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Question:
1. Recurrences[10%] Consider the recurrence T(n). [Bonus Question]
1, if n = 1
5, if n = 2
5T(n − 1) − 6T(n − 2), if n > 2 & n ∈ N.

Prove by strong induction that T(n) = 3n −2 for all n ≥ 1 and n ∈ N.2. Mathematical Induction[10%]. Prove using mathematical induction that 3 divides n 3 + 2n for every n ∈ Z +.
3. Relations [10%]. Find a relation R on the set Z that is
• Not reflexive.
• Not symmetric.
• Not transitive.

5. Relations [10%]. Let S = {{1}, {1, 2}, {1, 3}} be a set. Is there exists a totally ordered binary relation R on A. Prove your claim. Relations [10%] Define a binary relation R on Z as (x, y) ∈ R only if
| x − y |< 1.
• Is R Reflexive. Prove or disprove your claim.
• Is R Symmetric. Prove or disprove your claim.
• Is R Transitive. Prove or disprove your claim.
• Is R Antisymmetric. Prove or disprove your claim.
• Is R Comparable. Prove or disprove your claim.

6. Set Partition [10%] Let Z3 = {[0], [1], [2]} denote the set of equivalence classes modulo 3. Prove that [0] ∩ [1] = ∅, [0] ∩ [2] = ∅, and [1] ∩ [2] = φ.7. Number Theory[10%]. Prove that if n ∈ Z then n 2 ≡ 0 (mod 4) or

Number Theory[10%]. Prove that if a, b, c and m are integers such that m ≥ 2, c > 0, and a ≡ b (mod m), then ac ≡ bc (mod mc). Division Theorem [10%]. Prove that every prime number except 2 and 3 is of form 6k + 1 or 6k + 5 for some k ∈ Z. Hint note n = 6 and apply GCD [10%]. Let a, b, d ∈ Z such that d = GCD(a, b). Let d 0 ∈ Z be .

11. Number Theory[10%]. Prove that the fundamental theorem of arithmetic cannot be extended to negative integers.
12.CRT [10%] Solve the following system of congruences. [Bonus Question]
2x ≡ 5 (mod 7)
4x ≡ 2 (mod 5)
3x ≡ 9 (mod 11)

13. Fermat’s Little Theorem [10%]. Using Fermat’s little theorem, prove that if n is a positive integer, then 42 divides n Quadratic Residusity [10%]. Let p be an odd prime and 1 ≤ a ≤ p−1. Prove that if a ∈ QRp is a quadratic residue modulo p then a p−1 2 ≡ 1 (mod p).
Answer:
Prove by strong induction that T(n) = 3n − 2 n for all n ∈ N .
Solution. : Given : T0 = 0, T1 = 1
Base case :
T2 = 5T1 − 6T0 = 5 = 32 − 2 2
T3 = 5T2 − 6T1 = 19 = 33 − 2 3
Assumption :
T(n) = 3n − 2
n and T(n−1) = 3(n−1) − 2 (n−1)

Proving :
T(n+1) = 3(n+1) − 2 (n+1) T(n+1) = 5T(n) − 6T(n−1)
= 5(3n − 2 n ) − 6(3(n−1) − 2 (n−1) = 5(3n − 2 n
) − 2.3 n + 3.2 n = 3n .3 − 2 n .2 = 3(n+1) − 2 (n+1)

is the solution of the given recurrence
Problem 2. prove using mathematical induction that 3 divides n 3 + 2n for every n ∈ Z + .
Solution. : Given : P(n) : 3 divides n 3 + 2n 1 3 + 2(1) = 3 = 3(1) Hence 3 divides 3 Therefore, P(n) is valid for n = 1
(ii) Assume that p(m) : m3 + 2m Now prove that it is valid for m + 1
Proving :
(m + 1)3 + 2(m + 1) is divisible by 3
(m + 1)3 + 2(m + 1) = m3 + 3m2 + 3m + 1 + 2m + 2
= m3 + 2m + 3m2 + 3m + 3
= m3 + 2m + 3(m2 + m + 1)
= m3 + 2m + 3(a) again, let m2 + m + 1 = a
m3 + 2m is divisible by three ;3(a) is also divisible by 3 (by induction hypothesis)

Hence, the sum of m3+2m, 3(a) are both divisible by three. Also, 3 divides (m+1)3+2(m+1) Thus for any integer n, 3 divides n 3 + 2n
Problem 3. A set Z from a relation R that is: not transitive, not symmetric and not
reflexive.
Solution. : Given : u, v, w for R = (u,v),(v,w)
Not transitive :
(u, w) ∈/ R
Not symmetric :
(v, u) ∈/ R
Not reflexive :
(u, u) ∈/ R

Note that u is related to v and v is related to w.  A totally ordered binary relation R in A for a set S = (1, 1),(1, 2),(1, 3).
Solution. : Total order relation must fulfill 4 aspects:
1)Transitivity
2)Anti-symmetric
3)Reflexive
4)Comparability

S = (1, 1),(1, 2),(1, 3)
Absence of (2, 2),(3, 3) =⇒ non-reflexive
Absence of (2, 1) =⇒ anti-symmetry
Presence of (1, 1),(1, 2) =⇒ transitive
since the set is non reflexive, then there is no totally ordered binary relation R in A

Problem 5. Defining a binary relation R on Z for (x, y) and when | x − y |< 1
Solution. : R is reflexive :
| x − x |< 1
Thus xRx
R is symmetric :
For R x and y, when xRy, | x − y |< 1 and | y − x |< 1
yRx = xRy
R is transitive :
If xRy = yRz, then | x − y |< 1 and | y − z |< 1
| x − y | + | y − z | = | x − z |< 1
Thus xRz

Hence, Transitive R is anti-symmetric :For R x and y, when xRy, | x − y |< 1 and | y − x |< 1
y 6= x
Hence R is not anti-symmetric R is comparable :
Neither y ≥ x nor x ≥ y for R
Thus, R is incomparable

Prove for a set of equivalence classes of modulo 3 .
Solution. : Let : x ∈ [0] ∩ [1], then x ∈ [0] and x ∈ [1]
so, 3 | x and 3 | x − 1
so, 3 | x − (x − 1)
so, 3 | 1, contradicting
Thus, [0] ∩ [1] = ∅
Let : y ∈ [1] ∩ [2], then y ∈ [1] and x ∈ [2]
so, 3 | y − 1 and 3 | y − 2
so, 3 | (y − 1) − (y − 2)
so, 3 | 1, contradicting
Thus, [0] ∩ [2] = ∅
Let : z ∈ [1] ∩ [2]
so, 3 | z and 3 | z − 2
so, 3 | z − (z − 2)
so, 3 | 2, contradicting
Thus, [1] ∩ [2] = ∅
Problem 7. Prove that n ∈ Z , n
2 ≡ 1 (mod 4) or n
2 ≡ 0 (mod 4)
Solution. : Proof : Then, n is either even or odd
Case (i) : Assume n is odd
m ∈ Z such that n = 2m + 1
Thus, n
2 = 4m2 + 4m + 1
Hence : n
2 − 1 = 4(m2 + m)
So n
2 ≡ 1 (mod 4)
Case (ii) : Assume n is even
m ∈ Z such that n = 2m
Thus, n
2 = 4m2
Hence : 4 | m2
So n
2 ≡ 0 (mod 4)
Proved!
Problem 8. Solution. : It means that : a ≡ b (mod m) =⇒ m | a − b
Let c > 0 then mc | c(a − b)
Note that when (a | b also ac | bc)
=⇒ mc | c(a − b)
=⇒ mc | ca − cb)
=⇒ ca ≡ cb (mod mc)
Proved!
Problem 9. Prove for prime number exist as 6k + 5 or 6k + 1 apart from 2 and 3 for k ∈ Z

Solution. : Let : n > 3 and n be a prime
Division algorithm yields n = 6k = 2.3.k ; even
n = 6k +1;prime
n = 6k +2 = 2(3k + 1); even
n = 6k +3 = 3(2k + 1)
n = 6k +4 = 2(3k + 2); even
n = 6k +5;prime

Conclusion : None of these will represent a prime greater than 2 and 3 Proving : Thus p must be either 6k + 1 or 6k + 5. . Prove that d 0
| d when d = GCD (a, b) and when d
0 ∈ Z
Solution. : Given : Bezout’s identity
Proof :
∃x, y ∈ Z such that d = ax + by
Let d
0 be a C.D for a, b

Proved!
Solving a system of linear congruence  : If you are supposed to solve one by one
1: 2x ≡ 5 (mod 7) We begin by finding the inverse of 5 modulo 7
5.3 ≡ 15 ≡ 1 (mod 7)
So, 5(−1) ≡ 3 (mod 7)
3.2x ≡ 6x ≡ 1 (mod 7)
also find the inverse of 6 modulo 7
6.6 ≡ 36 ≡ 1 (mod 7)
Hence 6 is its own inverse and we have:
6.6x ≡ x ≡ 6 (mod 7)
So the reduced form is x ≡ 6 (mod 7)
2: 4x ≡ 2 (mod 5)
We have 3.2 ≡ 6 ≡ 1 (mod 5)
So, 2(−1) ≡ 3 (mod 5)
3.4x ≡ 12x ≡ 2x ≡ 1 (mod 5)
and;
3.2x ≡ x ≡ 3 (mod 5)
So the reduced form is x ≡ 3 (mod 5)
3: 3x ≡ 9 (mod 11)

We have 9.5 ≡ 45 ≡ 1 (mod 11)
So, 9(−1) ≡ 5 (mod 11)
5.3x ≡ 15x ≡ 4x ≡ 1 (mod 11)
and;
4.3 ≡ 12 ≡ 1 (mod 11)
So, 4(−1) ≡ 3 (mod 11)
3.4x ≡ x ≡ 3 (mod 11)
So the reduced form is x ≡ 3 (mod 11)
Combined:

If they are to be solved simultaneously, we will use the Chinese remainder theorem.
x ≡ 6 (mod 7)
x ≡ 3 (mod 5)
x ≡ 3 (mod 11)
Let’s start with
x = 5.11 + 7.11 + 7.5
These two terms should vanish when a remainder of x modulo 7, 5, 11 are to be considered.
Taking modulo 7, we have:
x ≡ 5.11 ≡ 55 ≡ 6 (mod 7)
Taking modulo 5, we have:
x ≡ 7.11 ≡ 77 ≡ 2 (mod 5)
We want to end with 3, thus we multiply the second term with 2 modulo 3 and then 3.
But, 2.3 ≡ 1 (mod 5)
Hence, our new choice of x is:
x = 5.11 + 7.11.3.3 + 7.5
Taking modulo 5, we end up with a remainder 3 as we desired
Taking modulo 11, we have:
x ≡ 35 ≡ 2 (mod 11)
We wanted to remain with 3, thus we multiply the third term with the inverse of 2 modulo
11 and by 3.
But, 2.6 ≡ 1 (mod 11)
Hence, our new choice of x is:
x = 5.11 + 7.11.3.3 + 7.5.6.3 = 1378
Hence, from CRT, our least positive solution is:
x ≡ 1378 (mod 7.5.11) reduced to
x ≡ 1378 (mod 385)
=⇒ x ≡ 223 (mod 385)
Thus the general solution to this system is:
x = 223 + 385n for n ∈Z
Problem 11. Prove that fundamental theorem cannot be satisfied with negative integers .
Solution. : Proof : Every positive integer n is a product of primes
From induction, n = 1 and n > 1
Example:
15 = 3 * 5

The same cannot be expounded to negative integers Proving :
15 6= (−3) ∗ (−5)
The integer is divided into a prime, composite and unit Suppose n = 2, 3, ..., k, consider k + 1. It is either a prime
If k + 1 is not prime, then, k + 1= ab
with definitions as 1 ¡ a k and 1 ¡ b k
a and b are finite product of primes
ab is a finite product (−3) ∗ (−5) = −15 6= 15 (finite product)
Solution. Since 42 has prime multiples of 2, 3, 7 we need to show that n 
can be divided by 2, 3 and 7.
From FLT:
if m = 2, 3, 7 then m | n
n
(m−1) ≡ 1(mod m)equation1
(m − 1) | 6 for m = 2, 3, 7
Thus, ( 6 m−1 ) is an integer
Raise both sides of equation 1 to ( 6 m−1
) th power 6 
n(m−1)(m−1) ≡
6 1 (m−1) (mod m)
n 6 ≡ 1 (mod m)
multiply both sides by n
n 7 ≡ n (mod m )
This means:
m | (n 7 − n) for m = 2, 3, 7
Since 2, 3, 7 are all prime (n
7 − n) is divided by 42.
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