Question 1
Since n is composite then we have an ab=n where if both factors a and b are considered to be > then their product n=ab would be > =n which is a contradiction thus either a or b must not > that is
But we know that a has a prime divisor (), so n has same divisor (2)
Question 2
a)
proved by contradiction of the well ordering principle.
Assume that is a strictly decreasing infinite sequence of natural numbers. Expressed as n1>n2>…nk
Then is a nonempty subset of natural numbers which has no smallest element. This contradicts the well-ordering principle.
Thus, there can be no strictly decreasing infinite sequences of natural numbers. In other words, every strictly decreasing sequence of natural numbers must be finite.
b)
suppose N be set of natural number of the form n-dp where q is a natural number. This set is nonempty since -dp can be made as large as possible.
Using well ordering property, N has a least set where and that r<d if not then n not a member of N. to show this suppose r≥d because it follows that and consequently q and r with 0≤ r <d.
To prove for uniqueness
Suppose we have two pairs (q,r) and
So that
We will show that the pairs are really the same
Therefore
The only multiple of d in the range is 0 thus we conclude that
Thus, the number q and r are unique as shown above
Question 3
a)
When n = 1 the result is clear, 13 = 12
Assume the result is true for n = k, that is
13 + 23 + 33 + 43 + ... k3 = (1 + 2 + 3 +...k)2
Let n = k + 1, then
13 + 23 + 33 + 43 + ... n3
= 13 + 23 + 33 + 43 + ... + (k+1)3
= (1 + 2 + 3 + ... + k)2 + (k+1)3
= (k(k+1)/2)2 + (k+1)3
= (k2 (k+1)2)/4 + (k+1)3
= (k+1)2/4 (k2 + 4k + 4)
= ((k+1)2 (k+2)2)/4
= ((k+1)(k+2)/2)2
= (1 + 2 + 3 + ... + (k+1))2
Thus 13 + 23 + 33 + 43 + ... n3 = (1+2+3+...n)2 for all positive intergers n.
b)
Question 4
a)
to show has no solutions we have
Thus
thus no solutions are found
Further test if
However, since 0 mod 6 = 1
Then does not have solutions
b)
Question 5
a)
if 12345678923
17= 13*1 remainder 4
Thus, remainder after division by 13 is 4
b)
12345678923^123
First check divisibility of 12345678923
=
=
Question 6
a)
to show has no solutions we have
Thus
thus no solutions are found
Further test if
However, since 0 mod 6 = 1
Then does not have solutions
Further testing for mod 5 we have;
x = 2
x = 3
thus 0 mod 5 has solutions
b)
if any two real numbers a and b, if ab = 0 then either a = 0 or b = 0 is true in that multiple of zero are always zero, this statement is true for mod 6 that is, if ab ≡ 0 (mod 6), is either a ≡ 0 (mod 6) or b ≡ 0 (mod 6), and considering 0mod(6)=0
then given a or b is =0 then ab≡0≡0 mod 6
thus, the case applies to mod 6.
c)
d)
e)
