Discrete Mathematics AMTH140- Assignment

Added on - 24 Feb 2020

  • 14


  • 1518


  • 104


  • 0


Trusted by +2 million users,
1000+ happy students everyday
Showing pages 1 to 4 of 14 pages
Discrete MathematicsAMTH140 Assignment 3STUDENT ID[Pick the date]
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, then12=3(4)+7(0)Step 2:Let n is an positive integer value for whichn12And hence,nwouldbewrittentheformofab,suchasn=3a+7bWhere,abarealsosomeintegervaluesforwhicha,b0.In this regards, two following conditions would be taken into account.1.Leta2and thus,3(a2)+7(b+1)=n+12.Leta1and thus,3a+7b=n127b91
b23(a+5)+7(b2)=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+1postages would bemade in either with 7₡ stampsand 3stamps as per induction step.(b)Frommathematical induction3+6+9+.............+3n=3n(n+1)2Step 1:LetP(n)=3+6+9.............+3n=3n(n+1)2Step 2:For n = 1LHS = 3RHS=31(1+1)2=3Here, LHS =RHS highlights thatP(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, assume2
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 ofa1,a2a3Valueofa1: 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
You’re reading a preview
Preview Documents

To View Complete Document

Click the button to download
Subscribe to our plans

Download This Document