Team 2 3 5

Introductions 

The material below is original. I have created it in the year 1972.

Originally, I applied my method right away to "team  2 3 5 7" while below I consider the simpler case, namely "team  2 3 5".


Near 1 fractions

In order to compute all three  log(2)  log(3)   log(5)   we will need three different near 1 fractions  (a+1)/a  such that  a  is a natural number, and  a*(a+1) is divisible by no prime different from  2 3 5. (Such three fractions should be in a sense independent, though they tend to be so).

Thus, let's find all such near 1 fractions  (a+1)/a. Then we will select the three fractions with the largest denominators.

It's clear that numerator  a+1  or  denominator  a  must be a power of  2  or. 3  or  5  while the "other floor" of the fraction can be divisible only by the other two (or one or none) of the primes  2  3  5, and bo no other prime.


Let's start with the consecutive powers of  2. Then we get:

2/1      and      3/2

4/3      and      5/4

9/8

16/15

Now it's the turn of 3:

3/2      and      4/3

9/8      and      10/9

81/80

Finally, consider 5:

5/4      and      6/5

25/24


According to a difficult theorem, there are no more near 1 fractions as the above.

Let's list the above fractions (without repetitions):

2/1    3/2    4/3    5/4    6/5    9/8    10/9    16/15    25/24    81/80

There are just ten of them.


Computing  log(2)  log(3)  log(5)

Let's concentrate on the last three fractions of the above ten:

16/15        25/24        81/80

For the sake of brevity, let's define:

X := log(2)        Y := log(3)        Z := log(5)

and

 A := log(16/15)        B := log(25/24)        C := log(81/80)

Thus, the following three equations hold:

 4*X - Y - Z       =  A
-3*X - Y + 2*Z  = B
-4*X + 4*Y - Z  = C

It's routine to solve the above system of three equations. By adding the first and the last equation we get an equation only two variables, in Y and Z. Next, we multiply the above first equation by 3, then the second eqution by 4, and we add the so obtained two equations thus obtaining another equation in the same two variables Y Z only:

 3*Y - 2*Z  =  A+C
-7*Y + 5*Z  = 3*A + 4*B

Now we can compute  Y  and  Y -- first, we multiply the first of the two above equation by 5, and the second one by 2; then we add the two resulting equations getting  Y.  Then we go back to the above two equations, we multiply the first equation by 7, and the second by 3, etc., obtaing. Z:

Y  =  5*(A+C) + 2*(3*A + 4*B)
Z  =  7*(A+C) + 3*(3*A + 4*B)
i.e.
Y  =  11*A +  8*B  + 5*C
Z   =  16*A + 12*B + 7*C

 Furthermore, from the first of the three equations above we get:

4*X  =  Y + Z + A  =  28*A + 20*B + 12*C
hence
X  =  7*A + 5*B  + 3*C

Replacing back  X Y Z A B C  by logarithms we obtain

log(2)  =   7*log(16/15)  +  5*log(25/24)  + 3*log(81/80)
log(3)  =  11*log(16/15) +  8*log(25/24)  + 5*log(81/80)
log(5)  =  16*log(16/15) + 12*log(25/24) + 7*log(81/80)

These three additive (logarithmic) equations admit also their equivalent multiplicative style:

2  =  (16/15)^7  * (25/24)^5  * (81/80)^3

3  =  (16/15)^11 * (25/24)^8  * (81/80)^5

5  =  (16/15)^16 * (25/24)^12 * (81/80)^7


Verification 

Since we have derived the above formulas for  2 3 5  we can be reasonably(?) sure that these formulas are correct. But what if you got these equations without any explanation? How can you verify their correctness?

One could reproduce the derivation. But that's not easy as we saw, and it takes rather a long time.

What's more, there could be a derivation error, the formulas might be not true.

Thus, how can one confirm the truth of our formulas?

Fortunately, in the give case the confirmation is easy. You compute the overall exponent of primes  2 3 5 as follow:

consider product   (16/15)^7  * (25/24)^5  * (81/80)^3  (it should represnt 2) --  we see that the contribution of  2  to power  (16/15)^7  is  (2^4)^7 = 2^28; and the contribution of  2  to  (25/24)^5  is  (2^(-3))^5 = 2^(-15);  finally, the contribution of  2  to  (81/80)^3  is  (2^(-4))^3 = 2^(-12).  It follows that the contribution of  2  to the whole product  (16/15)^7  * (25/24)^5  * (81/80)^3  is

2^28 * 2^(-15) * 2^(-12)  =  2^(28 - 15 - 12)   =   2^1  =  2

Next, the contribution of  3  to  (16/15)^7  * (25/24)^5  * (81/80)^3  is the oroduct of the contributions of  3  to  (16/15)^7,  which is  (3^{-1})^7 = 3^(-7); and of the contribution to  (25/24)^5, which is  (3^{-1))^5 = 3^(-5); and to (81/80)^3, which is  (3^4)^3 = 3^12.  Thus, the contribution of  3  to the whole product is:

3^(-7) * 3^(-5) * 3^12  =  3^(-7-5+12)  =  3^0  = 1

A similar calculation shpws that also the contribution of  5  to  (16/15)^7  * (25/24)^5  * (81/80)^3  is  1  as well.

This way we proved that:

(16/15)^7  * (25/24)^5  * (81/80)^3 = 2*1*1 = 2

i.e. we have verified the formula for  2.  The verifications for  3  and  5. are similar.







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