Solution to Problem 3: Statement P (n) is defined by 1 3 + 2 3 + 3 3 + ... + n 3 = n 2 (n + 1) 2 / 4STEP 1: We first show that p (1) is true.Left Side = 1 3 = 1Right Side = 1 2 (1 + 1) 2 / 4 = 1 hence p (1) is true. (There are lots of other possible counterexamples, too.). Exercises ; Building Mathematical Statements; Conditional Statements; Detachment and Syllogism; The Quality of Equality; Proofs; Algebraic Proofs; Congruence, Equality, and Geometry; Proof by Deduction; Proof by Induction; Induction in Action; Proof by Contradiction /Length 1622 What's the base case when we prove inductively that n – 2 is positive when n > 2? Use the Principle of Induction to prove the following results. No need to get fancy. Use the smallest possible n-value. proofbyinduction.net is part of ADA Maths , a Mathematics Databank When n = 3, we get 3 – 2 = 1, which is positive. :�RJ����=�+����f��"� Name one counterexample that shows he can't prove his general statement. ��T�`"�qF0UK��3k����m08�`2�A�uwX�u���{�ꁵ�DH��[I����Q��'���`&�|�9�O�p(G�vqœ��D�:�
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@�7b(�)����f���֛*��Q7"qOcD�%t���b]\ߖ@�v^�֜���Hi��w� ��x/ ��!�ƛ���1"�;j!�=zD$�� 3.3 Prove by induction on n that 13 divides 24n+2 +3n+2 for all natural numbers n. Choose the simplest positive integer you can think of, then plug it in for n. When n = 1, we get 31 + 1 = 4, which is divisible by 2. What's the base case when we prove inductively that 10n – 5 is divisible by 5 for all positive integers? %PDF-1.5 (Click on the green letters for the solutions.) It works fine when n = 1, but pay close attention to other possible values for n. When n = 2, we get 6(2) + 2 = 12 + 2 = 14, which is not divisible by 4. Exercise 4. Assume n is a natural number. Proof By Induction Questions, Answers and Solutions proofbyinduction.net aims to have the biggest database of proof by induction solutions on the internet! You want to mess up his proof, because you think he's wrong. %���� Are there any counterexamples that disprove the statement, "6n + 2 is always divisible by 4 when n is a positive integer"? �T@��-T��,a����v��;}��s���>}�tU]wA)����Jf(��yx�����IW�I5A��Qh� �ʐ�81T�[R�c�؛�P���?S���R�G�a�q�4l�����USS� �w�L+���5I"Ae��U�tt�l:j�߿�v�f�uu(����d�m �A�A�,F�UR�7,e����+�����K�6ٽ�+Ӗf�������ĨS�����*n9w��7\�rP�v�ي���kL�D�|'`-Q��>��|��pwn�Q*���Q�y��_���$�Jb
��[�U6����"/��܈V�T�m֮m�Mі�p�$ �Ւj�p���"L�b���Y7�u���o0\���@(�b�3IG����{8�ϡ�ԉ#3Lf�3�11�\��1Ύ2�t�uyP\K���أ�$Q�z4��[zNz�Lє̣6���P3�v눚��&�O�ԓt#Z��s�;���H�2�2��8�N���,��MH,�렰@�>-:����2��I7Yz^�:�\Lg����l��Z���x����Ot�����uE}��ڕ5է�A1\��lR-w?վ�P�`�R�F4L��k��qE������j�6W��V��{dE8��); �lsE�T�a�jJe;̏��xC�&�@ ��9�Ȣ�nc��n]�Y�s�9"?��e�ŗt��{da�,�r�j��^��Z�Qq��[�b�)�����5Hn��b�?u�v8[.>糍�. stream /Filter /FlateDecode Brandon is trying to prove that 3n + 1 is an even number whenever n is a positive integer. 3.2 (5.2 of [IMR]) Prove by induction that n3 2n for all integers n with n 10. A counterexample is just any n-value that makes the statement untrue. We can't use n = 1 this time, so what's the next simplest value we can plug in? >> 3 - PROOF BY INDUCTION - EXERCISES 3.1 (5.1 of [IMR]) Prove by induction that for all natural numbers n, n3 n is divisible by 3. 3 0 obj When n = 2, we get 3(2) + 1 = 6 + 1 = 7, which is not an even number. When n = 1, we get 101 – 5 = 5, which is divisible by 5. << Whole numbers are easier than fractions/decimals. (a) 5n −1 is divisible by 4, (b) 9n +3 is divisible by 4, (c) 3n > n2, (d) P n j=1 1 j2 ≤ 2− n. Exercise 5. What's the base case when we prove inductively that 3n + 1 is divisible by 2 for all positive integers? (There are lots of other possible counterexamples, too.). Inductively that 10n – 5 = 5, which is divisible by 2 for all positive integers of other counterexamples. Of other possible counterexamples, too. ) we ca n't use n = 1, we get –! N > 2 that 3n + 1 is divisible by 2 for all positive integers we 101! Of [ IMR ] ) prove by induction that n3 2n for all positive integers shows ca... Simplest value we can plug in so what 's the base case we! 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