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Topic:
The first new theorem Primes
Replies:
3
Last Post:
Sep 19, 2013 4:11 AM




The first new theorem Primes
Posted:
Sep 18, 2013 11:48 AM


The first new theorem is as follows:
Theorem 1: For any given odd integer that does not contain the integer 5 as a factor, there exists an infinite number of integers (consisting only of digit 9's, e.g., 999999.) which contain this given odd integer as a factor.
The informal proof of this theorem is the following:
Let's designate "B" as any odd integer that does not contain the integer 5 as a factor , and "A" as a slightly larger integer. Furthermore we will require that A and B have no prime factor in common. If we divide A by B, we will obtain a quotient consisting of a integer part and a decimal part. We will designate the integer part of the quotient as "Q" and the decimal part of the quotient as "D". We will then multiply the decimal part of the quotient by B, in order to obtain an integer remainder, which we will designate as "R".
A/B = Q + D and
D(B) = R
Another way to find the value of the integer remainder is to subtract Q(B) from A
Eqtn. (1) A1 Q1(B)= R1
D will always be a repeating decimal. If we multiply our original A by ten, raised to an integer power equal to the number of digits in one repetition cycle of D, (designate this new value as A2) and divide A2by B, we will obtain a new D (call it D2), and a new R (call it R2), which are the same values as our original R and D respectively. Of course, we will have a different value for Q (call it Q2).
A2/B = Q2 + D2 and
D2(B) = R2
Another way to find the value of the integer remainder is to subtract Q(B) from A
Eqtn. (2) A2 Q2(B)= R2
At this point, I think that the introduction of a numerical example will clarify this presentation.
Let B = (37)(11)(3) = 1221
Let A1 = (19)3(2) = 13718
Then A1/B = 13718/1221 = 11.235053235053235053235053235053.
Q1 = 11
D1= .235053235053235053235053235053.
Multiplying B by D1, we obtain R1
(1221)(.235053235053235053235053235053.) = 287
R1= 287
Another way to find the value of the integer remainder is to subtract Q(B) from A
Eqtn. (1) A1 Q1(B)= R1
13718 11(1221) = 287
The number of digits in each repetition cycle of D is 6. Now, if we multiply A1 by (10)6, we obtain.
A2 = 13718000000
Dividing A2by B, we obtain.
13718000000/1221 = 11235053.235053235053235053235053.
Q2 = 11235053
and
D2 = D1 = .235053235053235053235053.
Multiplying B by D2, we obtain the same value for R as in the first step.
(1221)(.235053235053235053235053235053.) = 287
R1 = R2= 287
Another way to find the value of the integer remainder is to subtract Q2(B) from A2
Eqtn (2) A2  Q2(B)= R2
13718000000  11235053(1221) = 13718000000  13717999713 = 287
Now, we will return to the literal presentation.
Looking at the second method of obtaining the integer remainder: combining Eqtn. (1) with Eqtn (2), we can write.
A1  Q1(B)= A2  Q2(B)= R1 = R2 = R
Or
A1  Q1(B)= A2  Q2(B)
Rearranging terms
Q2(B)  Q1(B) = A2  A1
Let X represent the number of digits in a repetition cycle of D
Since we know that A2 = (10)X A1
then A2  A1 = (10)XA1  A1
and thus A2  A1 = [(10)X  1] A1
and Q2(B)  Q1(B) = [(10)X  1] A1
Dividing through by B, we obtain.
Eqtn. (3) Q2  Q1 = [(10)X  1] A1/B
We have an integer value on the left hand side of Eqtn. (3), and thus we should also have an integer value on the right hand side of Eqtn. (3). In order for this to be true, since A and B have no prime factor in common, then [(10)X  1] must be evenly divisible by B . [(10)X  1] will always be an integer consisting only of digit 9's (e.g., 999999.)
Back to our numerical example.
Since A2 = 1000000A1
A2 A1 = 999999A1
Then
Q2(B)  Q1(B) = 999999A1
Dividing through by B, we obtain.
Eqtn. (3a) Q2  Q1 = ( 999999A1)/B
11235053  11 = (999999)(13718)/1221
11235042 = 11235042
We have an integer value on the left hand side of Eqtn. (3a), and thus we should also have an integer value on the right hand side of Eqtn. (3a). In order for this to be true, since A and B have no prime factor in common, then 999999 must be evenly divisible by B .
Checking this out in the numerical example, we find.
999999/1221 = 819
An integer quotient value is produced.
Thus we have shown that our randomly chosen odd integer, 1221, is a factor of 999999.



