Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.



Re: The first new theorem Primes
Posted:
Sep 18, 2013 7:31 PM


On Wednesday, September 18, 2013 10:48:15 AM UTC5, Brüder des Schattens Söhne des Lichts wrote: > 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.
If I understand the statement of your result correctly, then this is a theorem of Crelle. His version allows for concatenated blocks of digits, that is things like 321321321....
Don



