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The list of primes is infinite.
Let us assume that they are not infinite, but that the list
is finite.
Now we form the product of all the primes in the list and let
this product be x. Then x+1 is not dividable by any of the primes in the list.
So x+1 must be prime or be the product of primes not in the list. The
assumption that the primes is finite is therefore wrong and the primes must be
infinite many.
Some stuff the reader should know
== means congruent
So if x==y mod z it means that z divides x-y
So
9==1mod2 because 2 divides 9-1, the result being 4
Therefore 9==9mod x as x
divides 0 zero times for x<>0
Also a | b means a divides b
So 2 |
16 the result being 8
Also v | b+c means v divides (b+c).
a == b mod c means a is congruent to b mod c which means that c | a-b. So it means c divides a-b.
The following theorems we will proof.
Note that all variables used here are integer and unequal to 0
1. a == a mod c
2. If a == b mod c and d | c then a == b mod d
3. a == 0 mod a
4. If a == b mod c then b == a mod c
5. If a == b mod c then a + x == b + x mod c
Where a+x == b+x mod c means (a+x) == (b+x) mod c
6. If a == b mod c then ax == bx mod c
Where ax == bx mod c means (ax) == (bx) mod c
7. If a == b mod c and d == e mod c then a+d == b+e mod c
8. If a ==b mod c and d == e mod c then ad == be mod c
9. If a == nc mod c then a == 0 mod c
Proof 1
Prove that a == a mod c
We know that c | 0 and a-a=0 , so c | a-a , so a == a mod c
Proof 2
Prove that if a == b mod c and d | c then a == b mod d
If a == b mod c then (a-b)/c = k , so ck = a-b , but c = df , so dfk = a-b , so d = (a-b)/(fk) , so d | a-b
So a == b mod d
Proof 3
Prove that a == 0 mod a
a | a , so a | a-0 as a-0 = a , so a == 0 mod a
Proof 4
Prove that if a == b mod c then b == a mod c
We know that if a == b mod c then c | a-b , but a-b = -(b-a) and c | -(b-a) , so c | b-a and thus b == a mod c
Proof 5
Prove that if a == b mod c then a + x == b + x mod c
So if a == b mod c then c | a-b , so c | (a-b +x - x) , so c | (a+x) - (b+x)
Therefore a+x == b+x mod c
Proof 6
Prove that if a == b mod c then ax == bx mod c
If a == b mod c , then c | a-b , so c | x(a-b) , so c | ax - bx , so ax == bx mod c
Proof 7
Prove that if a == b mod c and d == e mod c then a+d == b+e mod c
If a == b mod c then cv = a-b and if d == e mod c then cw = d-e
So cv + cw = ch = a-b + d-e = (a+d) - (b+e) , so c | (a+d) - (b+e)
Therefore a+d == b+e mod c
Proof 8
Prove that if a ==b mod c and d == e mod c then ad == be mod c
If a == b mod c then ad == bd mod c
If d == e mod c then e == d mod c
So then be == bd mod c and thus ad - be == bd -bd mod c and so ad - be == 0 mod c
Therefore ad - be + be == be mod c and so ad == be mod c
Proof 9
Prove that if a == nc mod c then a == 0 mod c
If a == nc mod c then ck = a - nc , so ck + nc = a , so gc = a , so c | a , so c | a-0 , so a == 0 mod c
Examples using mod theory to solve some mathematical problems
Example 1: Prove that if c | a-1 then that c | an - 1
Now if c | a-1 then a == 1 mod c , so a*a == 1*1 mod c , so a2== 1 mod c
So a*a2== 1 mod c , so a3 ==1mod c
So we can go on with the process until we have an ==1mod c
This then means that c | an -1
Example 2: Prove all primes>3 can be written as 6n+1 or 6n-1
Now let us examine an integer number q>3
We can therefore write q as either 6n , 6n+1 , 6n+2 , 6n+3 , 6n+4 or 6n+5
So q = 6n , q-1 = 6n , q-2=6n , q-3=6n , q-4=6n , q-5=6n
So the possibilities for q are as follow.
q == 0 mod 6 , q == 1 mod 6 , q == 2 mod 6 , q == 3 mod 6 , q == 4 mod 6 , q == 5 mod 6
Now let us examine the above when q prime.
If q = 5 then q = 6*1 - 1 and if q = 7 then q = 6*1 + 1.
In general we see that if q prime that 6 | q for q == 0 mod 6 and this cannot be as q is prime and 6 is not
If q == 2 mod 6 with q prime we have q == 2 mod 2 as 2 | 6 . As 0== -2 mod 2 we have q+0 == 2 - 2 mod 2
So q == 0 mod 2 and so 2 | q for q == 2 mod 6 and this cannot be as q is prime>2
If q == 3 mod 6 we have q == 3 mod 3 as 3 | 6 and so q == 0 mod 3 and so 3 | q and this cannot be as q is prime>3
If q == 4 mod 6 with q prime we have q == 4 mod 2 as 2 | 6 and so q == 0 mod 2 and so 2 | q and this cannot be as proven already
So either q == 1 mod 6 which is q = 6n+1
or q == 5 mod 6 , so q == 5-6 mod 6 , so q = -1 mod 6 , so q = 6n-1
So if q prime>3 then q = 6n+1 or q=6n-1
Case proven
In the same way we can prove that any prime q>2 can be written as either 4n+1 or 4n-1
Prove that if p>=3 and p prime that there exists infinite primes p
in the form p==3mod4
Proof follows
Let us assume that there exist only a finite number of
primes in this form. The form being p = 4n+3
Now let us form the product of all these primes excluding the prime 3 and let
the biggest prime in the form 4n+3 = R .
Let us call m the products of all
primes in the form 4n+3 the result m.
Now we form z = 4m+3 where z is
uneven. Clearly none of the primes in m divides z. So if z is prime then z
==3mod4 and z>R and we have a contradiction, so that the list of primes in
the form 4n+3 must be infinite.
If z is not prime then it has at least two factors. The factors can only be
of the form 1mod4 or/and 3mod4
If the factors are both 1mod4 we have that z = (4a+1)(4b+1) = 4c+1 = 4m+3
So (4c-4m) = 2 , so c-m = 1/2 which cannot be as both m and c are integers.
If the factors are both 3mod4 we have that z = (4a+3)(4b+3) = 4c+9 = 4c+8+1 =
4d+1 which as we have seen above cannot be.
Therefore z must have a factor 4a+1 and a factor 4b+3 , so that z =
(4a+1)(4b+3) = 4c+3 = 4m+3 so that m=c which can be. Now either the factor 4b+3
is prime or it is not. If it is prime we are finished and this prime will be
bigger than R. If the factor 4b+3 is not prime we again use the arguments we
devised until we get a factor in the form 4k+3 which is prime. We will therefore
arrive at a contradiction sooner or later that the list of primes==3mod4 we
assume is finite , cannot be finite.
The list is therefore infinite.
Proof done.
If x and y have a certain common factor, then (x+y) and (x-y) have that same common factor
Let x = vb, and y = vc where (b,c)=1
Then the common factor of x and y is v
Now x+y = v(b+c) and x-y = v(b-c). So we see that v again is a common factor.
Did you know that e^(i*z) = cos(z) + i*sin(z)
Proof: Let T = cos(z) + i*sin(z) then dT/dz = -sin(z) + i*cos(z) = i*[cos(z) + i*sin(z)] = i*T
So dT/T = i*dz , so ln(T) = i*z + k , so T = e^(i*z + k)
At z = 0 we have T = cos(0)+i*sin(0) = 1 = e^(k) , so k = 0
and we have e^(i*z) = cos(z) + i*sin(z)
Mathematics is the Queen of the Sciences - FC
Gauss
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