I see it, I demonstrate it, but do I understand it?

Photo by Nick Fewings on Unsplash

Mathematical reasoning is not a simple calculation of which we foresee what it will give and of which we know in advance all the limits; it sometimes generates miracles. We are going to present eight small puzzles and their solutions that you will find unexpected, absurd or implausible.

They all make us discover a fairly simple property which achieves a sort of exploit which is a priori impossible, or which places us in front of an affirmation very far from anything we could imagine.

Miracle 1 — Do you know the root of 2?

Only head, so without using any computer, pencil, paper … or smartphone, find the 200th decimal digit after the decimal point of the number (1 + √2)¹⁰⁰⁰ . Yes, it is possible!

Reply

(1 + √2)¹⁰⁰⁰ + (1 — √2)¹⁰⁰⁰ is an integer, because when we expand, all terms with √2– cancel out, as with (1 + √2–)² + (1 –√2–)² = 1 + 2√2– + 2 + 1–2√2– + 2 = 6.

However (1 –√2)¹⁰⁰⁰ is very small since it is a number lower than 1/2 raised to the power of 1000. Knowing that 210 = 1024, we see that it is lower than 1/2¹⁰⁰⁰ = 1/1024¹⁰⁰ <1 / (10³) 100 = 1/10³⁰⁰.

The number of the statement is therefore the difference between an integer and a number less than 0.000 … 0001 with 300 zeros. All its digits between the decimal point and the 300th digit after the decimal point are therefore 9s.

This riddle comes from a problem column by Elwyn Berlekamp and Joe Buhler that appeared in the Fall 1999 issue of the Emissary newsletter, a publication of msri, the Mathematical Sciences Research Institute at the University of California at Berkeley.

Miracle 2 — The sum of the numbers

The number magician tells you: “Take any whole number made up of increasing digits in the broad sense, and the last two digits of which are different (eg 1,333,456,778). Multiply it by 9. ”She adds,“ I know the sum of the numbers in the result. “ Why ?

Reply

Because the sum of the digits is always 9. With our example: 9  1 333 456 778 = 12 001 111 002. The sum of the digits is indeed 9. This is unexpected, because the starting number can be very long, and so the one obtained by multiplying by 9 also.

The demonstration of this strange and astonishing result is based on the method of subtraction learned in school. Let’s start with a 6-digit integer N (the method extends to any number of digits): N = a1 a2 a3 a4 a5 a6

with a1 ≤ a2 ≤ a3 ≤ a4 ≤ a5 ≤ a6.

The number 9 × N is equal to 10 × N — N. It is therefore the result of the following subtraction:

a1 a2 a3 a4 a5 a6 0
- a1 a2 a3 a4 a5 a6
— — — — — — — –
b1 b2 b3 b4 b5 b6 b7

10 × N
- NOT
— — —
9 × N

The number b7 is 10 — a6, because there is a carry.

The number b6 is equal to a6 + 1 — a5, because the previous carry has been carried over. The hypothesis a6> a5 ensures that there is no new carry to introduce and take into account for the next column.

The number b5 equals a5 — a4, because a5 ≥ a4 ensures that there is no new carry to be introduced. Likewise, b4 is a4 — a3, b3 is a3 — a2, b2 is a2 — a1, and b1 is a1. The sum of the digits of 9N is therefore:

(10 — a6) + (a6 — a5–1) + (a5 — a4) + (a4 — a3) + (a3 — a2) + (a2 — a1) + a1 = 10–1 = 9.

The discovery of this beautiful little result seems due to Felix Lazebnik of the University of Delaware (Mathematics Magazine, vol. 87 (3), pp. 212–221, 2014).

Miracle 3 — An exceptional number?

The magician of numbers wants to amaze you a second time. She places a transparent glass on the table in which a crumpled scarf has been sunk. She gives you a paper and a pencil and prompts you to choose a number N of four digits not all equal (for example, the number 3,333 is not allowed). She asks :

“Sort the digits of N in ascending order, that gives you a four-digit number X; sort the digits of N in descending order, this gives you a four digit Y number. Calculate Z = Y — X. Repeat from Z the same operations. Do this until it becomes unnecessary, the resulting Z giving back Z. “

In a matter of seconds you are doing what was asked and so you have a four-digit Z result. The sorceress then pulls the scarf from the glass and unfolds it. It says 6,174 there. This is indeed the Z you found. Why ?

Reply

Whatever the initial number N, we always end up falling on 6 174! I don’t know any proof by reasoning of this result, but since it is a verifiable property by a finite computation of trying all possible Ns, I wrote a small program and made sure that we actually always get Z = 6,174. This computer proof is a little disappointing, but leaves no room for doubt.

Here is some more information about this quirk of four-digit numbers which today has no deep or general explanation. This exceptional property of the number 6 174 could be the simple fruit of chance.

• No number other than 6,174 falls on itself when the required calculation is carried out.

  • The maximum number of steps of calculation to arrive at 6 174 is 7. This is for example what happens for 1 400, which gives successively 4086, 8 172, 7 443, 3 996, 6 264, 4 176 , 6,174.
  • The average number of calculation steps before reaching 6,174 is 4.7. This number of steps takes all possible values between 0 (for 6 174) and 7. The number XN of four-digit integers requiring N calculation steps is given by the following list: X₀ = 1, X₁ = 356, X₂ = 519, X₃ = 2,124, X₄ = 1,124, X₅ = 1,379, X₆ = 1,508, X₇ = 1,980.

The same miracle occurs when we start from a number of three digits not all equal. This invariably leads to 495, again without being able to do anything but see it. We are in a situation where we know an unexpected result, which we prove by the force of calculation, but which basically we do not understand.

The mystery is all the more profound that, if one tries the same thing starting with numbers of five digits, it does not work any more: all the numbers lead to cycles (it cannot be otherwise), but not necessarily the same.

This formidable property of 6 174 and 495 was discovered by the Indian mathematician Dattatreya Kaprekar (Scripta Mathematica, vol. 15, pp. 244–245, 1949). Details can be found at https://plus.maths.org/content/mysterious-number-6174.

Miracle 4 — The 229 Power of Lenstra

As for the first miracle, you must answer just by reasoning, without using a computer, paper, pencil … or smartphone: knowing that 229 has 9 all different digits, what is the missing digit?

You don’t have to be a prodigy calculator who actually calculates 229. Even if it seems impossible, the information in the statement allows you to find the answer.

Reply

For n varying from 1 to infinity, the number 2n modulo 9 (that is to say by counting as when we do a “proof by 9”, see the box opposite) is successively worth 2, 4, 8, 7 (because 16 gives 1 + 6 = 7), 5 (because 32 gives 3 + 2 = 5), 1 (because 64 gives 10, which gives 1), then again 2, 4, 8, 7, 5 , 1 cyclically. Since 29 = 6 × 5–1, we deduce that 229 modulo 9 is worth 5. We also know that all the digits, except one of them, are present. If they were all present, the value of 229 modulo 9 would be that of the sum 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45, or 0 modulo 9. Only the deletion of 4 leads to a total which equals 5 modulo 9. It is therefore the number 4 which is missing. This can be easily verified, this time with a machine, since 229 = 536 870 912.

The problem is due to the famous Dutch mathematician Hendrik Lenstra, a specialist in integer factorization algorithms.

Miracle 5 — An Incredible Strategy

You have in front of you two envelopes containing two different amounts of money. You open one and look at its contents. Now you have to choose either to keep the contents of the envelope you have opened or to take the other envelope without knowing the contents. There is a strategy that allows you to make the right choice, that is, to leave with the envelope containing the most, in more than 50% of the cases.

It seems absurd since you will only open one envelope and know nothing about the other. The astonishing thing is that such a strategy does exist!

Reply

One way to do this is to:

1) Uniformly choose a real number x between 0 and 1. For example, we spin a lottery wheel with a perimeter of 1 meter and which has a mark on this perimeter; once the wheel has stopped, the distance x between the mark and the stopping point of the wheel is measured on the perimeter.

2) Calculate y = x / (1 — x), which will be a number between 0 and infinity.

3) If the open envelope contains more than y euros, keep it; if not, take the other.

To demonstrate that this method gives more than a 50% chance of making the right choice, suppose that the envelopes contain the sums a and b respectively, with a <b, and calculate the probability of taking the correct envelope (the one that contains b) . Consider two cases.

Case 1. The number y is between a and b.

We will be sure to make the right choice, because if we took the right envelope, we will not change it, and if we took the wrong one, we will change it.

Using the equivalence of the relations y = x / (1 — x) and x = y / (y + 1), we know that the number y is between a and b if and only if x is between a / ( a + 1) and b / (b + 1). Since x is drawn with a uniform probability between 0 and 1, the probability c that x is between a / (a ​​+ 1) and b / (b + 1) is the length of the interval [a / (a ​​+ 1 )), b / (b + 1)]. In other words, c = b / (b + 1) — a / (a ​​+ 1).

Case 2. The number y is not between a and b.

The probability of having made the right choice is then equal to 1/2. Indeed :

- if a and b are both less than y, then we will have chosen an envelope at random and we will change it, which will lead to an envelope chosen at random, which will therefore be the best with a probability equal to 1/2.

- if a and b are both greater than y, then we will have chosen an envelope at random and we will not change it. As before, we will still have made the right choice with a probability equal to 1/2.

The probability of making the right choice when y is not between a and b is therefore 1/2. As the probability for y of not being in the interval [a, b] is equal to 1 — c, by combining the two cases we find that the probability of making the right choice is c + (1 — c) / 2 = 1/2 + c / 2 = 1/2 + (b / (b + 1) — a / (a ​​+ 1)) / 2, which is strictly greater than 1/2.

For a = 10 and b = 100, for example, the probability of keeping the best envelope is equal to:

1/2 + 1/2 × (100/101–10/11) = 54.05%.

This disturbing miracle seems to have been discovered several times in various forms. It is found mentioned, for example, by the American mathematician David Blackwell (1919–2010) and by the American information theorist and probability specialist Thomas Cover (1938–2012).

Miracle 6 — Will the beetles fall?

A family of 20 beetles sits on a wooden ruler in a horizontal position, the length of which is exactly 1 meter. They are placed randomly on the ruler, each facing right or left, at random. Each beetle advances at 1 centimeter per second. When two beetles meet, they both turn around. It is assumed that the U-turn is instantaneous and that the beetles are of negligible length. When a beetle gets to one end of the ruler, it falls to the ground.

Without knowing any more precise data, and this is the miracle, we can be sure that all the beetles will be on the ground in less than an hour, and that it is still true if, instead of 20 beetles, there are has 30 or 100. Why? The answer is given in the box above.

Miracle 7 — The Universal String Theorem

Some math results are totally baffling and seem almost absurd. This is the case with the “universal string theorem” of which I admit that, even after having read the proof, I cannot really understand the reason for it. We present it in the form of a riddle.

We consider a continuous function f of the interval [0, 1] in the set of real numbers, such that f (0) = f (1). For example f (x) = x (1 — x). We call a horizontal chord of f of length L, any horizontal segment of length L connecting two points of the graph of f. The abscissas of these two points are x and x + L and we have f (x) = f (x + L).

Depending on the function, some lengths of horizontal ropes are possible and others are not. Thus, for f (x) = x (1 — x) or sin (πx), all lengths L between 0 and 1 are possible (of course, for the functions f as defined above, we are certain that they have a chord of length 1 because, by hypothesis, f (0) = f (1)). But for the function h (x) = sin (2πx), there is no horizontal chord of length strictly between 1/2 and 1. Other lengths L than length 1 are always possible whatever the continuous function. f satisfying f (0) = f (1). Which ones?

Reply

The result is astounding: for any positive integer n, and whatever the function f verifying the hypotheses, there will be a horizontal chord of length L = 1 / n. Moreover, for the numbers L which are not inverses of integers, there exist functions satisfying the hypotheses and having no chord of length L.

The positive part of the theorem was discovered in 1806 by the physicist André-Marie Ampère. For the negative part, we had to wait for the French mathematician Paul Lévy (1886–1971), who demonstrated it in 1934.

Let’s prove the positive part. We consider, whatever the integer n> 1, the function g (x) = f (x + 1 / n) — f (x), which is defined for any x between 0 and 1–1 / n. We have :

g (0) + g (1 / n) + g (2 / n) + … + g (1–1 / n) = f (0) — f (1) = 0. We deduce that the function g cannot be strictly positive everywhere. Nor can it be strictly negative everywhere. According to the intermediate value theorem (according to which a continuous function which is equal to y at one point and z at another passes through all the values ​​between y and z), g therefore takes the value 0. In other words, there exists at minus a value x₀ such that g (x₀) = 0, i.e. f (x₀) = f (x₀ + 1 / n), and there is therefore a horizontal chord of length 1 / n.

Paul Lévy’s function, which shows that these are the only chord lengths required, is gL (x) = sin2 (πx / L) — x sin2 (π / L), which has no chord of length L if L is not the inverse of an integer. The study that shows this is not very easy, but by having your computer draw the function gL (x) — gL (x + L) for x between 1 and 1 — L, you will see that it does not is never zero.

A discussion of this little-known theorem can be found in the book A Primer of Real Functions, by Ralf Boas (4th edition, The Mathematical Association of America, 1996).

Miracle 8 — Period 3 involves chaos

Let’s finish, this time without giving any demonstrations because they are too complicated, with an incredible result on the cycles of continuous functions.

We consider a continuous function from [0, 1] to [0, 1]. For any x between 0 and 1, we consider the following numerical sequence: x, f (x), f (f (x)), f (f (f (x))), …, f (f (. .. (x) …)).

If this sequence falls on the initial number x, we say that we have a cycle. The smallest integer for which this occurs is the “order” of the cycle. For example, the function f (x) = 1 — x is such that any x  1/2 generates a cycle of order 2: starting from x = 1/3, we have the cycle 1/3 → 2/3 → 1/3.

We have the following astonishing property, discovered several times, in particular by Tien-Yien Li and James Yorke (“Period three implies chaos”, The American Mathematical Monthly, vol. 82 (10), pp. 985–992, 1975): if fa has a cycle of order 3, then it has cycles of any order. But the reverse is false: for all k ≠ 3, there exist functions having a cycle of order k and not having a cycle of order 3.

The theorem was made more precise in 1964 by the Ukrainian mathematician Oleksandr Charkovski, who discovered a very subtle and totally unexpected relationship of necessity between the orders of cycles. Order 3 implies order 5 which implies 7, which implies 9, etc. But also, the 6 implies the 10, which implies the 14, etc. This system of involvement is defined by the table above which must be read line by line, from top to bottom.

This table cannot be simplified, because there are for example functions having cycles of order 10 (therefore 14, 18, etc.), but no cycles of order 6, 3, 5, 7, … for details, see Chung-Wu Ho and Charles Morris, “A graph-theoretic proof of Sharkovsky’s theorem on the periodic points of continuous functions”, Pacific Journal of Mathematics, vol. 96 (2), pp. 361–370, 1981 (http://bit.ly/2haC9LH) and the article Sharkovskii’s theorem on Wikipedia.

Who would have imagined that the mere fact of being a continuous function of an interval within itself hid this mysterious painting?

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