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Assignment 3 Of MTH202 (Fall 2012)

 

                                                                                              Maximum Marks: 15                                                                                      

                                                                                                     Due Date: 15-01-2013

 

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Q#01:  Use mathematical induction to prove that  for every integer  with . (Note that this inequality is false for  and .)                   Marks = 6

 

 

Q#02:  Find the GCD of  and  using Division Algorithm.       Marks = 4

 

 

Q#03:  Define a sequence  by the formula , for all integers . Show that this sequence satisfies the recurrence relation , with  and  for all integers .                                      Marks = 5

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Replies to This Discussion

Please Discuss here about this assignment.Thanks

Our main purpose here discussion not just Solution

All students must share points and also your problems.... so it can discussed and solution can prepare..

Dont wait for solution just participate in Discussions becoz after discussions a correct solution will prepare...

maine question 1 kiya hai ye hai

question 1 ka solution:

BASIC STEP:

p(4) = 2^n                       

 2^4 =16 < 24 = 4!

 

INDUCTIVE STEP:

 

For this step we assume that p(k) is true for posiive integer k with k>=4

Assume:

2^k < k!

2^k+1 < (k+1)!

2^k+1 = 2.2^k

 

          <  2.k!

  

          < (k+1)k!

 

          = (k+1)!

 

So p(k+1) is true when p(k) is true.

HENCE PROVED.....



solution of question 2:

Let a= 4566891  and b=182  . Also, let's introduce the variable "r" for the remainder

Let's evaluate a/b  . It turns out that a/b  = 25092 remainder 147. What we're are interested in is the remainder.

So let  . Now assign the value of "b" to "a" and the value of "r" to "b".

So now a=182 ,b=147  and r=147

 

Now we must ask ourselves: "Does b=0?". Since b=147 (which is NOT zero), we must start all over and keep going until "b" equals zero.


Let's evaluate a/b  . It turns out that a/b= 182/147  = 1 remainder 35. What we're are interested in is the remainder.

So let  . Now assign the value of "b" to "a" and the value of "r" to "b".

So now a=147 ,b=35  , and r=35

 

Now we must ask ourselves: "Does b=0?". Since b=35 (which is NOT zero), we must start all over and keep going until "b" equals to zero


Let's evaluate a/b  . It turns out that a/b= 147/35  = 4 remainder 7. What we're are interested in is the remainder.

So let  . Now assign the value of "b" to "a" and the value of "r" to "b".

So now a=35 ,b=7 , and  r=7

 

Now we must ask ourselves: "Does b=0?". Since b=7 (which is NOT zero), we must start all over and keep going until "b" equals zero.

 

Let's evaluate a/b  . It turns out that a/b = 35/7  = 5 remainder 0. What we're are interested in is the remainder.

So let  . Now assign the value of "b" to "a" and the value of "r" to "b".

So now a=7 ,b=0  , and  r=0

Since the value of b is now zero, we can stop the process.
Now the value of "a" in the last row is the GCD of the numbers "a" and "b"
So this means that GCD(4566891,182)=7.  

.

good work

thnx

 

welcome 

great work Sw8 & Sal8y

thnx

nice job

 

plzzz give some idea for the question no 3 ........................... and thnku soo much Sw8 & Salty for ur help

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