That is, we claim that. /Filter /FlateDecode 3. Prove by induction: For every n>=1, 2 | f3n ( i.e. For every integer n >= 1, is divisible by at least n distinct primes. Another form of Mathematical Induction is the so-called Strong Induction described below. (12) Use induction to prove that n 3 − 7n + 3, is divisible by 3, for all natural numbers n. Solution (13) Use induction to prove that 10 n + 3 × 4 n+2 + 5, is divisible by 9, for all natural numbers n. Solution. Proof. We first note that for n=1, this just says that 8 | 8 which is clearly true. Here we prove not just that Sn is a perfect square, but that it is a particular perfect square, namely n2. 9k -1 = 8t for some integer t. We claim that the result is true for the next larger integer, k+1. (fg)' = f'g + fg'] you wanted to prove to someone that for every integer n >= 1, the derivative of is . More generally, For every k >= 1, and n >= 1, if .                         Example Induction Proofs. Suppose that having just learned the product rule for derivatives [i.e. The left-hand side of this expression is an assertion. Then if we show both of (i) and (ii) below, then P(n) is true for all n >= 1. Does the series converge or diverge? For n=1 this just says that 7 | 7 which is true. For n=1, the expression has the value . x2n = f2n-1 - xf2n. Let fn be the nth Fibonacci number, Now the result is easily seen to be true in the case k=1, since 1 is a perfect square. Prove, using induction, that all binomial coefficients are integers. Proof: For n=1 this asserts that - which is certainly true. An integer n is a perfect square if it is the square of some other integer. stream It's true for n=1, that's pretty clear. Then P(n) is true for all integers n >= 1. Now suppose I wanted to prove the following theorem. We will be finished if we can show that . In this exercise we will determine some highly interesting properties of the Fibonacci numbers. xڭX�r7}�WlߤI���3�L��Sg�$Mԙt?��&�Ė]����{㚫K���� 88 x:�|2 VJ�M>fL�9dZjb��&�����a��@`4�L��K �݋�v1�n��b�-?�����ef����9�…�F��1Rec� �&X��_-�%�z�f�a���yx���%_�J��%(jyF��*�D�� �O*��_\��0|s5'P�}�Z����G~�13���b��9�����t�}\��O��۰XL7�մzp:�wN��h�t�J�2h�s��T����)b��$���"����CShT`0���ԈX @�7b(�)����f���֛*��Q7"qOcD�%t���b]\ߖ@�v^�֜���Hi��w� ��x/ ��!�ƛ���1"�;j!�=zD$�� And you see we can keep on going this way - do you see the pattern? Suppose that m is a fixed integer and . 2 4. Since and since multiplying corresponding sides of congruences is still a congruence, . Show that if x is the root of 1- x - x2 so that x2= 1 - x, then for every integer n >= 1, Once we show this we will be finished. We argue by induction. Then, f3(k+1) = f3k+3 = f3k+2 +f3k+1 = f3k+1 + f3k + f3k+1 = 2f3k+1 + f3k = 2(f3k+1 + m ). Proof by Induction. The Principle of Mathematical Induction. 6. Consequently, we know that the sequence must converge to some limit t. Problem: Find the exact value of t. Note that by (A) your answer should be a number somewhere between 1 and 2. Now that I know it's true for n=5, I can show you that it is true for n=6 like this: and so the formula works for n=6, too. Hard to argue with that! This simple observation can serve as the foundation of some very significant mathematical arguments. So if r >0, then r must belong to A(n, m). Definition. The proof isn't going anywhere.       (a). We will argue by induction(1). Let n and m be positive integers, and let Well, we seem to be stuck. For appropriate values of n and k. It is a useful exercise to prove the recursion relation (you don’t need induction). 3. 2. The Fibonacci numbers are defined by the recurrence relation. And now the induction proof works! Don't say, "assume that the result holds for all n," or anything equivalent to it. The Well-Ordering Principle simply states that every non-empty subset of the positive integers has a smallest element. And now we might say that it is fairly evident that we can continue this process and so it is true that for every integer n >= 1.       r = n - qd = n - q(an + bm) = (qa+1)n - (qb)m. Let f n be the nth Fibonacci number, Prove by induction: For every n>=1, 2 | f 3n ( i.e. 5. 8. Hence the result follows by induction. Our argument would be almost the same as before except that at the very end. 6. Some Induction Exercises. Now assume that for some integer k, . How might you go about doing this? We will argue by induction. E. Show that satisfies the conditions of the lemma in (D). We claim that . 43. But, 9k+1 -1 = 9( 9k -1 ) + 8 = 9( 8t ) + 8 = 8( 9t +1 ) and so 9k+1 -1 is a multiple of 8, and so 8 | 9k+1 -1. For convenience, let Sn = 1 + 3 + 5 + 7 + ... + 2n-1. But, 8k+1 - 1 = 8 ( 8k - 1 ) + 7 = 8(7t) + 7 = 7( 8t +1 ). But that's not very satisfying. 3.3 Prove by induction on n that 13 divides 24n+2 +3n+2 for all natural numbers n. A1, A2, ..., An. Now we must show that f3(k+1) is even. Well, Sk+1 = Sk + (2k+1) = t2 + 2k +1. By the Well-ordering Principle, A(n, m) does have a smallest element. 7. ��T�`"�qF0UK��3k����m08�`2�A�uwX�u���{�ꁵ�DH��[I����Q��'���`&�|�9�O�p(G�vqœ��D�:� f���P��[#f�t*ظ_���Hf`���s��� ��,�.���:��u�b��� l7� ����O���P�� S��m�̦z�Y,�&b�b�� �?��@؀��VN�=H�l�E��Ag�!�;��eƐ|�ه1`2�X*?             8 | 9k+1 - 1 9( 9k -1 ) + 8,       9k+1 - 1 = 9( 9k -1 ) + 8 (Both sides are algebraic expressions. Guess a formula for the value of an and use induction to prove you are right. If n and m are relatively prime, then so are n and n + m, and so are n and n - m. Exercise: (Let fn denote the nth Fibonacci number.). Let an denote the number of subsets of {1, 2, 3, ... n} (including the empty set Consider another example. Prove by induction that the sum 1 + 3 + 5 + 7 + ... + 2n-1 is a perfect square. 5. n2 + n + 41 is prime for n=1,2...40, but it is not prime for all n >= 1. /Length 1622       (b). Prove by induction, x 1 + x 2 (1+x 1) + x 3 (1+x 1) (1+x 2) + … +[x n (1+x 1)…(1+x n-1)] = (1+x 1) (1+x 2) …(1+x n) – 1. Show an = 2an-1. We will first show that d is a common divisor of m and n. By the divison algorithm, there exist integers q and r with 0 <= r < d and n = qd + r. But then, Show that n lines in general position divide the plane into regions. This is just to let everybody know what we are up to so they will know what to expect in the forthcoming argument. To verify your guess you will need to use the strong form of induction. Here are a few examples. 3. It is almost impossible to prove this assertion without proving much more. 4. 45* Prove the binomial theorem using induction. But his last expression reduces to . Prove by induction that the sum 1 + 3 + 5 + 7 + ... + 2n-1 (i.e. "8 divides 9k+1 - 1, " while the right-hand side is an algebraic expression. Let Dn denote the number of ways to cover the squares of a 2xn board using plain dominos. I can also check it directly for n =2, 3 ,4 and 5. 2. See Exercise (19) for an example that shows that the basis step is needed in a proof by induction. And so we have 8k+1 - 1 is a multiple of 7 and so, 7 | 8k+1 - 1. Thus, there is some integer m such that . Corollary. (a) Show that ∀n ∈ N, ( ) 2 n n n m2 +1−m =a m +1−b , where a n, b n ∈ Z, m ∈ N. (b) Show that for any n ∈ N, we can find a natural number N with ( m 1 m) N 1 N n 2 + − = + − . Now what?? For any positive integers n and m, the smallest element of A(n, m) is gcd(m, n) - the greatest common divisor of n and m. Proof. (Its proof is an exercise for you). For n=1 this says that f3 = 2 is even - which it is. Then Sn+1 = Sn + (2n+1). Theorem.       Then for every integer n >= 1, . �T@��-T��,a����v��;}��s���>}�tU]wA)����Jf(��yx�����IW�I5A��Qh� �ʐ�81T�[R�c�؛�P���?S���R�G�a�q�4l�����USS� �w�L+���5I"Ae��U�tt�l:j�߿�v�f�uu(����d�m �A�A�,F�UR�7,e����+�����K�6ٽ�+Ӗf�������ĨS�����*n9w��7\�rP�v�ي���kL�D�|'`-Q��>��|��pwn�Q*���Q�y��_���$�Jb ��[�U6����"/��܈V�T�m֮m�Mі�p�$ Proof. (ii). Prove or Disprove: For each n >= 1, fn and fn+2 are relatively prime. Also equivalent to the Principle of Induction is the Well-Ordering Principle. But look what happens if we try to prove the stronger result that Sn= n2. Now suppose that for some integer k >= 1, . Lemma. I could say something like, "so, see, I can continue just like this and prove the result for one integer after another." Using this fact, prove: Lemma. How many ways are there to cover the squares of a 2xn chessboard with dominos? 44. 1. Theorem. Prove by induction: For any n>=1, 7 | 8n - 1. (This is pretty simple - you don't need induction here.) Hence, n = qd and this means that d divides n. Similarly, d divides m. So we have shown that d is a common divisor of n and m. On the other hand if t is any common divisor of m and n then say n = kt and m = st, then d = an + bm = akt + bst = (ak+bs)t. So d is a multiple of t. Hence d is at least as large as any other common divisor of m and n. Thus d is the largest of the common divisors of n and m. This completes the proof. Now assume that Sk is a perfect square, say Sk = t2 for some t. Then we must show that Sk+1 is also a perfect square.

proof by induction exercises

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proof by induction exercises 2020