1) Primes are positive integers greater than 1 that are only divisible by 1 and themselves. The fundamental theorem of arithmetic states that every positive integer can be uniquely expressed as the product of primes. 2) Euclid's proof shows there are infinitely many primes. Euclid numbers form a sequence where each term is the sum of the previous terms plus 1, and the early terms are prime. However, not all Euclid numbers are prime. 3) The largest power of a prime p that divides n! is given by the sum of the number of times p divides the numbers from 1 to n in their prime factorizations. This can be determined from the number of 1s in the binary representation

1. Number Theory
Primes: A positive integer p is called prime if it has just two divisors, namely 1 and p.
By convention, 1 is not prime, so the sequence of primes starts out like this:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41,
Primes are of great importance, because they are the fundamental building blocks of all
the positive integers. Any positive integer n can be written as a product of primes,
1 2
1
m
m k
k
n p p p p

   ,where 1 2 mp p p    (1)
For example, 12 2 2 3 ; 11011 7 11 11 13 ; 11111 41 271        
Here the products denoted by  are analogous to sums denoted by  . There is only one
way to write n as a product of primes in non-decreasing order. This statement is called
the Fundamental Theorem of Arithmetic.
We can prove the Fundamental Theorem of Arithmetic by contradiction. Suppose, we
have two factorization of an integer number, n
1 2 1 2m nn p p p q q q   , 1 2 1 2andm np p p q q q      
where p’s and q’s are all prime. We will prove that 1 1p q . If not, we can assume that
1 1p q , making 1p smaller than all q’s. Since 1p and 1q are prime their gcd must be 1.
Hence, Euclid’s self-certifying algorithm gives us integer a and b such that
1 1
1 2 3 1 2 3 2 3
1 2 3 2 3
1 2 3 1 2 3 2 3
1 2 3 2 3 2 3
1
( )
n n n
n n
n m n
n m n
ap bq
ap q q q bq q q q q q q
ap q q q bn q q q
ap q q q bp p p p q q q
p aq q q bp p p q q q
 
  
  
  
  
  
 
  
  
Since 1p divides left hand side, hence 1p should divide right hand side 2 3 nq q q . But 1p is
not divisible by 2 3 nq q q according to our assumption. Thus 1p and 1q must be equal.
Similarly we can show that, 2 2 3 3, , , m np q p q p q   .
Every positive integer can be written uniquely in the form
pn
p
n p  where each 0pn   (2)
Formula (2) represents n uniquely, so we can think of the sequence 2 3 5( , , , )n n n  as a
number system for positive integers. For example, prime exponent representation of 12 is
(2 , 1 , 0 , 0 , . . .) and the prime exponent representation of 18 is (1 , 2 , 0 , 0 , . . .). To
multiply two number, we simply add their representations. In other words,
for allp p pk mn k m n p     (3)
This implies that
for allp pm n m n p   (4)
and it follows immediately that,
p1 can't divide ...
np is the power of the prime p in the unique prime factorization of the integer n
example:
100 = 2^2 * 3^0 * 5^2 * 7^0 * 11^0 * ...
352 = 2^5 * 3^0 * 5^0 * 7^0 * 11^1 * ...
11 = 1 * 11
43 = 1 * 43
for prime integers:
***
MUST
SEE
THIS
PROOF
***
GCD (p1, q1) = 1 because both p1, q1 primes
Euclid's theorem: The GCD of two numbers
can always be written as a linear combination
of the two numbers ==> so, ap1 + bq1 = 1 for
integers a, b
uniquely
END
Start

2. gcd( , ) min( , ) for allp p pk m n k m n p    (5)
( , ) max( , ) for allp p pk lcm m n k m n p    (6)
For example, since 2 1
12 2 3  and 1 2
18 2 3  , we can get their gcd and lcm by taking the
min and max of common exponents:
min(2,1) min(1,2) 1 1
gcd(12,18) 2 3 2 3 6    
max(2,1) max(1,2) 2 2
(12,18) 2 3 2 3 36lcm     
Prime Examples: Euclid proved that, there are infinitely many primes. The proof is
as follows: Suppose, there are only finitely many primes, say k of them 2,3,5, , kp .
Then we should consider the number, 2 3 5 1kM p      .
None of k primes can divide M, because each divides M-1. Thus there must be some other
prime that divides M; perhaps M itself is a prime. This contradicts our assumption that,
2,3,5, , kp are the only primes, so there must be infinitely many prime.
Euclid Number: 1 2 1 1n ne e e e   , when 1n   (7)
The sequence starts out
1
2
3
1 1 2
2 1 3
2 3 1 7
e
e
e
  
  
   
4 2 3 7 1 43e     
these are all prime. But the next case, 5e is 1807 13 139  . However, Euclid numbers are
all relatively prime to each other; that is
gcd( , ) gcd(1, ) gcd(0,1) 1m n me e e   , where m n .
Recurrence (7) can be simplified by removing three dots. If 1n  , we have
2
1 2 2 1 1 1 1 11 ( 1) 1 1n n n n n n ne e e e e e e e e            
The numbers of the form 2 1p
 (where p is a prime), is called Mersenne numbers. The
Mersenne prime known occur for p  2,3,5,7,13,17,19,31,61,89,107,127,521,607,1279,
2203,2281,3217,4253,4423,9689,9941,11213,19937,21701,23209,44497,86243,110503,
132049,216091,756839 and 859433. The number 2 1n
 cannot possibly prime if n is not
prime, because 2 1km
 has 2 1m
 as a factor. We can prove it by following way:
2 3 ( 1) 2 1
1 2 2 2 2
2 1
km
m m m k m
m
 
     


Left hand side of the equation is integer, which refers to right hand side is also an integer.
That means, 2 1km
 is divisible by 2 1m
 . Thus 2 1km
 is not a prime number.
But, 2 1p
 is not always a prime when p is prime. For example, 11
2 1 2047 23 89    is
the smallest such nonprime.
e0 = 1 and e1=e0+1=2 .... now compute e2, e3, e4, so on ...
Q: Write the Prime Exponent Representation of GCD(m,n), LCM(m,n) and PRODUCT(m,n). Also, Provide Concrete Numeric
Example.
Q: What do you understand by Euclid number/ Mersenne number? Explain with Concrete examples.
e2 = e0*e1 + 1 = 1*2+1=3
e3 = e0*e1*e2+1=1.2.3 + 1 = 7
e4 = 1.2.3.7 + 1 = 43
en = e0 * e1 * e2 * ... ... ... en-1 + 1
Short Q: Prove or Disprove that Euclid number is always Prime: Ans: Disprove by a counter example (e5 = 1807 = 13*139)
Short Q: Prove or Disprove that Mersenne number is always Prime: Ans: Disprove by Counter example (p=11 => 2^11 - 1
= 2047 = 23 *89)
e5 = 1.2.3.7.43 + 1 = 1807
Prove that: Any integer of the form 2^p - 1 can never be a prime number, given that p is a composite number.

3. Number Theory
Factorial Factors: The factorials:
1
! 1 2
n
k
n n k

      , integer 0n 
We can find out factorial of an integer n using following recursive formula:
0! 1
! ( 1)!n n n  , for 0n 
Here are the first few values of the factorial function.
n 0 1 2 3 4 5 6 7 8 9 10
!n 0 1 2 6 24 120 720 5040 40320 362880 3628800
Because of it’s exponential nature it’s very difficult to find a closed form of factorial. To
approximate n! more accurately for large n we can use Stirling’s formula:
! 2
n
n
n n
e

 
 
 
We would like to determine, for any given prime p, the largest power of p that divides n! ;
that is we want the exponent of p in n!’s unique factorization. We denote this number by
( !)p n , and we start our investigations with the small case p = 2 and n = 10.
1 2 3 4 5 6 7 8 9 10 Powers of 2
Divisible by 2 X X X X X 105
2
 
 
Divisible by 4 X X 102
4
 
 
Divisible by 8 X 101
8
 
 
Powers of 2 0 1 0 2 0 1 0 3 0 1 8
For general n this method gives,  2
1
!
2 4 8 2k
k
n n n n
n

       
                  

This sum is actually finite, since the summand is zero when 2k
n . Therefore it has
only lg n   nonzero terms, and it’s computationally quite easy. For instance, if n = 100,
 2
100 100 100 100 100 100
100! 50 25 12 6 3 1 97
2 4 8 16 32 64

           
                                  
Let’s look at the binary representation of all numbers to find out what’s going on.
2
2
2
2
100 (1100100) 100
100 (110010) 50
2
100 (11001) 25
4
100 (1100) 12
8
 
   
 
   
 
   
 
2
2
2
100 (110) 6
16
100 (11) 3
32
100 (1) 1
64
   
 
   
 
   
 
Find
e2(10!)
= 8

4. We merely drop the least significant bit from one term to get the next. The binary
representation shows us how to derive another formula,
2 2( !) ( )n n v n   , where 2 ( )v n is the number of 1’s in the binary representation of n.
Generalizing our findings to an arbitrary prime p, we have
2 3
1
( !)p k
k
n n n n
n
p p p p


       
           
       

We can find out the upper bound of ( !)p n by simply removing the floor from the
summand and then summing an infinite geometric progression.
2 3 2
1 1
( !) 1p
n n n n
n
p p p p p p

 
        
 
 
1
1
1 1
1 ;[ <1]
1
1
1
n
p p p
n p
p p
n p
p p
n
p


 
  
 
 
  
 
 
  
 



For example, when p = 2 and n = 100, this inequality says that 97 < 100. Thus the upper
bound 100 is not only correct but also close to the true value 97.
☺Good Luck☺
Prove
that ...

5. Number Theory
Relative Primality: When gcd( , ) 1m n  , the integers m and n have no prime factors
in common and we say that they are relatively prime.
m _|_ n  m, n are integers and gcd( , ) 1m n 
A fraction /m n is in lowest terms if and only if m _|_ n. Since we reduce fractions to
lowest terms by casting out common factor of numerator and denominator,
/ gcd( , )m m n _|_ / gcd( , )n m n .
There is a beautiful way to construct the set of all nonnegative fractions /m n with m _|_ n,
called Stern-Brocot tree. The idea is to start with the two fractions
0 1
,
1 0
 
 
 
and then to
repeat the following operation as many times as desired:
Insert
m m
n n


between two adjacent fractions
m
n
and
m
n


. The new fraction
m m
n n


is
called mediant of
m
n
and
m
n


. For example, the first step gives us one new entry.
0 1 1
, ,
1 1 0
and the next gives two more:
0 1 1 2 1
, , , ,
1 2 1 1 0
. The next gives four more:
0 1 1 2 1 3 2 3 1
, , , , , , , ,
1 3 2 3 1 2 1 1 0
and then we will get 8, 16 and so on. The entire array can be
regarded as an infinite binary tree structure whose top levels look like this:
0 1
1 0
1
1
1 2
2 1
1 2 3
3 3 2
3
1
1 2 3 3 4 5 5 4
4 5 5 4 3 3 2 1
1 2 3 3 4 5 5 4 5 7
5 7 8 7 7 8 7 5 4 5
8 7 7 8 7 5
5 4 3 3 2 1
Level 0
Level 1 from 1 to 1
Level 2 from 1/2 to 2/1
Level 3 from 1/3 to 3/1
Level 4 from 1/4
to 4
Level 5
from 1/5 to 5
Level 1
a
a
Level 2
Level 3
(Or, m and n are Relatively Prime)

6. If /m n and /m n  are consecutive fractions at any stage of the construction, we have
1m n mn   . We can prove it by induction.
Basis: Initially,
0 1
and
1 0
m m
n n

 

. Thus 1 1 0 0 1m n mn      
Hypothesis: Let, 1m n mn   is true for two consecutive fractions / and /m n m n  at
any stage of the Stern-Brocot tree.
Induction: Consider, we have new mediant ( ) / ( )m m n n   between / and /m n m n .
( ) ( ) 1
( ) ( ) 1 (Proved)
m m n m n n mn m n mn mn m n mn
m n n m m n m n m n mn m n m n mn
              
                    
If
m m
n n



and if all values are nonnegative, it’s easy to verify that
m m m m
n n n n
 
 
 
.
1
( ) ( ) ( )
m m m mn m n mn mn m n mn
n n n n n n n n n n n n
        
   
      
.
Again,
1
( ) ( ) ( )
m m m m n m n mn m n m n mn
n n n n n n n n n n n n
             
   
          
.
Though 0n  and 0n  . we can write,
1 1
0 and 0
( ) ( )n n n n n n
 
   
which concludes
the prove
m m m m
n n n n
 
 
 
.
Is there any positive fraction /a b with a _|_ b possibly omitted from Stern-Brocot tree?
Let, /a b is a positive fraction. Thus,
0 1
1 0
m a m
n b n
 
      
. The construction forms
( ) / ( )m m n n   and there are three cases. Either ( ) / ( ) /m m n n a b    and we win;
or ( ) / ( ) /m m n n a b    and we can set ,m m m n n n     ;
or( ) / ( ) /m m n n a b    and we can set ,m m m n n n       . The process can’t go
infinitely because of the conditions
0 0 (1) and 0 0 (2)
a m m a
an bm bm an
b n n b

          

  .
 (1) ( ) (2) ( ) ( ) ( )( )
( ) ( )
( )( )
m n m n m n an bm m n bm an m n m n
am n bmm ann bmn bmm amn bm n ann m n m n
am n bmn amn bm n m n m n
m n a b mn a b m n m n
a b m n mn m n m n
a b m
                     
                    
            
          
         
    ; [ 1]n m n m n mn     
Either m or n or morn increases at each step, so we must win after at most (a + b) steps.
m/n, m+m'/n+n', m'/n'
Proof 1:
Proof 2:
Proof 3:
m/n < a/b < m'/n' and the inequality gradually comes closer and closer to equality to a/b as
SKIP the
Calculation
(i.e., Move to the
Right SubTree)
(Move to Left Subtree)
*Important *
*Important*

7. The Farey series of order N, denoted by NF , is the set of all reduced fractions between 0
and 1 whose denominators are N or less, arranged in increasing order. For example, if
6N  we have 6
0 1 1 1 1 2 1 3 2 3 4 5 1
, , , , , , , , , , , , .
1 6 5 4 3 5 2 5 3 4 5 6 1
F 
We can obtain NF in general by starting with 1
0 1
,
1 1
F  and then inserting mediants
whenever it’s possible to do so. To obtain NF from 1NF  , we simply insert the fraction
( ) /m m N between consecutive fractions / and /m n m n  of 1NF  whose denominators
sum equals to N. For example, it’s easy to obtain 7F from the elements of 6F by inserting
1 2 6
, , ,
7 7 7
 according to the stated rule:
7
0 1 1 1 1 2 1 2 3 1 4 3 2 5 3 4 5 6 1
, , , , , , , , , , , , , , , , , , .
1 7 6 5 4 7 3 5 7 2 7 5 3 7 4 5 6 7 1
F 
When N is prime, 1N  new fractions will appear, but otherwise we will have fewer than
1N  factors, because this process generates only numerators that are relatively prime to
N. NF is a subtree of the Stern-Brocot tree, obtained by pruning off unwanted branches.
It follows that 1m n mn   whenever / and /m n m n  are consecutive elements of a
Farey series.
Let’s used the letter L and R to stand for going down to the left or right branch as we
proceed from the root of the Stern-Brocot tree to a particular fraction; then a string of L’s
and R’s uniquely identifies a place in the tree. For instance, LRRL means that we go left
from
1
1
down to
1
2
, then right to
2
3
, then right to
3
4
, then left to
5
7
. We can consider
LRRL to represent
5
7
. Every positive fraction gets represented in this way as a unique
string of L’s and R’s.
Suppose, we are given a string of L’s and R’s, we have to find out what fraction
corresponds to it in a Stern-Brocot tree. For example, f (LRRL) =
5
7
. We can maintain a
2 2 matrix to find out such fractions.
( )
n n
M S
m m
 
   
A step to left replaces n by n n and m by m m ; hence
1 1 1 1
( ) ( )
0 1 0 1
n n n n n
M SL M S
m m m m m
       
               
Similarly, when we turn right then, we replace n by n n and m by m m .
1 0 1 0
( ) ( )
1 1 1 1
n n n n n
M SR M S
m m m m m
        
                

8. Therefore, we can define L and R as 2 2 matrices.
1 1 1 0
and
0 1 1 1
L R
   
    
   
For example,
1 1 1 0 1 0 1 1
( )
0 1 1 1 1 1 0 1
M LRRL LRRL
    
      
    
2 1 1 1 3 4
1 1 1 2 2 3
    
     
    
The ancestral fraction that enclose LRRL =
2 3 5
3 4 7



.
Consider, we are given positive integers m and n with m _|_ n, we have to find out the
string of L’s and R’s that corresponds to /m n in Stern-Brocot tree. We can do it using
‘binary search’ on Stern-Brocot tree:
:S I
while / ( )m n f S do
if / ( )m n f S then (output(L); :S SL )
else (output(R); R SR ) .
This outputs the desired string of L’s and R’s. For example, if given / 5/ 7m n  , then the
algorithm works according to the following way:
Pass 1 : m/n = 5/7 < f(S) = 1, Output : L.
1 1 0 1 1
0 1 1 1 2
L
  
   
 
Pass 2 : m/n = 5/7 > f(L) = 1/2 , Output : LR.
1 1 1 0 2 1 1 1 2
0 1 1 1 1 1 2 1 3
LR
     
       
    
Pass 3 : m/n = 5/7 > f(LR) = 2/3 , Output : LRR.
2 1 1 0 3 1 2 1 3
1 1 1 1 2 1 3 1 4
LRR
     
       
    
Pass 4 : m/n = 5/7 < f(LRR) = 3/4 , Output : LRRL.
3 1 1 1 3 4 2 3 5
2 1 0 1 2 3 3 4 7
LRRL
     
       
    
.
There is another algorithm available to find out L’s and R’s of particular fraction in
Stern-Brocot tree using following property
( ) ( ), where
( ) ( ), where
m m n
f RS f S m n
n n
m m
f LS f S m n
n n m

   
   

Determine the fraction corresponding to the sequence: LRRL without drawing the Stern Brocot tree
read the
Algorithm
of the
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9. i.e. we can transform the binary search algorithm to the following matrix-free procedure.
while m n do
if m < n then (output(L); :n n m  )
else (output(R); :m m n  )
For instance, given m/n = 5/7, we have successively
Pass 1 Pass 2 Pass 3 Pass 4
m = 5 5 5 – 2 = 3 3 – 2 = 1 1
n = 7 7 – 5 = 2 2 2 2 – 1 = 1
output : L R R L
☺Good Luck☺
Write an Algorithm to generate the the L-R sequence to locate a given Fractional value in the Stern-Brocot tree.
Demonstrate the algorithm using an example.
*include these 2 lines in the algo ==> if (gcd(m,n) > 1) then output ("Not Present in Tree, because m and n are not relatively prime"); return;
else if (m=n==1) output ("Found at Root!"); return;
Q: Determine the level of the Stern Brocot tree that contains the fraction 5/7
Ans: Find the LR sequence as above: LRRL ... this means level no. = 4+1 = 5