0, t − z ≥ 0, then yt − xz = yt − xt + xt − xz = (y − x)t + x(t − z) ≥ 0. So xz ≤ yt. Suppose that 0 ≤ x < y and 0 ≤ z < t. By multiplying the inequality x < y by z ≥ 0, we obtain xz ≤ yz. By multiplying the inequality z < t by y > 0, we obtain yz < yt. So xz < yt. 7) Let x, y ∈ R. If 0 ≤ x ≤ y, then 0 ≤ xx ≤ yy by the part (6), so x2 ≤ y 2. If 0 ≤ x < y, then 0 ≤ xx < yy by the part (6), so x2 < y 2. 8) Let x, y ∈ R. Suppose that 0 < x < y. Since x > 0 and y > 0, then xy > 0. 1 1st method: By multiplying the inequality x < y by xy > 0, we obtain: x y 1 1 < i.e., <. xy xy y x 2nd method: Since xy > 0 and y − x > 0, then 1 1 y−x − = > 0. x y xy 1 1 So y < x.  c Dr. Ali Ayad & Dr. Ali Fares 9 1 1 Suppose that x < y < 0, then 0 < −y < −x, so −x < −y by the above 1 1 discussion. Hence y < x. 9) Let x, y ∈ R such that x > 0 and y > 0. We have: (x + y)2 x2 + y 2 − 2xy (x − y)2   1 1 (x + y) + −4= −4= = ≥ 0. x y xy xy xy   So (x + y) x1 + y1 ≥ 4. 2 Exercise 3 xy x+y 1) Show that ≤ for any strictly positive real numbers x, y. x+y 4 2) Deduce that, for any strictly positive real numbers a, b and c, we have: ab bc ac a+b+c + + ≤. a+b b+c a+c 2 Solution 1) Let x and y be two strictly positive real numbers. We have: x+y xy (x + y)2 − 4xy x2 + y 2 − 2xy (x − y)2 − = = = ≥ 0. 4 x+y 4(x + y) 4(x + y) 4(x + y) xy x+y So ≤. x+y 4 2) Let a, b and c be three strictly positive real numbers. By using the part (1), we obtain: ab bc ac a+b b+c a+c + + ≤ + + a+b b+c a+c 4 4 4 2a + 2b + 2c a+b+c = =. 2 4 2 Definition 1.1.1 1) A part I of R is said to be an interval of R if for all a, b ∈ I and x ∈ R, we have: a ≤ x ≤ b ⇒ x ∈ I. 2) A sub-interval of an interval I of R is every interval of R contained in I. Let a, b ∈ R such that a ≤ b. We denote by: [a, b] the closed interval [a, b] = {x ∈ R, a ≤ x ≤ b}.  c Dr. Ali Ayad & Dr. Ali Fares 10 ]a, b[ the open interval ]a, b[= {x ∈ R, a < x < b}. ]a, b] the left-open interval ]a, b] = {x ∈ R, a < x ≤ b}. [a, b[ the right-open interval [a, b[ = {x ∈ R, a ≤ x < b}. [a, +∞[ the closed interval [a, +∞[ = {x ∈ R, x ≥ a}. ]a, +∞[ the open interval ]a, +∞[ = {x ∈ R, x > a}. ] − ∞, b] the closed interval ] − ∞, b] = {x ∈ R, x ≤ b}. ] − ∞, b[ the open interval ] − ∞, b[ = {x ∈ R, x < b}. In the exercise 16 below, we prove that they are the only intervals of R. Exercise 4 Let a, b, x ∈ R such that a < b. Show that x ∈ [a, b] if and only if there exists t ∈ [0, 1] such that x = ta + (1 − t)b. Solution x−b N.C. Put t = , then we have x = ta + (1 − t)b. As x ∈ [a, b], then a ≤ x ≤ b, so a−b a − b ≤ x − b ≤ 0. Dividing this double inequality by a − b < 0, we obtain 0 ≤ t ≤ 1, i.e., t ∈ [0, 1]. S.C. Since 1 − t ≥ 0 and b − a ≥ 0, then (1 − t)(b − a) ≥ 0, and therefore x = a + (1 − t)(b − a) ≥ a. In the other side, since t ≥ 0 and b − a ≥ 0, then t(b − a) ≥ 0, and therefore x = b − t(b − a) ≤ b. Hence x ∈ [a, b]. 2 1.2 The absolute value function Definition 1.2.1 For every x ∈ R, we define the absolute value of x, denoted |x| by:  +x if x ≥ 0, |x| = −x if x ≤ 0. Proposition 1.2.1 For any x, y ∈ R, the following properties are satisfied: 1) |x| = max{−x, +x}. 2) x = ±|x|, and therefore − |x| ≤ x ≤ + |x|.  c Dr. Ali Ayad & Dr. Ali Fares 11 3) |x| ≥ 0. 4) |x| = 0 ⇔ x = 0. 5) For any a ∈ R+ , we have: |x| = a ⇔ (x = a or x = −a). 6) |x − y| = x − y if x ≥ y and |x − y| = y − x if x ≤ y. 7) |−x| = |x|. 8) |xy| = |x| |y|. 9) |x + y| ≤ |x| + |y|. This inequality is called the triangular inequality. Equality holds if and only if x and y have the same sign. 10) |x|2 = x2. 11) |x| = |y| ⇔ x2 = y 2. 12) |x| ≤ |y| ⇔ x2 ≤ y 2. Example 1.2.1 Let’s solve, in R, the equation |x − 1| = 5. We have: |x − 1| = 5 ⇔ (x − 1 = 5 or x − 1 = −5) ⇔ (x = 6 or x = −4). So the set of solutions of this equation is S = {−4, 6}. Let’s solve, in R, the equation |x − 1| = x − 5. First, if x0 ∈ R is a solution of this equation, then x0 − 5 ≥ 0 since |x0 − 1| ≥ 0, so x0 ∈ [5, +∞[. Hence we look for solutions of this equation in the interval [5, +∞[. In the other side, h i |x − 1| = x − 5 ⇔ x − 1 = x − 5 or x − 1 = −(x − 5) ⇔ (−1 = −5 or 2x = 6) ⇔ x = 3. As 3 ∈ / [5, +∞[, then this equation has no solutions in R. Exercise 5 Let x ∈ R and a > 0. 1) Prove that |x| ≤ a ⇔ x ∈ [−a, a]. 2) Deduce that |x| > a ⇔ x ∈ ] − ∞, −a[ ∪ ]a, +∞[. Solution 1) N.C.  c Dr. Ali Ayad & Dr. Ali Fares 12 If x ≥ 0, then |x| = x, so 0 ≤ x ≤ a, therefore x ∈ [−a, a]. If x ≤ 0, then |x| = −x, so −x ≤ a, therefore x ≥ −a, therefore −a ≤ x ≤ 0, hence x ∈ [−a, a]. S.C. If x ≥ 0, then x ∈ [0, a], so |x| = x ≤ a. If x ≤ 0, then x ∈ [−a, 0], so x ≥ −a, therefore −x ≤ a, hence |x| = −x ≤ a. 2) This equivalence can be deduced from that of the part (1) by using the fact that if P and Q are two propositions, then (P ⇔ Q) ⇔ (kP ⇔ kQ). 2 Exercise 6 1) Show that |a − b| ≤ |a − c| + |c − b| for all a, b, c ∈ R. 2) Show that |x| − |y| ≤ |x − y| for all x, y ∈ R. Solution 1) Let a, b, c ∈ R. By using the triangular inequality, we obtain: |a − b| = (a − c) + (c − b) ≤ |a − c| + |c − b|. 2) Let x, y ∈ R. 1st method: By using the part (1) for a = x, b = 0 and c = y, we obtain: |x| = |x − 0| ≤ |x − y| + |y − 0| = |x − y| + |y|. So |x| − |y| ≤ |x − y|. Similarly, by using the part (1) for a = y, b = 0 and c = x, we obtain: |y| = |y − 0| ≤ |y − x| + |x − 0| = |x − y| + |x|. So |x| − |y| ≥ −|x − y|. Therefore −|x − y| ≤ |x| − |y| ≤ |x − y|. Hence |x| − |y| ≤ |x − y| by the exercise 5. 2nd method: Put a = |x| − |y| and b = x − y. Since a2 − b2 = (|x| − |y|)2 − (x − y)2 = (|x|2 + |y|2 − 2|x||y|) − (x2 + y 2 − 2xy) = (x2 + y 2 − 2|xy|) − (x2 + y 2 − 2xy) = 2(xy − |xy|) ≤ 0 (since xy ≤ |xy|), then a2 ≤ b2 , so |a| ≤ |b| by the part (12) of the proposition 1.2.1. Hence |x| − |y| ≤ |x − y|. 2  c Dr. Ali Ayad & Dr. Ali Fares 13 Exercise 7 Solve, in R, each of the following equations and inequations: 1) |x2 + 4x − 5| = x2 + 4x − 5. 2) |x2 + 4x − 5| ≤ x2 + 4x − 5. 3) |x2 + 4x − 5| < x2 + 4x − 5. 4) |x2 + 4x − 5| ≥ x2 + 4x − 5. 5) |x2 + 4x − 5| > x2 + 4x − 5. 3x − 1 6) + 2 = 1. 2x − 1 Solution 1) For any a ∈ R, |a| = a if and only if a ≥ 0. So |x2 + 4x − 5| = x2 + 4x − 5 ⇔ x2 + 4x − 5 ≥ 0 ⇔ x ∈ ] − ∞, −5] ∪ [1, +∞[. So the set of solutions of this equation is S = ] − ∞, −5] ∪ [1, +∞[. 2) For any a ∈ R, |a| ≤ a if and only if a ≥ 0. So |x2 + 4x − 5| ≤ x2 + 4x − 5 ⇔ x2 + 4x − 5 ≥ 0 ⇔ x ∈ ] − ∞, −5] ∪ [1, +∞[. So the set of solutions of this inequation is S = ] − ∞, −5] ∪ [1, +∞[. 3) For any a ∈ R, |a| < a is false. Then this inequation has no solutions. 4) For any a ∈ R, |a| ≥ a. Then the set of solutions of this inequation is R. 5) For any a ∈ R, |a| > a if and only if a < 0. Then |x2 + 4x − 5| > x2 + 4x − 5 ⇔ x2 + 4x − 5 < 0 ⇔ x ∈ ] − 5, 1[. So the set of solutions of this inequation is S = ] − 5, 1[. 6) We have: 3x − 1 3x − 1 3x − 1 +2 =1 ⇔ +2=1 or + 2 = −1 2x − 1 2x − 1 2x − 1 ⇔ 3x − 1 = −2x + 1 or 3x − 1 = −6x + 3 2 4 ⇔ x= or x =. 5 9 So the set of solutions of this equation is 5 , 9. 2 2 4 Exercise 8 Let x, y ∈ R. Prove that: c Dr. Ali Ayad & Dr. Ali Fares 14 √ p 1) |x| + |y| ≤ 2 x2 + y 2. √ √ p 2) x + y ≤ 2(x + y) if x, y ∈ R+. 3) x + y + |x − y| x + y − |x − y| max(x, y) = and min(x, y) =. 2 2 4) For any a ∈ R+ , if |x + y| ≤ a and |x − y| ≤ a, then |x| + |y| ≤ a. (Hint: we may distinguish four cases according to the signs of x and y). Solution 1) We have: h√ p i2     2 x2 + y 2 − (|x| + |y|)2 = 2 |x|2 + |y|2 − |x|2 + |y|2 + 2|xy| = |x|2 + |y|2 − 2|xy| = (|x| − |y|)2 ≥ 0. h√ p i2 √ p So (|x| + |y|)2 ≤ 2 x2 + y 2. Hence |x| + |y| ≤ 2 x2 + y 2 , i.e., |x| + √ p |y| ≤ 2 x2 + y 2. √ √ 2) Suppose that x, y ∈ R+. Put a = x and b = y, then 2 2 p a = x and b = y. By √ √ √ √ the part (1), |a| + |b| ≤ 2 a2 + b2 , so x + y ≤ 2(x + y). 3) If x ≤ y, then max(x, y) = y, min(x, y) = x and |x − y| = y − x. So x + y + |x − y| x+y+y−x = = y = max(x, y) 2 2 x + y − |x − y| x+y−y+x = = x = min(x, y). 2 2 If x ≥ y, then max(x, y) = x, min(x, y) = y and |x − y| = x − y. So x + y + |x − y| x+y+x−y = = x = max(x, y) 2 2 x + y − |x − y| x+y−x+y = = y = min(x, y). 2 2 4) Let a ∈ R+. We distinguish the following four cases: Suppose that x ≥ 0 and y ≥ 0. Then x + y ≥ 0, so |x| + |y| = x + y = |x + y| ≤ a.  c Dr. Ali Ayad & Dr. Ali Fares 15 Suppose that x ≥ 0 and y < 0, then we put y 0 = −y > 0. As |x + y 0 | = |x − y| ≤ a, then, by using the first case, we obtain: |x| + |y| = |x| + |y 0 | ≤ a. Suppose that x < 0 and y ≥ 0, then we put x0 = −x > 0. As |x0 + y| = | − x + y| = |x − y| ≤ a, then, by using the first case, we obtain: |x| + |y| = |x0 | + |y| ≤ a. Suppose that x < 0 and y < 0. Then x + y < 0, so |x| + |y| = −x − y = −(x + y) = |x + y| ≤ a. 2 Exercise 9 Let a, b ∈ R. 1) Show that if a ≤ b + ε for all ε > 0, then a ≤ b. 2) Deduce that if x ∈ R such that |x| ≤ ε for all ε > 0, then x = 0. Solution a−b 1) Suppose that a > b. Put ε = 2 > 0, then a ≤ b + ε by using the hypothesis. So a−b a+b a+a a≤b+ = < = a, 2 2 2 which is impossible. Hence a ≤ b. 2) Let x ∈ R such that |x| ≤ ε for all ε > 0, Put a = |x| and b = 0, then a ≤ b + ε for all ε > 0. So a ≤ b, i.e., |x| ≤ 0 by the part (1). But |x| ≥ 0, then |x| = 0, and therefore x = 0. 2 1.3 Supremum and infimum Definition 1.3.1 Let A be a nonempty subset of R. 1) We say that A is bounded from above in R if there exists M ∈ R such that x ≤ M for all x ∈ A. In this case, we say that A is bounded from above by M and that M is an upper bound of A. 2) We say that A is bounded from below in R if there exists m ∈ R such that m ≤ x for all x ∈ A. In this case, we say that A is bounded from below by m and that m is a lower bound of A. 3) We say that A is bounded in R if A is bounded both from above and from below, i.e., if there exist m, M ∈ R such that m ≤ x ≤ M for all x ∈ A.  c Dr. Ali Ayad & Dr. Ali Fares 16 4) We say that an element a of A is a greatest element (or a maximum) of A if x ≤ a for all x ∈ A. 5) We say that an element a of A is a least element (or a minimum) of A if a ≤ x for all x ∈ A. Remark 1.3.1 Let A be a nonempty subset of R. A is bounded in R if and only if there exists a real number M ≥ 0 such that |x| ≤ M for all x ∈ A. A greatest element of A is every upper bound of A in R that belongs to A. A least element of A is every lower bound of A in R that belongs to A. If A has a greatest element, then it is unique, denoted by max A. If A has a least element, then it is unique, denoted by min A. Example 1.3.1 1) The set N is bounded from below in R by 0, but not bounded from above in R. Moreover, 0 is the least element of N. On the other hand, N has no greatest element. 2) The sets Z, Q and R are not bounded from above, neither bounded from below in R. Moreover, they have no least and greatest elements. 3) Every nonempty finite subset of R is bounded and has least and greatest elements. Definition 1.3.2 Let A be a nonempty subset of R. 1) We say that a real number a is a supremum of A if it satisfies the following two condi- tions: i) a is an upper bound of A. ii) If M ∈ R is an upper bound of A, then a ≤ M. In other words, a real number a is a supremum of A if and only if a is the least upper bound of A in R. 2) We say that a real number a is an infimum of A if it satisfies the following two condi- tions: i) a is a lower bound of A. ii) If m ∈ R is a lower bound of A, then m ≤ a. In other words, a real number a is an infimum of A if and only if a is the greatest lower bound of A in R.  c Dr. Ali Ayad & Dr. Ali Fares 17 Remark 1.3.2 Let A be a nonempty subset of R. If A has a supremum, then it is unique, denoted by supR (A) or simply sup A. If A has an infimum, then it is unique, denoted by inf R (A) or simply inf A. If A has a greatest element, then A has a supremum with sup A = max A. If A has a least element, then A has an infimum with inf A = min A. If A has a supremum in A, i.e., sup A ∈ A, then A has a greatest element with max A = sup A. If A has an infimum in A, i.e., inf A ∈ A, then A has a least element with min A = inf A. Exercise 10 Let A (resp. B) be a nonempty subset having supremum and infimum in R. 1) Show that if A ⊆ B, then sup A ≤ sup B and inf A ≥ inf B. 2) Prove that if A ∩ B has supremum and infimum in R, then n o n o sup(A ∩ B) ≤ min sup A, sup B and inf (A ∩ B) ≥ max inf A, inf B. Verify that these two inequalities are strict for A = [0, 1] ∪ {−2, 2} and B = [0, 1] ∪ {−3, 3}. 3) Prove that A ∪ B has supremum and infimum in R and that n o n o sup(A ∪ B) = max sup A, sup B and inf (A ∪ B) = min inf A, inf B. Solution 1) Let x ∈ A. Since A ⊆ B, then x ∈ B, so x ≤ sup B. Therefore sup B is an upper bound of A in R. Hence sup A ≤ sup B since sup A is the least upper bound of A in R. Let x ∈ A. Since A ⊆ B, then x ∈ B, so x ≥ inf B. Therefore inf B is a lower bound of A in R. Hence inf A ≥ inf B since inf A is the greatest lower bound of A in R. 2) Suppose that A ∩ B has supremum and infimum in R. Since A ∩ B ⊆ A and A ∩ B ⊆ B, then, by the part (1), sup(A ∩ B) ≤ sup A and sup(A ∩ B) ≤ sup B, inf (A ∩ B) ≥ inf A and inf (A ∩ B) ≥ inf B.  c Dr. Ali Ayad & Dr. Ali Fares 18 So n o n o sup(A ∩ B) ≤ min sup A, sup B and inf (A ∩ B) ≥ max inf A, inf B. Strict inequalities: Let A = [0, 1] ∪ {−2, 2} and B = [0, 1] ∪ {−3, 3}. Since 2 (resp. −2) is an upper bound (resp. lower bound) of A belonging to A, then sup A = 2 and inf A = −2. Similarly, since 3 (resp. −3) is an upper bound (resp. lower bound) of B belonging to B, then sup B = 3 and inf B = −3. In the other side, A ∩ B = [0, 1], since 1 (resp. 0) is an upper bound (resp. lower bound) of A ∩ B belonging to A ∩ B, then sup(A ∩ B) = 1 and inf (A ∩ B) = 0. Hence n o sup(A ∩ B) = 1 < 2 = min sup A, sup B n o inf (A ∩ B) = 0 > −2 = max inf A, inf B. n o n o 3) Put M = max sup A, sup B and m = min inf A, inf B. Let x ∈ A ∪ B, then x ∈ A or x ∈ B, so x ≤ sup A or x ≤ sup B. Therefore x ≤ M. Hence M is an upper bound of A ∪ B. In the other side, let M 0 ∈ R be an upper bound of A ∪ B, then x ≤ M 0 for all x ∈ A ∪ B. In particular, since A ⊆ A ∪ B and B ⊆ A ∪ B, then x ≤ M 0 for all x ∈ A and x ≤ M 0 for all x ∈ B. Therefore M 0 is an upper bound of A and B, so sup A ≤ M 0 and sup B ≤ M 0 , hence M ≤ M 0. Thus M is the least upper bound of A ∪ B, i.e., sup(A ∪ B) = M. Let x ∈ A ∪ B, then x ∈ A and x ∈ B, so x ≥ inf A and x ≥ inf B. Therefore x ≥ m. Hence m is a lower bound of A ∪ B. In the other side, let m0 ∈ R be a lower bound of A ∪ B, then x ≥ m0 for all x ∈ A ∪ B. In particular, since A ⊆ A ∪ B and B ⊆ A ∪ B, then x ≥ m0 for all x ∈ A and x ≥ m0 for all x ∈ B. Therefore m0 is a lower bound of A and B, so inf A ≥ m0 and inf B ≥ m0 , hence m ≥ m0. Thus m is the greatest lower bound of A ∪ B, i.e., inf (A ∪ B) = m. 2 Exercise 11 Let   1 1 A= + , such that m, n ∈ Z \ {0}. m n Prove that A admits supremum and infimum in R to be determined. 1 1 Solution Let n ∈ Z \ {0}. If n ≥ 1, then 0 ≤ n ≤ 1, and if n ≤ −1, then −1 ≤ n ≤ 0. In the two cases, we obtain: 1 −1 ≤ ≤ 1. n So, for all m, n ∈ Z \ {0}, we have: 1 1 −2 ≤ + ≤ 2. m n  c Dr. Ali Ayad & Dr. Ali Fares 19 So A is bounded from above by 2 and from below by −2. In the other side, as 1 1 1 1 2= + ∈A and −2= + ∈ A, 1 1 −1 −1 then 2 (resp. −2) is the greatest (resp. least) element of A, i.e., max A = 2 (resp. min A = −2). Hence A admits a supremum and an infimum in R with sup A = max A = 2 and inf A = min A = −2. 2 Theorem 1.3.1 (Supremum criterion) Let M ∈ R and A be a nonempty subset of R. Then M = sup A if and only if the following two properties are satisfied: i) M is an upper bound of A. ii) For any ε > 0, there exists a ∈ A such that M − ε < a ≤ M. Proof N.C. Since M = sup A, then M is an upper bound of A. In the other side, suppose, by contradiction, that the property (ii) is not satisfied, then there exists ε > 0 such that x ≤ M − ε for all x ∈ A (we have not x > M since M is an upper bound of A). So M − ε is an upper bound of A. As M = sup A is the least upper bound of A, then M ≤ M − ε, therefore ε ≤ 0, which is impossible. Hence the property (ii) is satisfied. S.C. Let’s show that M is the least upper bound of A. Indeed, let M 0 be another upper bound of A in R. Suppose, by contradiction, that M > M 0 , then put ε = M − M 0 > 0. By the property (ii), there exists a ∈ A such that M − ε < a ≤ M , and therefore M 0 < a, which is impossible since M 0 is an upper bound of A, therefore M ≤ M 0. Hence M is the least upper bound of A, i.e., M = sup A. 2 Exercise 12 Let   n A= , n∈N. n+1 1) Show that A is bounded in R. 2) Prove that sup A = 1 and inf A = 0. n Solution For every n ∈ N, put un = n+1. Then A = {un , n ∈ N}. n 1) For all n ∈ N, n ≤ n + 1, so un = n+1 ≤ 1. Hence A is bounded from above in R and 1 is an upper bound of A in R. In the other side, for all n ∈ N, un ≥ 0. So A is bounded from below in R and 0 is a lower bound of A in R. Thus A is bounded in R.  c Dr. Ali Ayad & Dr. Ali Fares 20 2) For sup A, we will use the supremum criterion. First, 1 is an upper bound of A in R. Let ε > 0. Let’s find n0 ∈ N such that 1 − ε < un0 ≤ 1. (∗) Since un0 = n0n+10 = 1− 1 n0 +1 , then the double inequality (∗) is equivalent to the following inequality: 1 < ε. n0 + 1 This is equivalent to 1 n0 > − 1. ε Such n0 exists since the set of natural numbers N is infinite. Hence sup A = 1. Since 0 is a lower bound of A in R and 0 ∈ A (since u0 = 0), then 0 is the least element of A, i.e., min A = 0. Hence inf A = min A = 0. 2 Theorem 1.3.2 (Infimum criterion) Let m ∈ R and A be a nonempty subset of R. Then m = inf A if and only if the following two properties are satisfied: i) m is a lower bound of A. ii) For any ε > 0, there exists a ∈ A such that m ≤ a < m + ε. Proof N.C. Since m = inf A, then m is a lower bound of A. In the other side, suppose, by contradiction, that the property (ii) is not satisfied, then there exists ε > 0 such that x ≥ m + ε for all x ∈ A (we have not x < m since m is a lower bound of A). So m + ε is a lower bound of A. As m = inf A is the greatest lower bound of A, then m ≥ m + ε, therefore ε ≤ 0, which is impossible. Hence the property (ii) is satisfied. S.C. Let’s show that m is the greatest lower bound of A. Indeed, let m0 be another lower bound of A in R. Suppose, by contradiction, that m < m0 , then put ε = m0 − m > 0. By the property (ii), there exists a ∈ A such that m ≤ a < m + ε, and therefore a < m0 , which is impossible since m0 is a lower bound of A, therefore m ≥ m0. Hence m is the greatest lower bound of A, i.e., m = inf A. 2 Exercise 13 Let A = ]0, 1]. Prove that sup A = 1 and inf A = 0. Solution For all x ∈ A, x ≤ 1. So 1 is an upper bound of A in R. In the other side, since 1 ∈ A, then 1 is the greatest element of A i.e., max A = 1. Hence sup A = max A = 1.  c Dr. Ali Ayad & Dr. Ali Fares 21 For inf A, we will use the infimum criterion. First, for all x ∈ A, x ≥ 0 (since x > 0). So 0 is a lower bound of A in R. Let ε > 0. Let’s find x0 ∈ A such that 0 ≤ x0 < 0 + ε. ε If ε ≤ 1, then we take x0 = 2 ≤ 12 < 1, so x0 ∈ A. 1 If ε > 1, then we take x0 = 2 < 1, so x0 ∈ A. Hence inf A = 0. 2 Remark 1.3.3 We can generalize the exercise 13 in the following way: If A is one of the intervals [a, b], [a, b[, ]a, b], ]a, b[ (where a < b), then sup A = b and inf A = a. If A is one of the intervals [a, +∞[ or ]a, +∞[, then A is not bounded from above in R and inf A = a. If A is one of the intervals ] − ∞, b] or ] − ∞, b[, then A is not bounded from below in R and sup A = b. Theorem 1.3.3 (Supremum criterion axiom) Every nonempty bounded from above part of R has a supremum. Every nonempty bounded from below part of R has an infimum. Exercise 14 Let a ∈ Q and A = {r ∈ Q, r < a}. Prove that A admits a supremum, but A has no greatest element. Solution Since a − 1 ∈ Q and a − 1 < a, then a − 1 ∈ A, so A 6= ∅. Moreover, as r ≤ a for all r ∈ A, then A is bounded from above in R by a. Hence A is a nonempty bounded from above part of R, so A admits a supremum. In the other side, if A has a greatest element b, then b ∈ A, so b ∈ Q and b < a. Put r = a+b 2 ∈ Q, then r < a, so r ∈ A. Hence r ≤ b since b = max A, which is impossible since b < r. Thus A has no greatest element. 2 Exercise 15 1) Let A and B be two nonempty and bounded from above parts of R. Let AB = {ab, such that a ∈ A, b ∈ B}. Show that if A, B ⊆ ]0, +∞[, then AB is nonempty bounded from above and sup(AB) = (sup A)(sup B).  c Dr. Ali Ayad & Dr. Ali Fares 22 2) Give an example of two nonempty bounded from above parts A and B of R such that AB is not bounded from above. Solution 1) Since A and B are two nonempty and bounded from above parts of R, then sup A and sup B exist. As A 6= ∅ and B 6= ∅, then there exist a ∈ A and b ∈ B, so ab ∈ AB. Hence AB 6= ∅. Let x ∈ AB, then there exist a ∈ A and b ∈ B such that x = ab. As 0 < a ≤ sup A and 0 < b ≤ sup B, then x = ab ≤ (sup A)(sup B). So AB is bounded from above by (sup A)(sup B), and then sup(AB) exists and sup(AB) ≤ (sup A)(sup B). In the other side, if a ∈ A and b ∈ B, then ab ∈ AB, so ab ≤ sup(AB). But b > 0, then sup(AB) a≤ , ∀ a ∈ A. b sup(AB) So b is an upper bound of A. Therefore sup(AB) sup A ≤. b But b > 0 and sup A > 0, then sup(AB) b≤ , ∀ b ∈ B. sup A sup(AB) So sup A is an upper bound of B. Therefore sup(AB) sup B ≤. sup A Hence (sup A)(sup B) ≤ sup(AB). Thus sup(AB) = (sup A)(sup B). 2) Take, for example, A = B = ] − ∞, 1] ⊆ R, they are nonempty and bounded from above by 1. Let’s show that AB = R. Indeed, AB ⊆ R. If x ∈ R, then we consider the two cases: if x ≤ 1, then x = x × 1 ∈ AB since x ∈ A and 1 ∈ B. √ √ √ if x > 1, then x > 1, so − x < −1. Therefore − x ∈ A = B and therefore √ √ x = (− x)(− x) ∈ AB. Hence R ⊆ AB. Thus AB = R is not bounded from above. 2 Exercise 16 The goal of this exercise is to prove that the only intervals of R are the intervals of the form: ∅, R, [a, +∞[, ]a, +∞[, ] − ∞, b], ] − ∞, b[, [a, b], ]a, b], [a, b[ and ]a, b[. c Dr. Ali Ayad & Dr. Ali Fares 23 1) Show that if I is an interval of R such that I is not bounded from below, neither bounded from above in R, then I = R. 2) Show that if I is a nonempty interval of R such that I is bounded from below, but not bounded from above in R, then there exists a ∈ R such that I = [a, +∞[ or I = ]a, +∞[. 3) Show that if I is a nonempty interval of R such that I is bounded from above, but not bounded from below in R, then there exists b ∈ R such that I = ] − ∞, b] or I = ] − ∞, b[. 4) Show that if I is an interval of R such that I is bounded from above and bounded from below in R, then there exist a, b ∈ R such that I = [a, b] or I = ]a, b] or I = [a, b[ or I = ]a, b[. Solution 1) Let I be an interval of R such that I is not bounded from below, neither bounded from above in R. Let x ∈ R. As x is not a lower bound (resp. upper bound) of I, then there exists a ∈ I (resp. b ∈ I) such that a ≤ x (resp. x ≤ b). As I is an interval of R and a ≤ x ≤ b with a, b ∈ I, then x ∈ I. So R ⊆ I, but I ⊆ R, then I = R. 2) Let I be a nonempty interval of R such that I is bounded from below, but not bounded from above in R. Then I admits an infimum a = inf I. As a is a lower bound of I, then a ≤ x for all x ∈ I, so I ⊆ [a, +∞[. Suppose that a ∈ I. Let x ∈ [a, +∞[, then x ≥ a. As x is not an upper bound of I (since I is not bounded from above), then there exists b ∈ I such that x ≤ b. As I is an interval of R and a ≤ x ≤ b with a, b ∈ I, then x ∈ I. So [a, +∞[ ⊆ I. Hence I = [a, +∞[. Suppose that a ∈ / I, then I ⊆ ]a, +∞[. Conversely, let x ∈ ]a, +∞[, then x > a. As x is not a lower bound of I (since a is the greatest lower bound of I), then there exists a0 ∈ I such that a0 ≤ x. Moreover, as x is not an upper bound of I (since I is not bounded from above), then there exists b ∈ I such that x ≤ b. As I is an interval of R and a0 ≤ x ≤ b with a0 , b ∈ I, then x ∈ I. So ]a, +∞[ ⊆ I. Hence I = ]a, +∞[. 3) Let I be a nonempty interval of R such that I is bounded from above, but not bounded from below in R. Then I admits a supremum b = sup I. As b is an upper bound of I, then x ≤ b for all x ∈ I, so I ⊆ ] − ∞, b]. Suppose that b ∈ I. Let x ∈ ]−∞, b], then x ≤ b. As x is not a lower bound of I (since I is not bounded from below), then there exists a ∈ I such that a ≤ x. As I is an interval of R and a ≤ x ≤ b with a, b ∈ I, then x ∈ I. So ] − ∞, b] ⊆ I. Hence I = ] − ∞, b]. Suppose that b ∈ / I, then I ⊆ ] − ∞, b[. Conversely, let x ∈ ] − ∞, b[, then x < b. As x is not an upper bound of I (since b is the least upper bound of I), then there exists b0 ∈ I such that x ≤ b0. Moreover, as x is not a lower bound of I (since I is not bounded from below), then il existe a ∈ I such that a ≤ x. As I  c Dr. Ali Ayad & Dr. Ali Fares 24 is an interval of R and a ≤ x ≤ b0 with a, b0 ∈ I, then x ∈ I. So ] − ∞, b[ ⊆ I. Hence I = ] − ∞, b[. 4) Let I be an interval of R such that I is bounded from above and bounded from below in R. Then I admits an infimum a = inf I and a supremum b = sup I. As a is a lower bound and b is an upper bound of I, then a ≤ x ≤ b for all x ∈ I, so I ⊆ [a, b]. In each of the following four cases, we apply the same technique of the parts (2) and (3): If a, b ∈ I, then we get I = [a, b]. If a ∈ / I and b ∈ I, then we get I = ]a, b]. If a ∈ I and b ∈ / I, then we get I = [a, b[. If a ∈ / I, then we get I = ]a, b[. 2 / I and b ∈ 1.4 Archimedean principle Proposition 1.4.1 (Archimedean principle: R is Archimedean) For any a > 0 and b ∈ R, there exists n ∈ N∗ such that na > b. Proof If a ≥ b, then it is sufficient to take n = 2 since 2a > a ≥ b. Suppose that a < b. Suppose, by contradiction, that na ≤ b for all n ∈ N∗. Then the set A = {na, n ∈ N∗ } is a nonempty bounded from above (by b) part of R. So A admits a supremum. Put M = sup A. By the supremum criterion, for ε = a > 0, there exists n0 ∈ N∗ such that M − a < n0 a ≤ M. So M < (n0 + 1)a. Put m0 = n0 + 1 ∈ N∗ , then M < m0 a with m0 a ∈ A, this is in contraddiction with the fact that M is an upper bound of A. Hence there exists n ∈ N∗ such that na > b. 2 1 Corollary 1.4.1 For any ε > 0, there exists n ∈ N∗ such that n < ε. Proof Let ε > 0. Put a = ε > 0 and b = 1 ∈ R. By the Archimedean principle, there 1 exists n ∈ N∗ such that na > b, so nε > 1. Hence n < ε. 2 Exercise 17 1 Let a and b be two real numbers such that a ≤ b + n for all n ∈ N∗. Show that a ≤ b.  c Dr. Ali Ayad & Dr. Ali Fares 25 1 Solution Let ε > 0. By the corollary 1.4.1, there exists n ∈ N∗ such that n < ε. So 1 a≤b+ ≤ b + ε. n Hence a ≤ b by the exercise 9. 2 Exercise 18 Show that the set Z is not a bounded from above subset of R. Solution Suppose that Z is a bounded from above subset of R. As R is complete, then Z has a supremum. Let M = sup Z. Put ε = 1 > 0, then, by the supremum criterion, there exists n ∈ Z such that M − 1 < n ≤ M. So M < n + 1. But n + 1 ∈ Z, then n + 1 ≤ M (since M is an upper bound of Z). Therefore M < M , which is impossible. Thus Z is not bounded from above in R. 2 Exercise 19   2 ∗ 1) Consider the set A = ,n∈N. n Prove that sup A = 2 and inf A = 0.   1 n ∗ 2) Consider the set B = + (−1) , n ∈ N. n Prove that sup B = 23 and inf B = −1. Solution 2 1) For every n ∈ N∗ , put un = n , then A = {un , n ∈ N∗ } 2 Let n ∈ N∗. If n = 1, then u1 = 2 ≤ 2. If n ≥ 2, then un = n ≤ 1 ≤ 2. So 2 is an upper bound of A. Moreover, as u1 = 2, then 2 ∈ A. Hence 2 is the greatest element of A i.e., max A = 2. Hence sup A = max A = 2. 2 For all n ∈ N∗ , un = n ≥ 0. So 0 is a lower bound of A. In the other side, let ∗ ε > 0, let’s find n0 ∈ N such that 0 ≤ un0 < 0 + ε. Since 2ε > 0, then, by the Archimedean principle, there exists n0 ∈ N∗ such that 1 n0 < 2ε , so n20 < ε, i.e., un0 < ε. Hence inf A = 0 by the infimum criterion. 1 2) For every n ∈ N∗ , put vn = n + (−1)n , then B = {vn , n ∈ N∗ }. Let n ∈ N∗. If n is even, then n ≥ 2 and (−1)n = 1, so 1 1 3 vn = +1≤ +1=. n 2 2  c Dr. Ali Ayad & Dr. Ali Fares 26 If n is odd, (−1)n = −1, so 1 3 vn = −1≤0≤. n 2 3 Hence 2 is an upper bound of B. Moreover, as v2 = 12 + 1 = 32 , then 32 ∈ B. 3 Hence 2 is the greatest element of B i.e., max B = 23. Hence sup B = max B = 3 2. For any n ∈ N∗ , as (−1)n ≥ −1, then 1 1 vn = + (−1)n ≥ − 1 ≥ −1. n n So −1 is a lower bound of B. In the other side, let ε > 0, let’s find n0 ∈ N∗ such that −1 ≤ vn0 < −1 + ε. 1 By the Archimedean principle, there exists n1 ∈ N∗ such that n1 < ε. If n1 is odd, then we take n0 = n1 , so 1 1 vn0 = vn1 = + (−1)n1 = − 1 < −1 + ε. n1 n1 1 1 1 If n1 is even, then we take n0 = n1 + 1 ∈ N∗ is odd, so n0 = n1 +1 < n1 and 1 1 1 vn0 = + (−1)n0 = −1< − 1 < −1 + ε. n0 n0 n1 Hence inf B = −1 by the infimum criterion. 2 Exercise 20 Consider the set 3n2 + 4   A= un , such that un = and n ∈ N. n2 + 1 1) Show that 3 ≤ un ≤ 4 for all n ∈ N. 2) Determine the supremum and the infimum of A. Solution 1) Let n ∈ N. We have: 3n2 + 4 3(n2 + 1) + 1 1 un = = =3+. n2 + 1 n2 + 1 n2 + 1 1 Since n2 +1 ≥ 0, then un ≥ 3. 1 Since n2 +1 ≤ 1, then un ≤ 4.  c Dr. Ali Ayad & Dr. Ali Fares 27 2) By the part (1), the nonempty subset A of R is bounded from above by 4 and bounded from below by 3, so sup A and inf A exist in R. Since u0 = 4, then 4 ∈ A, so 4 is the greatest element of A, i.e., max A = 4. Hence sup A = max A = 4. Let ε > 0. Let’s find n0 ∈ N such that 3 ≤ un0 < 3 + ε. 1 Indeed, by the Archimedean principle, there exists n0 ∈ N∗ such that n0 < ε. Since n20 + 1 > n20 ≥ n0 , then n21+1 < n10 , and therefore 0 1 1 un0 = 3 + 0, there exists n0 ∈ N∗ such that 1 n0 < inf3 A. By using (∗), we obtain: 3 un0 ≤ < inf A, n0 which is impossible. Hence inf A = 0. 2 c Dr. Ali Ayad & Dr. Ali Fares 28 Exercise 22 Let A and B be two nonempty bounded subsets of R. Set: n o A + B = a + b, such that a ∈ A and b ∈ B. 1) Show that A + B is a nonempty bounded subset of R, and that sup(A + B) = sup A + sup B and inf (A + B) = inf A + inf B 2) Let   2 2 ∗ E= + , such that m, n ∈ N. m n Prove that E is a nonempty bounded subset of R, and determine sup E and inf E. Solution 1) Since A and B are two nonempty bounded subsets of R, then sup A, sup B, inf A and inf B exist. Since A 6= ∅ and B 6= ∅, then there exist a0 ∈ A et b0 ∈ B, so x0 = a0 + b0 ∈ A + B. Hence A + B 6= ∅. In the other side, let x ∈ A + B, then there exist a ∈ A and b ∈ B such that x = a + b. Since inf A ≤ a ≤ sup A and inf B ≤ b ≤ sup B, then inf A + inf B ≤ x ≤ sup A + sup B. So A + B is bounded from above by sup A + sup B and bounded from below by inf A + inf B. Hence A + B is a nonempty bounded subset of R. Then sup(A + B) and inf (A + B) exist and sup(A + B) ≤ sup A + sup B and inf A + inf B ≤ inf (A + B). In order to prove the equalities, we distinguish the following two methods: 1st method: Let b ∈ B. For any a ∈ A, a + b ∈ A + B, so a + b ≤ sup(A + B), and then a ≤ sup(A + B) − b. So sup(A + B) − b is an upper bound of A, and then sup A ≤ sup(A + B) − b since sup A is the least upper bound of A. So, for all b ∈ B, b ≤ sup(A + B) − sup A. Then sup(A + B) − sup A is an upper bound of B. Hence sup B ≤ sup(A + B) − sup A since sup B is the least upper bound of B. Hence sup A + sup B ≤ sup(A + B). Thus sup(A + B) = sup A + sup B. c Dr. Ali Ayad & Dr. Ali Fares 29 Let b ∈ B. For any a ∈ A, a + b ∈ A + B, so a + b ≥ inf (A + B), and then a ≥ inf (A + B) − b. So inf (A + B) − b is a lower bound of A, and then inf A ≥ inf (A + B) − b since inf A is the greatest lower bound of A. So, for all b ∈ B, b ≥ inf (A + B) − inf A. Hence inf (A + B) − inf A is a lower bound of B, and then inf B ≥ inf (A + B) − inf A since inf B is the greatest lower bound of B. Hence inf A + inf B ≥ inf (A + B). Thus inf (A + B) = inf A + inf B. 2nd method: We will use the supremum and infimum criteria. Indeed, ε Let ε > 0. Since 2 > 0, then, by the supremum criterion, there exist x ∈ A and y ∈ B such that ε ε sup A − < x ≤ sup A and sup B − < y ≤ sup B. 2 2 Taking the sum of these two double inequalities, we obtain x + y ∈ A + B with (sup A + sup B) − ε < x + y ≤ (sup A + sup B). Hence sup(A + B) = sup A + sup B by the supremum criterion. ε Let ε > 0. Since 2 > 0, then, by the infimum criterion, there exist x ∈ A and y ∈ B such that ε ε inf A ≤ x < inf A + and inf B ≤ y < inf B +. 2 2 Taking the sum of these two double inequalities, we obtain x + y ∈ A + B with (inf A + inf B) ≤ x + y < (inf A + inf B) + ε. Hence inf (A + B) = inf A + inf B by the infimum criterion. 2) E = A + A where   2 ∗ A= , such that n ∈ N. n By the part (1) of the exercise 19, the set A is nonempty bounded in R with sup A = 2 and inf A = 0. By the part (1) above, E is nonempty bounded in R with sup E = sup A + sup A = 4 and inf E = inf A + inf A = 0. 2  c Dr. Ali Ayad & Dr. Ali Fares 30 1.5 The integer part function Definition 1.5.1 For every x ∈ R, denote E(x) (or bxc), the unique integer m ∈ Z such that m ≤ x ≤ m + 1, and call it the integer part of x. The function E is called the integer part function. For example, E(1, 3) = 1, E(3) = 3, E(π) = 3, E(−2, 6) = −3. Proposition 1.5.1 1) For any x ∈ R, E(x) ≤ x < E(x) + 1. 2) For any x ∈ R, n o E(x) = max m ∈ Z, m ≤ x. 3) E(x) = x if and only if x ∈ Z. 4) E(E(x)) = E(x). 5) The function E is increasing on R, i.e. if x, y ∈ R such that x ≤ y, then E(x) ≤ E(y). 6) If k ∈ Z and x ∈ R such that k ≤ x, then k ≤ E(x). 7) E(x) ≥ 0 if and only if x ≥ 0. 8) For any k ∈ Z and x ∈ R, E(x + k) = E(x) + k. Exercise 23 By using the integer part function, give another proof of the Archimedean prin- ciple (Proposition 1.4.1). Solution Let a > 0 and b ∈ R. If b < 0, then a > b, so it is sufficient to take n = 1 ∈ N∗.       If b ≥ 0, then put n = E ab +1, so n ∈ N∗ since E ab ≥ 0. Since b a b (since a > 0). 2 Exercise 24 For every x ∈ R, put:     1 x f (x) = E , g(x) = E and s(x) = g(x) − g(−x). x2 + 1 x2 + 1 1) Calculate f (x) for all x ∈ R. 2) Calculate g(x) for all x ∈ R. (Hint: we may use the signs of the trinomials x2 − x + 1 and x2 + x + 1).  c Dr. Ali Ayad & Dr. Ali Fares 31 3) Deduce s(x) for all x ∈ R. 4) Deduce that, for all x ∈ R, −x      x |x| = x E −E. x2 + 1 x2 + 1 Solution 1 1) Let x ∈ R. If x = 0, then f (0) = E(1) = 1. If x 6= 0, then 0 < x2 +1 < 1, so   1 f (x) = E x2 +1 = 0. 2) Since the common discriminant of the trinomials x2 − x + 1 and x2 + x + 1 is ∆ = −3 < 0 and the common coefficient of x2 is 1 > 0, then x2 − x + 1 > 0 and x2 + x + 1 > 0 for all x ∈ R. So x < x2 + 1 and −x < x2 + 1 for all x ∈ R. Hence x −1 < < 1, ∀ x ∈ R. x2 + 1 x  x  If x ≥ 0, then 0 ≤ < 1, so g(x) = E = 0. x2 +1 x2 + 1 x   If x < 0, then −1 < 2 < 0, so g(x) = E x2x+1 = −1. x +1 3) Let x ∈ R. If x = 0, then s(x) = g(0) − g(0) = 0. If x > 0, then s(x) = g(x) − g(−x) = 0 − (−1) = 1. If x < 0, then s(x) = g(x) − g(−x) = −1 − 0 = −1. 4) Let x ∈ R. By the part (3), s(x) is the sign of x. So −x      x |x| = xS(x) = x E −E. 2 x2 + 1 x2 + 1 Exercise 25 Let n ∈ N∗ and x ∈ R. 1) Show that 0 ≤ E(nx) − nE(x) ≤ n − 1. 2) Deduce that   E(nx) E = E(x). n Solution 1) Since E(x) ≤ x < E(x) + 1, then, multiplying by n, we obtain: nE(x) ≤ nx < nE(x) + n. (∗)  c Dr. Ali Ayad & Dr. Ali Fares 32 Since E is an increasing function on R, then   E nE(x) ≤ E(nx).   As nE(x) ∈ Z, then E nE(x) = nE(x). So nE(x) ≤ E(nx). Hence 0 ≤ E(nx) − nE(x). In the other side, as E(nx) ≤ nx and nx < nE(x) + n (by (∗)), then E(nx) < nE(x) + n. So E(nx) − nE(x) < n. But E(nx) − nE(x) ∈ Z, then E(nx) − nE(x) ≤ n − 1. 2) Dividing the double inequality in the part (1) by n, we obtain: E(nx) n−1 0≤ − E(x) ≤ < 1. n n So   E(nx) E − E(x) = 0. n As E(x) ∈ Z, then     E(nx) E(nx) E − E(x) = 0, and therefore E = E(x). 2 n n Exercise 26 1) Let λ ∈ R and A be a nonempty bounded subset of R. Consider the set: λA = {λx, x ∈ A}. a) Show that  λ sup A if λ > 0 sup(λA) = λ inf A if λ < 0 b) Deduce that  λ inf A if λ > 0 inf (λA) = λ sup A if λ < 0 2) Consider the sets −π π       1 sin k A= , n ∈ N and B= , n ∈ N , where k ∈ ,. 2n 2n 2 2 Find sup A, inf A, sup B and inf B. c Dr. Ali Ayad & Dr. Ali Fares 33 Solution 1) a) Since A is a nonempty bounded subset of R, then sup A and inf A exist. As A 6= ∅, then ∃ x0 ∈ A, so λx0 ∈ λA. Hence λA 6= ∅. Case λ > 0: Let y ∈ λA, then ∃ x ∈ A such that y = λx. As x ≤ sup A and λ > 0, then y = λx ≤ λ sup A. So λ sup A is an upper bound of λA. Hence λA is a nonempty and bounded from above subset of R, so sup(λA) exists and sup(λA) ≤ λ sup A. In order to prove the equality, we distinguish the following two methods: 1st method: Let x ∈ A, then λx ∈ λA, so λx ≤ sup(λA). But λ > 0, then sup(λA) x≤. λ sup(λA) So λ is an upper bound of A. Hence sup(λA) sup A ≤ , and therefore λ sup A ≤ sup(λA) λ Thus sup(λA) = λ sup A. 2nd method: Put M = λ sup A. This method uses the supremum criterion. Indeed, let ε > 0, let’s find y ∈ λA such that M − ε < y ≤ M. ε Since λ > 0, then, by the supremum criterion, there exists x ∈ A such that ε sup A − < x ≤ sup A. λ Multiplying this double inequality by λ > 0 and by taking y = λx ∈ λA, we obtain M − ε < y ≤ M. Thus sup(λA) = M = λ sup A. Case λ < 0: Let y ∈ λA, then there exists x ∈ A such that y = λx. As x ≥ inf A and λ < 0, then y = λx ≤ λ inf A. So λ inf A is an upper bound of λA. Hence λA is a nonempty and bounded from above subset of R, so sup(λA) exists and sup(λA) ≤ λ inf A. In order to prove the equality, we distinguish the following two methods: c Dr. Ali Ayad & Dr. Ali Fares 34 1st method: Let x ∈ A, then λx ∈ λA, so λx ≤ sup(λA). But λ < 0, then sup(λA) x≥. λ sup(λA) So λ is a lower bound of A. Hence sup(λA) inf A ≥ , and therefore λ inf A ≤ sup(λA) λ Thus sup(λA) = λ inf A. 2nd method: Put M = λ inf A. This method uses the supremum criterion. Indeed, let ε > 0, let’s find y ∈ λA such that M − ε < y ≤ M. Since − λε > 0, by the infimum criterion, there exists x ∈ A such that ε inf A ≤ x < inf A −. λ Multiplying this double inequality by λ < 0 and by taking y = λx ∈ λA, we obtain M − ε < y ≤ M. Thus sup(λA) = M = λ inf A. b) Let B = λA. 1 Case λ > 0: Put β = − λ < 0. By the part (a), sup(βB) = β inf B and sup(−A) = − inf A, so 1 inf (λA) = inf B = sup(βB) = −λ sup(−A) = λ inf A. β 1 Case λ < 0: Put β = λ < 0. By the part (a), sup(βB) = β inf B, so 1 inf (λA) = inf B = sup(βB) = λ sup A. β 2) For all n ∈ N, 21n ≤ 1. Then 1 is an upper bound de A. Moreover, as 1 = 210 ∈ A, then 1 is the greatest element of A, so sup A = max A = 1. In the other side, for all n ∈ N, 21n ≥ 0, so 0 is a lower bound of A. Hence A is a nonempty and bounded from below subset of R, so inf A exists and inf A ≥ 0. In order to prove the equality, we distinguish the following two methods: 1st method: Suppose that inf A 6= 0, then inf A > 0. By the Archimedean principle, for ε = inf A > 0, there exists n0 ∈ N∗ such that n10 < inf A. As 2n0 > n0 , then 1 1 1 < < inf A with ∈ A, 2n0 n0 2n0 which is impossible. Hence inf A = 0.  c Dr. Ali Ayad & Dr. Ali Fares 35 2nd method: This method uses the infimum criterion. Indeed, let ε > 0, let’s find n0 ∈ N such that 0 ≤ 2n10 < 0 + ε, i.e., 2n0 > 1ε (or n0 > − ln ln ε 2 ). It is sufficient to take   n ln ε o n0 = max 0, E − +1. ln 2 Hence inf A = 0. In the other side, let λ = sin k. Then B = λA. If k = 0, then λ = 0, so B = {0}, and therefore sup B = inf B = 0. If k ∈ 0, π2 , then λ > 0, so, by the part (1),   sup B = λ sup A = λ × 1 = sin k and inf B = λ inf A = λ × 0 = 0.  −π  If k ∈ 2 , 0 , then λ < 0, so, by the part (1), sup B = λ inf A = λ × 0 = 0 and inf B = λ sup A = λ × 1 = sin k. 2 1.6 Density Theorem 1.6.1 ( Q is dense in R) Between any two real numbers, there exists always a rational number. In other words, for any x, y ∈ R such that x < y, there exists r ∈ Q such that x < r < y. Proof Let x, y ∈ R such that x < y, then y − x > 0, then there exists n0 ∈ N∗ such that 0 < n10 < y − x by the corollary 1.4.1.   1st method: By the Archimedean principle, there exists n1 ∈ N∗ such that n1 n10 ≥ y. Let n1 be the smallest integer such that n n1 0 ≥ y. Put r = n1n−1 0 ∈ Q, then r < y (by the choice of n1 ). In the other side, if x ≥ r, then −x ≤ −r, so   n1 1 1 y−x≤y−r = y− + ≤ < y − x, n0 n0 n0 by using the fact that y − n n0 1 ≤ 0, which is impossible. Hence x < r, and therefore x < r < y. 2nd method: Put n1 = E(n0 x), then n1 ≤ n0 x < n1 + 1. So n1 n1 + 1 n1 1 ≤x< = + < x + (y − x) = y. n0 n0 n0 n0 Put r = n1 +1 n0 ∈ Q, then x < r < y. 2  c Dr. Ali Ayad & Dr. Ali Fares 36 Exercise 27 1) Let A be a subset of Z having a supremum. Put M = sup A. (a) Show that M ∈ Z. (b) Deduce that M ∈ A and that max A = sup A. 2) Let a, b ∈ R such that a − b > 1. Show that there exists k ∈ Z such that b < k < a. (Hint: we may use the set A = {n ∈ Z, n < a}). 3) Deduce that between any two real numbers, there exists always a rational number (Theorem 1.6.1). Solution 1) (a) Suppose that M ∈ / Z, then E(M ) < M. Put ε = M − E(M ) > 0. By the supremum criterion, there exists a ∈ A ⊆ Z such that M −ε 1. Consider the set A = {n ∈ Z, n < a}. Since A is a nonempty and bounded from above subset (by a) of R, then A admits a supremum. Put k = sup A. By the part (1), k ∈ A, so k ∈ Z and k < a. In the other side, as E(b) + 1 ∈ Z and E(b) + 1 ≤ b + 1 < a, then E(b) + 1 ∈ A, so E(b) + 1 ≤ sup A = k. Hence b < E(b) + 1 ≤ k. Thus b < k < a. 3) Let x, y ∈ R such that x < y, then y − x > 0. By the Archimedean principle, there exists n ∈ N∗ such that n(y − x) > 1, i.e., ny − nx > 1. Put a = ny and b = nx, then a − b > 1. By the part (2), there exists k ∈ Z such that b < k < a, then nx < k < ny, so x < n k k < y with n ∈ Q. 2 Exercise 28 √ Consider the set of rational numbers between 0 and 2: i √ h n √ o A = 0, 2 ∩ Q = x ∈ Q, 0 < x < 2. Prove that A admits a supremum and an infimum in R to be determined. √ Solution Since 1 ∈ Q and 0 < 1 < 2, then 1 ∈ A, so A 6= ∅. For any x ∈ A, we have √ √ 0 ≤ x ≤ 2. So A is bounded from above by 2 and bounded from below by 0. Hence A is bounded in R, and then A admits a supremum and an infimum in R.  c Dr. Ali Ayad & Dr. Ali Fares 37 Let ε > 0. Let’s find x ∈ A such that √ √ 2 − ε < x ≤ 2. √ √ If √2 − ε < 1 (i.e., ε > 2 − √1), then we take x = 1 ∈ A. If 2 − ε ≥ 1 (i.e., 0 < ε ≤ 2 − 1), then, since Q is dense in R, there exists r ∈ Q such that √ √ 2 − ε < r < 2. √ √ √ x = r ∈ A since 0 < 2 − ε < r < 2. In this case, we take Hence sup A = 2 by the supremum criterion. Let ε > 0. Let’s find x ∈ A such that 0 ≤ x < 0 + ε. If ε > 1, then we take x = 1 ∈ A. If 0 < ε ≤ 1, then, since Q is dense in R, there exists r ∈ Q such that 0 < r < ε. √ In this case, we take x = r ∈ A since 0 < r < ε ≤ 1 < 2. Hence inf A = 0 by the infimum criterion. 2 Theorem 1.6.2 ( R \ Q is dense in R) Between any two real numbers, there exists always an irrational number. In other words, for any x, y ∈ R such that x < y, there exists t ∈ R \ Q such that x ≤ t ≤ y. Proof Let x, y ∈ R such that x < y. Put a = √x ∈ R and b = √y ∈ R, then a < b. 2 2 Since Q is dense in R (theorem 1.6.1), there exists r ∈ Q such that a < r < b. So √ x < r 2 < y. √ √ Put t = r 2, then x < t < y. If t ∈ Q, then 2 = rt ∈ Q, which is impossible, so t ∈ R \ Q. 2 Exercise 29 1) Does the product of two irrational numbers is an irrational number ? 2) Show that if a ∈ R such that a2 is irrational, then a is irrational. Is the converse always true ? π π 3) Deduce that cos( 12 ) and sin( 12 ) are irrational numbers. Solution √ √ √ 1) No. For example, 2 is irrational, but 2 2 = 2 is rational.  c Dr. Ali Ayad & Dr. Ali Fares 38 2) Let a ∈ R such that a2 is irrational. If a ∈ Q, then there exist p ∈ Z, q ∈ N∗ such 2 that a = pq , so a2 = pq2 with p2 ∈ Z and q 2 ∈ N∗. Hence a2 ∈ Q, which is impossible. Thus √ a∈/ Q, i.e., √ a is irrational. The converse is not always true. Indeed, for example, 2∈ / Q, but ( 2)2 = 2 ∈ Q. 3) Since √ 1 + cos( π6 )   2 π 1 3 cos = = + , 12 2 2 4 √ π π / Q, then cos2

and 3∈ 12 ∈ ∈ / Q. So cos / Q by the part (2). In the other side, 12 as     √ 2 π 2 π 1 3 sin = 1 − cos = − , 12 12 2 4 √ / Q, then sin2 12 π π / Q by the part (2). 2

and 3 ∈ ∈ / Q. So sin 12 ∈ Exercise 30 (A little arithmetic) 1) Let m, n ∈ N∗. (a) Let a ∈ Z and b ∈ N∗ such that a and b are relatively prime. Show that if r = ab ∈ Q is a root of the equation xn = m, then a divides m and b = 1. (Hint: we may use Gauss’ lemma: if a, b, c ∈ Z such that a divides bc and a is coprime with b, then a divides c). (b) Deduce that the only possible rational solutions of the equation xn = m are integers. √ 2) Deduce that p is an irrational number for any prime number p. √ √ 3) Show that if p and q are two prime numbers, then p + q is an irrational √ √ number (Hint: we may consider the number p − q). Solution 1) (a) Suppose that r = ab is a root of the equation xn = m. Then r n = m, so an bn = m, and then an = mbn. As a divides an , then a divides mbn. As a is coprime with bn , then a divides m by Gauss’ lemma. In the other side, as b divides mbn , then b divides an , so b is the positive gcd of b and an , therefore b = 1 since b and an are relatively prime. (b) By t