Lec10_Induction - TDS1191 Discrete Structures Lecture 10...

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Discrete Structures 1 TDS1191 Discrete Structures Lecture 10 Induction Multimedia University Trimester 2 Session 2011/2012 [Lecture 10][Induction]
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Discrete Structures 2 Contents Why do we need proof by induction? What is a proof by induction? Examples Exercises [Lecture 10][Induction]
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Discrete Structures 3 [Lecture 10][Induction] Introductory example Can we conclude that we can reach every rung? Picture adopted from (1)
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Discrete Structures 4 We know two things: 1. We can reach the first rung of the ladder. 2. If we can reach a particular rung of the ladder, then we can reach the next rung. So: By (1), we know that we can reach the first rung. Because of (1): By (2), we can also reach the second rung (the rung after the first). Applying (2) again, we can reach the third rung. Reapplying (2) for sometimes, we always can reach the next rung (the rung after the rung that we are currently hold to). The answer is YES!! [Lecture 10][Induction]
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Discrete Structures 5 Can you? (Take note, of course Gauss had nothing than such an old child’s slate and chalk at that time. Introductory example II Around 1784 a teacher asks his pupils to sum up all numbers from 1 . . 100 as he wanted to take a short nap in front of the class. After a few minutes the small boy Gauss came to the teacher’s desk and presented the correct solution. Carl Friedrich Gauss (1777 - 1855) [Lecture 10][Induction]
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6 Gauss had to compute f(100) for f(n) = S(n) and what did he do? 1 + 2 + 3 + 4 + . ..... ..... + 97 + 98 + 99 + 100 1 + 100 = 101 2 + 99 = 101 3 + 98 = 101 50 + 51 = 101 . Gauss got 50 pairs each having a sum of 101 and hence 50 * 101 = 5050 f(n) = k = 1 + 2 + 3 + . .. + n = (n/2) * (n+1) Generally, the formula is given as: As you see, the subject matter of the problems can vary widely. It can include algebra, geometry, and many other topics. 1 [Lecture 10][Induction]
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Lec10_Induction - TDS1191 Discrete Structures Lecture 10...

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