THE COURSEWORK RESIT

Running head: COURSEWORK RESIT
COURSEWORK RESIT
Name of the Student
Name of the University
Author Note

COURSEWORK RESIT1
1.
Given statement: If n is an odd number, then 4n-1 is prime.
Given, n is odd so let, n = 2m + 1 (where m = any positive whole number)
Thus, 4n-1 = 4*(2m+1) – 1 = 8m + 3.
Now, if 8m +3 is prime then (8m+3) mod m ≠ 0 for all m.
By, inspection it is found that when m = 3 then (8m+3) mod 3 = 27 mod 3 = 0. Hence,
(8m+3) is not prime.
Hence, the statement “If n is an odd number, then 4n-1 is prime” is false.
Converse statement: if 4n-1 is prime then n is an odd number.
By inspection, when n = 2 then 4n – 1 = 7 which is a prime number. Thus n can also be even
when 4n-1 is prime.
Thus the statement “if 4n-1 is prime then n is an odd number” is false.
2.
Given, there are 12 natural numbers in set A. Thus by using the pigeonhole principle when
divided by 11 then at least two of the 12 numbers must produce same remainder as the
numbers are different.
Let, the two numbers are ai and a j.
Hence, ai = 11k + r
a j = 11m + r
Where, k, m and r any integer.

COURSEWORK RESIT2
Thus ai−a j=11 k +r −11m−r = 11(k-m)
Thus ai−a j is divisible by 11.
Hence, by using the pigeonhole principle it proved that A contains at least two number ai and
a j such that ai−a j is divisible by 11.
3.
Given equation 5x – 12 = 0 in Z13.
Hence, it is required to find an element x such that
(5x-12) mod 13 = 0
Since, 0 = (m*13) mod 13 hence, 5x -12 = 13m
Now, xε {0,1,2,3,4,5,6,7,8,9,10,11,12 }
Hence, 5x -12 ε {−12 ,−7 ,−2,3,8,13,18,23,28,33,38,43,48 }
In the set there is only one number matching to the form 13m which is 13 that is for x = 1.
Hence, the solution of 5x – 12 = 0 in Z13 is x = 1.
Given, x^2 – x – 1 = 0 in Z11.
Thus it required to find an element x such that
(x^2 – x – 1) mod 11 = 0
Since, 0 = (m*11) mod 11 hence, (x^2 – x – 1) = 11m
Now, xε {0,1,2,3,4,5,6,7,8,9,10 }
Hence, (x^2 – x – 1) ε {−1 ,−1,1,5,11,19,29,41,55,71,89}