Download PUTNAM TRAINING MATHEMATICAL INDUCTION and more Lecture notes Calculus in PDF only on Docsity!
PUTNAM TRAINING
MATHEMATICAL INDUCTION
(Last updated: December 7, 2021)
Remark. This is a list of exercises on mathematical induction. —Miguel A. Lerma
Exercises
- Prove that n! > 2 n^ for all n ≥ 4.
- Prove that for any integer n ≥ 1, 2^2 n^ − 1 is divisible by 3.
- Let a and b two distinct integers, and n any positive integer. Prove that an^ − bn^ is divisible by a − b.
- The Fibonacci sequence 0, 1 , 1 , 2 , 3 , 5 , 8 , 13 ,... is defined as a sequence whose two first terms are F 0 = 0, F 1 = 1 and each subsequent term is the sum of the two previous ones: Fn = Fn− 1 + Fn− 2 (for n ≥ 2). Prove that Fn < 2 n^ for every n ≥ 0.
- Let r be a number such that r +1/r is an integer. Prove that for every positive integer n, rn^ + 1/rn^ is an integer.
- Find the maximum number R(n) of regions in which the plane can be divided by n straight lines.
- We divide the plane into regions using straight lines. Prove that those regions can be colored with two colors so that no two regions that share a boundary have the same color.
- A great circle is a circle drawn on a sphere that is an “equator”, i.e., its center is also the center of the sphere. There are n great circles on a sphere, no three of which meet at any point. They divide the sphere into how many regions?
- We need to put n cents of stamps on an envelop, but we have only (an unlimited supply of) 5/c and 12/c stamps. Prove that we can perform the task if n ≥ 44.
- A chessboard is a 8 × 8 grid (64 squares arranged in 8 rows and 8 columns), but here we will call “chessboard” any m × m square grid. We call defective a chessboard if one of its squares is missing. Prove that any 2n^ × 2 n^ (n ≥ 1) defective chessboard can be tiled (completely covered without overlapping) with L-shaped trominos occupying exactly 3 squares, like this. 1
- This is a modified version of the game of Nim (in the following we assume that there is an unlimited supply of chips). Two players arrange several piles of chips in a row. By turns each of them takes one chip from one of the piles and adds at will as many chips as he or she wishes to piles placed to the left of the pile from which the chip was taken. Assuming that the game ever finishes, the player that takes the last chip wins. Prove that, no matter how they play, the game will eventually end after finitely many steps.
- Call an integer square-full if each of its prime factors occurs to a second power (at least). Prove that there are infinitely many pairs of consecutive square-fulls.
- Prove that for every n ≥ 2, the expansion of (1 + x + x^2 )n^ contains at least one even coefficient.
- We define recursively the Ulam numbers by setting u 1 = 1, u 2 = 2, and for each subsequent integer n, we set n equal to the next Ulam number if it can be written uniquely as the sum of two different Ulam numbers; e.g.: u 3 = 3, u 4 = 4, u 5 = 6, etc. Prove that there are infinitely many Ulam numbers.
- Prove Bernoulli’s inequality, which states that if x > −1, x 6 = 0 and n is a positive integer greater than 1, then (1 + x)n^ > 1 + nx.
- Prove that two consecutive Fibonacci numbers are always relatively prime.
- Let an be the following expression with n nested radicals:
an =
Prove that an = 2 cos (^2) nπ+.
Solutions
- We prove it by induction. The basis step corresponds to n = 4, and in this case certainly we have 4! > 24 (24 > 16). Next, for the induction step, assume the inequality holds for some value of n ≥ 4, i.e., we assume n! > 2 n, and look at what happens for n + 1: (n + 1)! = n! (n + 1) > 2 n(n + 1) > 2 n^ · 2 = 2n+^. 6 by induction hypothesis Hence the inequality also holds for n + 1. Consequently it holds for every n ≥ 4.
- For the basis step, we have that for n = 1 indeed 2^2 ·^1 − 1 = 4 − 1 = 3 is divisible by
- Next, for the inductive step, assume that n ≥ 1 and 2^2 n^ − 1 is divisible by 3. We must prove that 22(n+1)^ − 1 is also divisible by 3. We have 2 2(n+1)^ − 1 = 2^2 n+2^ − 1 = 4 · 22 n^ − 1 = 3 · 22 n^ + (2^2 n^ − 1). In the last expression the last term is divisible by 3 by induction hypothesis, and the first term is also a multiple of 3, so the whole expression is divisible by 3 and we are done.
- By induction. For n = 1 we have that a^1 − b^1 = a − b is indeed divisible by a − b. Next, for the inductive step, assume that an^ − bn^ is divisible by a − b. We must prove that an+1^ − bn+1^ is also divisible by a − b. In fact: an+1^ − bn+1^ = (a − b) an^ + b (an^ − bn). On the right hand side the first term is a multiple of a − b, and the second term is divisible by a − b by induction hypothesis, so the whole expression is divisible by a − b.
- We prove it by strong induction. First we notice that the result is true for n = 0 (F 0 = 0 < 1 = 2^0 ), and n = 1 (F 1 = 1 < 2 = 2^1 ). Next, for the inductive step, assume that n ≥ 1 and assume that the claim is true, i.e. Fk < 2 k, for every k such that 0 ≤ k ≤ n. Then we must prove that the result is also true for n + 1. In fact: Fn+1 = Fn + Fn− 1 < 2 n^ + 2n−^1 < 2 n^ + 2n^ = 2n+1^ , 6 by induction hypothesis and we are done.
- We prove it by induction. For n = 1 the expression is indeed an integer. For n = 2 we have that r^2 + 1/r^2 = (r + 1/r)^2 − 2 is also an integer. Next assume that n > 2 and that the expression is an integer for n − 1 and n. Then we have ( rn+1^ +
rn+
rn^ +
rn
r +
r
rn−^1 +
rn−^1
hence the expression is also an integer for n + 1.
- By experimentation we easily find:
1
2 1
2
3
4
1
2 3
4
5 6 7
1
2 3
4
5 7 8 6 9 10 11
Figure 1. Plane regions.
n 1 2 3 4... R(n) 2 4 7 11... A formula that fits the first few cases is R(n) = (n^2 + n + 2)/2. We will prove by induction that it works for all n ≥ 1. For n = 1 we have R(1) = 2 = (1^2 + 1 + 2)/2, which is correct. Next assume that the property is true for some positive integer n, i.e.: R(n) =
n^2 + n + 2 2
We must prove that it is also true for n + 1, i.e.,
R(n + 1) =
(n + 1)^2 + (n + 1) + 2 2
n^2 + 3n + 4 2
So lets look at what happens when we introduce the (n + 1)th straight line. In general this line will intersect the other n lines in n different intersection points, and it will be divided into n + 1 segments by those intersection points. Each of those n + 1 segments divides a previous region into two regions, so the number of regions increases by n + 1. Hence: R(n + 1) = S(n) + n + 1. But by induction hypothesis, R(n) = (n^2 + n + 2)/2, hence:
R(n + 1) =
n^2 + n + 2 2
n^2 + 3n + 4 2
QED.
- We prove it by induction in the number n of lines. For n = 1 we will have two regions, and we can color them with just two colors, say one in red and the other one in blue. Next assume that the regions obtained after dividing the plane with n lines can always be colored with two colors, red and blue, so that no two regions that share a boundary have the same color. We need to prove that such kind of coloring is also possible after dividing the plane with n + 1 lines. So assume that the plane divided by n lines has been colored in the desired way. After we introduce the (n + 1)th line we need to recolor the plane to make sure that the new coloring still verifies that no
-
2 n+
6
?
2 n+ 6
?
2 n
- 2 n
Figure 2. A 2n+1^ × 2 n+1^ defective chessboard.
first n piles, since by induction hypothesis the game with n piles eventually ends. So sooner or later one player must take a chip from the (n + 1)th pile. It does not matter how many tokes he or she adds to the other n piles after that, it is still true that the players cannot keep taking chips only from the first n piles forever, so eventually someone will take another chip from the (n + 1)th pile. Consequently, the number of chips in that pile will continue decreasing until it is empty. After that we will have only n piles left, and by induction hypothesis the game will end in finitely many steps after that.^1
- The numbers 8 and 9 are a pair of consecutive square-fulls. Next, if n and n + 1 are square-full, so are 4n(n + 1) and 4n(n + 1) + 1 = (2n + 1)^2.
- For n = 2, 3 , 4 , 5 , 6 we have: (1 + x + x^2 )^2 = 1 + 2x + 3x^2 + 2x^3 + x^4 (1 + x + x^2 )^3 = 1 + 3x + 6x^2 + 7x^3 + · · · (1 + x + x^2 )^4 = 1 + 4x + 10x^2 + 16x^3 + · · · (1 + x + x^2 )^5 = 1 + 5x + 15x^2 + 30x^3 + · · · (1 + x + x^2 )^6 = 1 + 6x + 21x^2 + 50x^3 + · · · In general, if (1 + x + x^2 )n^ = a + bx + cx^2 + dx^3 + · · · , then (1 + x + x^2 )n+1^ = a + (a + b)x + (a + b + c)x^2 + (b + c + d)x^3 + · · · , (^1) An alternate proof based on properties of ordinal numbers is as follows (requires some advanced set-
theoretical knowledge.) Here ω = first infinite ordinal number, i.e., the first ordinal after the sequence of natural numbers 0, 1 , 2 , 3 ,.... Let the ordinal number α = a 0 + a 1 ω + a 2 ω^2 + · · · + an− 1 ωn−^1 represent a configuration of n piles with a 0 , a 1 ,... , an− 1 chips respectively (read from left to right). After a move the ordinal number representing the configuration of chips always decreases. Every decreasing sequence of ordinals numbers is finite. Hence the result.
hence the first four coefficients of (1 + x + x^2 )n+1^ depend only on the first four coeffi- cients of (1 + x + x^2 )n. The same is true is we write the coefficients modulo 2, i.e., as “0” if they are even, or “1” if they are odd. So, if we call qn(x) = (1 + x + x^2 )n^ with the coefficients written modulo 2, we have
q 1 (x) = 1 + 1x + 1x^2 q 2 (x) = 1 + 0x + 1x^2 + 0x^3 + 1x^4 q 3 (x) = 1 + 1x + 0x^2 + 1x^3 + · · · q 4 (x) = 1 + 0x + 0x^2 + 0x^3 + · · · q 5 (x) = 1 + 1x + 1x^2 + 0x^3 + · · · q 6 (x) = 1 + 0x + 1x^2 + 0x^3 + · · · We notice that the first four coefficients of q 6 (x) coincide with those of q 2 (x), and since these first four coefficients determine the first four coefficients of each subsequent polynomial of the sequence, they will repeat periodically so that those of qn(x) will always coincide with those of qn+4. Since for n = 2, 3 , 4 , 5 at least one of the first four coefficients of qn(x) is 0 (equivalently, at least one of the first four coefficients of (1 + x + x^2 )n^ is even), the same will hold for all subsequent values of n.
- Let Um = {u 1 , u 2 ,... , um} (m ≥ 2) be the first m Ulam numbers (written in increasing order). Let Sm be the set of integers greater than um that can be written uniquely as the sum of two different Ulam numbers from Um. The next Ulam number um+ is precisely the minimum element of Sm, unless Sm is empty, but it is not because um− 1 + um ∈ Sm.
- By induction. For the base case n = 2 the inequality is (1 + x)^2 > 1 + 2x, obviously true because (1 + x)^2 − (1 + 2x) = x^2 > 0. For the induction step, assume that the inequality is true for n, i.e., (1 + x)n^ > 1 + nx. Then, for n + 1 we have
(1 + x)n+1^ = (1 + x)n(1 + x) > (1 + nx)(1 + x) = 1 + (n + 1)x + x^2 > 1 + (n + 1)x , and the inequality is also true for n + 1.
- This can be proved easily by induction. Base case: F 1 = 1 and F 2 = 1 are in fact relatively prime. Induction Step: we must prove that if Fn and Fn+1 are relatively prime then so are Fn+1 and Fn+2. But this follows from the recursive definition of the Fibonacci sequence: Fn + Fn+1 = Fn+2; any common factor of Fn+1 and Fn+2 would be also a factor of Fn, and consequently it would be a common factor of Fn and Fn+ (which by induction hypothesis are relatively prime.)
- Note that an can be defined recursively like this: a 1 =
2, and an+1 =
2 + an for n ≥ 1. We proceed by induction. For n = 1 we have in fact a 1 =
2, and 2 cos π 4 = 2 · √^12 =
