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DMTH137 - Discrete Mathematics AMTH140- Assignment

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Macquarie University

   

Discrete Mathematics (DMTH137)

   

Added on  2020-02-24

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DMTH137 - Discrete Mathematics AMTH140- Assignment, based on the above discussion, it can be said that strong induction has formed the respective proposition for all the respective positive integers. Thus, n+1₡postages would be made either with 7₡stampsand 3₡stamps as per the induction step.

DMTH137 - Discrete Mathematics AMTH140- Assignment

   

Macquarie University

   

Discrete Mathematics (DMTH137)

   Added on 2020-02-24

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Discrete MathematicsAMTH140 Assignment 3STUDENT ID[Pick the date]
DMTH137 - Discrete Mathematics AMTH140- Assignment_1
Question 1(a)Any postage of at least 12 ₡ would be obtained using 3₡ and 7₡ stamps. Mathematical induction – “In normal mathematical induction, it has been assumed that if thebase case is true then some number n would be true when the case is also true for n+1 (Youse,2011) .” Step 1: Let n = 12, then 12=3(4)+7(0)Step 2: Let n is an positive integer value for which n≥12And hence, nwouldbewritten∈theformofa∧b,suchasn=3a+7bWhere, a∧barealsosomeintegervaluesforwhicha,b≥0. In this regards, two following conditions would be taken into account. 1.Let a≥2and thus, 3(a−2)+7(b+1)=n+12.Let a≤1 and thus, 3a+7b=n≥127b≥91
DMTH137 - Discrete Mathematics AMTH140- Assignment_2
b≥23(a+5)+7(b−2)=n+1Hence, it can be said that inductive argument to (n+1) is true for all the k values which is smallerthan n i.e. k <n. If n = 12, the condition is true because 12₡ would be made from - four 3₡ stamps. If n = 13, the condition remains true because 13₡ would be made from – one 7₡ stamps andtwo 3₡ stamps. If n = 14, the condition remains true because 14₡ would be made from – two 7₡ stamps.Therefore, based on above discussion, it can be said that strong induction has been formed therespective proposition for all the respective positive integers. Thus, n+1₡ postages would bemade in either with 7₡ stamps and 3₡stamps as per induction step. (b)From mathematical induction 3+6+9+.............+3n=3n(n+1)2Step 1: Let P(n)=3+6+9.............+3n=3n(n+1)2Step 2: For n = 1LHS = 3RHS=3∗1(1+1)2=3Here, LHS =RHS highlights that P(n) is true for n =1. Step 3: Let P (k) would be true and hence, P(k+1) would also be true (Youse, 2011). Now, assume 2
DMTH137 - Discrete Mathematics AMTH140- Assignment_3
P(k)=3+6+9+.........+3k=3k(k+1)2....(1)It is true for k. Now, for k+1P(k+1)=3+6+9+...+3(k+1)=3(k+1)((k+1)+1)23+6+9+.......+(3k+3)=3(k+1)(k+2)2.....(2)From equation (1)Put k = k+1P(k+1)=3+6+9+......+3(k+1)=3(k+1)((k+1)+1)23+6+9+.......+(3k+3)=3(k+1)(k+2)2It can be seen that P(k) would become same as P(k+1) and thus, P (k+1) would be true whenP(k) is true. The conclusion can be drawn based on the principle of mathematics induction that P(n) wouldbe true for all the n values , where n is a positive nature number. Question 2 (a)Value of a1,a2∧a3Valueofa1: when n = 1, There are only two disks with same size which would move directly tothe direction of desired pole and hence, a1=2.3
DMTH137 - Discrete Mathematics AMTH140- Assignment_4

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