Outside the high school program / Logic and paradox 13 exercises (including 2 corrected)

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AB 12 1. Continuity axiom E.2285 Consider a point X moving from A to B describing the entire segment. We break down the point’s path as follows : We’ll say that this point arrives at the first step, when it has traveled half of [ AB ] : we’ll note X 1 this point. It will be at the second stage when it has covered half of the remaining path, i.e. half of [ X 1 B ] : this point will be noted X 2 . The third step will be half [ X 2 B ] : half the distance restante. . . And so on. Consider X 0 as the starting point A . 1 Place the points X 1 , X 2 , X 3 and X 4 on the straight line below. 2 It will now be assumed that distance AB measures 1 me-ter. a Give the distance X 0 X 1 , X 1 X 2 , X 2 X 3 et X 3 X 4 . b By extrapolation, give the measurements of X 4 X 5 , X 5 X 6 , X 6 X 7 and X 7 X 8 . c For n ∈ N , make a guess as to the distance X n − 1 X n . 3 We propose to study the infinite sum of terms : S = 1 2 + 1 4 + 1 8 + 1 16 + · · · To do this, for n ∈ N , consider the sum to n terms : S n = 1 2 + 1 4 + 1 8 + · · · + 1 2 n a Write, without calculating, the sums S 3 , S 4 and S 5 . b Establish the formula : S n +1 = 1 2 + 1 2 · S n c Give a reason why the numbers S n approach the value 1 when the number of steps becomes very large. We’ll say that S n admits a limit when n tends to + ∞ . d Noting S the limit value of S n when n tends to + ∞ : we’ll note lim n ↦→ + ∞ S n = S . By passing to the limit of the formula 2 , we obtain the equality: S = 1 2 + 1 2 · S . Find the value of S . We’ve just highlighted one of the properties of the continuum hypothesis : Any segment can be cut into an infinity of segments all of non-zero length. E.2286 Definition: (from Le Petit Robert) dichotomy: Division, binary subdivision (between two elements that are clearly separated and opposed) Assuming the thesis of continuity, any object is infinitely di-visible. We know that √ 2 ∈ [1 ; 2] because : 1 2 < 2 2 < 2 2 Two numbers and their squares are arranged in the same order. 1 < 2 < 2 Implementation of the algorithm: At step 0 (the initial step) , we know that 2 is contained in the interval [1 ; 2] To move on to the next step, we consider the middle of this interval 1 ; 5 . Then, we compare 1 ; 5 and 2 to deter-mine whether 2 is contained in the interval [1 ; 1 ; 5] or the interval [1 ; 5 ; 2] . This gives us a new interval corresponding to step 1 . We repeat this process an infinite number of times. Let [ a n ; b n ] be the interval corresponding to step n and u n its length. 1 a Give the value of u 0 , u 1 , and u 2 . b Place points a 0 , b 0 , a 1 , b 2 , a 3 , and b 4 on the graduated line below. 2 a Give the value of u n based on n . b Give a range for b n − a n for n =10 . Deduce the accuracy of the approximation of 2 by u 10 3 What happens to the value of u n when n approaches + ∞ ? (We say that n tends toward infinity.) . In other words, what is the value of lim n ↦→ + ∞ u n ? https://chingmath.fr sacados/2285 AB sacados/2286 12
A0T0tt0 A1T1tt1 tt2 E.2287 Achilles and the tortoise (paradox of Zeno of Elea) If the tortoise has a head start on Achilles, Achilles can never catch up, no matter how fast he runs ; as Achilles runs to reach the point from which the tor-toise started, the tortoise advances in such a way that Achilles will never be able to cancel this advance It is assumed that the tortoise is one meter ahead, that Achilles runs at a speed of 2 m = s and the tortoise at a speed of 1 m = s . We note t 0 the starting instant, A 0 Achilles’ initial position and T 0 that of the tortoise. t 1 is the instant when Achilles reaches the Turtle’s previous position. Let’s note A 1 this position, the Turtle continuing forward is in T 1 . And so on. Based on the fact that Achilles’ and the tortoise’s journey can be cut into as many ends (assumption of continuity) , we be-lieve that Achilles will never catch up with the tortoise since he will always have part of the way to go. What’s more, all these little segments, put end to end, add up to what? In other words, Achilles will catch up with the tortoise ad infini-tum? 1 Place on the last graduated straight line the point A 2 and T 2 . 2 Give the values of T 0 − A 0 , T 1 − A 1 , T 2 − A 2 . 3 We accept the fact that, at any study rank, we have : u n = T n − A n = 1 2 n . We’re going to calculate how far from the Achilles point the participants will meet. We note : S n = u 0 + u 1 + u 2 + · · · + u n a Expand : 1 − 1 2 × S n b Deduct : S n = 1 − 1 2 n +1 1 − 1 2 c Give the limit of S n when n tends to infinity which we note : lim n ↦→ + ∞ u 0 + u 1 + · · · + u n ou lim n ↦→ + ∞ + ∞ k =0 u n Note: Zeno’s paradoxes were reported by Aristotle in his book on physics (Book VI) We could also have considered the two functions that to time associates the position of Achilles and the tortoise on the straight line and determine the point of intersec-tion. E.2284 The continuum hypothesis assumes that any part of the line can be cut indefinitely and that each part has a non-zero length. This assumption runs counter to the conception of matter as composed of atoms ; In quantum theory, light is seen as a particle (a photon) , but also as a continuous vibratory phe-nomenon (a wave) . Infinitesimal calculus is the branch of mathematics that stud-ies the infinitely far and the infinitely near. (paradox of Zeno of Elea) ˇThere is no motion, as the mobile must reach the middle of its path before reaching finı This paradox is supposed to defeat continuity: you can’t di-vide a length indefinitely. The Concept of continuity is the opposite of the discrete (atomistic theory) . 2. Unclassified financial years E.334 Consider an integer X formed by four digits : X = x 3 x 2 x 1 x 0 . For example, 1579 verifies : x 3 = 1 ; x 2 = 5 ; x 1 = 7 ; x 0 = 9 1 Which of the following two sentences is true? If x 0 is zero then X is divisible by 5. If X is divisible by 5 then x 0 is zéro 2 Which of the following two sentences is true? For x 0 to be 0, it is necessary for X to be divisible by 5. For x 0 to be 0. it is sufficient that X is divisible by 5. We then say (delete as appropriate) : ˇ X is divisible by 5 ı is a condition Nécessaire Suffisante so that ˇ x 4 is worth 0 ı. 3 Which of the following two sentences is true? For X to be divisible by 5, it is necessary that x 0 be 0. For X to be divisible by 5, it is sufficient that x 0 is 0. We then say (delete as appropriate) : ˇ x 0 is 0 ı is a condition Nécessaire Suffisante so that ˇ X is divisible by 5. ı https://chingmath.fr sacados/2287 A0T0tt0 A1T1tt1 tt2 sacados/2284 chapExoCorrec/334 sacados/334
4 We consider When P is both a necessary and a sufficient condi-tion for Q , we’ll say that these two properties are equivalent and we’ll note : P ⇐⇒ Q Complete the dotted lines to obtain the following equiv-alence : ˇ X is divisible by 5 ı is equivalent to ˇ x 0 is . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ı E.301 The sentence ˇ I’ll stick you if, and only if, you talk ı breaks down into two sentences : I’ll stick you if you talk I’ll stick you only if you talk Translate each of these two sentences into a sentence written in the form ˇ If . . . Then . . . ı (a change of tense may be useful for translation) E.1868 For each of the following pairs of state-ments, say whether : A is required for B A is sufficient for B A is equivalent to B A B 1 The triangle ABC is right-angled AB 2 = AC 2 + BC 2 2 2 x + 5 = 3 x = − 1 3 −→ AI = −→ IB I middle of [ AB ] 4 x 0 x 3 5 a = 5 a 2 = 25 E.1888 P ⇒ Q reads ˇ P implies Q ı Means that if we have the property P is true then Q will be true. Its reciprocal is Q ⇒ P 1 Here are some properties representing well-known theo-rems. Give the names of these theorems : a AB 2 + CB 2 = AC 2 = ⇒ ABC is rectangular at A . b ( CB ) == ( MN ) M ∈ [ AB ] N ∈ [ AC ] = ⇒ AB AM = AC AN = BC MN c M middle of [ AB ] N ∈ [ AC ] ( MN ) == ( BC ) = ⇒ N middle of [ AC ] 2 Say whether the reciprocal of these theorems is true E.1968 1 Place 4 points A , B , C and D in the plane. 2 Consider the point M verifying the relation: −−→ AM = 3 · −→ AC + 5 · −−→ CB + 2 · −−→ BD a Show that : −−→ AM =3 ·−−→ AB +2 ·−−→ CD b Place the point M . 3 Let N be the point in the plane verifying the relation: −−→ AN = 5 · −→ AC + 2 · −−→ CB − 3 · −−→ AD a Express the vector −−→ AN in terms only of the vectors −−→ AB and −−→ CD . b Place the point N in the plane. E.2728 Answer the following questions with true or false : a Si x < 2 then x < 3 b Si x < 3 then x < 2 c Si x 3 then x < 3 d Si x < 3 then x 3 e Si x =2 then 2 x +3=7 f Si 2 x + 3 = 7 then x = 2 g Si 2 x − 5 then x < 3 h Si x < 3 then 2 x − 5 < 2 E.36 1 Noting that 10 x =9 x + x , show that the property ˇ 10 n +1 is multiple of 9 ı, depending on n , is hereditary. 2 For all that, justify that this property is always false? E.114 1 Two square-shaped rice cakes are available: the first has side 10 cm and the second 15 cm . We want to make two wafers with the same side whose sum of their areas is equal to the sum of the first two wafers. What is this length? 2 Suppose now that the two patties are respectively for length of their sides a and b . Establish then that an av-erage wafer will have side : a 2 + b 2 2 https://chingmath.fr chapExoCorrec/301 sacados/301 sacados/1868 sacados/1888 A faire et finir sacados/1968 sacados/2728 Extrait du manuel "Ressourcres pour la classe de seconde" - dgesco 2009 sacados/36 sacados/114
9AB678345012 E.2012 A recycling company recovers a batch of digicodes, each with a keypad identical to the one shown opposite. Each of these digicodes has been programmed to operate with a code made up of two signs chosen from the twelve on this keypad. For example A 0 , BB , 43 are possible codes. To restore such a digicode to working order, you need to find its code. To facilitate such a search, a number R has been inscribed on the casing of each digicode, depending on the code. This number was obtained as follows : The code is considered as a number written in base 12 : A is the digit ten and B the digit 11. The number R inscribed on the case is the remainder of the Euclidean division of the code, converted to base 10, by 53. R is therefore a number written in base 10 and such that 0 R 53 . 1 How many possible codes are there? 2 It is assumed that the code for a digicode is AB . a Write in base 10 the number whose writing in base 12 and ( AB ) douze . b Determine the number R inscribed on the housing of this digicode. 3 On the housing of a digicode is inscribed the number R equal to 25. Show that (21) douze can be the code for this digicode. 4 Consider the function f , taken from an algorithm, where the argument R is a natural number: Function f(R) L ← empty list n ← 0 As long as 53n+R 143 Add the value of 53n+R at the end of the list L n ← n+1 End As long as Return L a What list is returned by the function f when the func-tion f is called with the value R =25 as argument? b It is assumed that the number R written on the case of a digicode is 25 . What are the three possible codes for this digicode? 5 Say whether the following statement is true or false. If the assertion is considered false, provide proof. Assertion : whatever the value of R the function f finds three codes among which is the secret code. https://chingmath.fr sacados/2012 9AB678345012