# [seqfan] Relativistic addition of fractions (by young Einstein)

Tomasz Ordowski tomaszordowski at gmail.com
Sat Oct 2 16:03:57 CEST 2021

```SOLVING THE MYSTERY FOR INSIDERS

Let n,x,y be natural numbers.

For n > 0, 1/n = 1/(n+x) + 1/(n+y) if and only if xy = n^2.
So the number of solutions x,y with x <= y is (d(n^2)+1)/2 for n > 0.
Cf. A018892 - OEIS <https://oeis.org/A018892> (see the name).

For n > 0, arctan(1/n) = arctan(1/(n+x)) + arctan(1/(n+y)) if and only if
xy = n^2+1.
So the number of solutions x,y with x < y is d(n^2+1)/2 for n > 0.
Cf. A147810 - OEIS <https://oeis.org/A147810> (see the last comment).

For n > 1, artanh(1/n) = artanh(1/(n+x)) + artanh(1/(n+y)) if and only if
xy = n^2-1.
So the number of solutions x,y with x < y is d(n^2-1)/2 for n > 1.
Cf. A129296 - OEIS <https://oeis.org/A129296> (see the formula). Q.E.D.

Thanks for the attention and patience!
Disappointed with no response,
T. Ordowski

niedz., 26 wrz 2021 o 15:24 Tomasz Ordowski <tomaszordowski at gmail.com>
napisał(a):

>
> P.S. The relationship with the hyperbolic function tanh(x)
> and its inverse artanh(x) should be noted at the end.
> The method used in SR. Physicists know this well.
>
> The relativistic formula w = (u+v)/(1+uv) = tanh(artanh(u)+artanh(v)) with
> c = 1.
> The equation artanh(1/x) + artanh(1/y) = artanh(1/z) in natural numbers n
> > 1.
>  is equivalent to the equation (x-1)(y-1)(z+1) = (x+1)(y+1)(z-1). Nice!
>
> I waited for someone to notice this on the SeqFan forum.
>
> T. Ordowski
>
> sob., 25 wrz 2021 o 09:24 Tomasz Ordowski <tomaszordowski at gmail.com>
> napisał(a):
>
>>
>> SUPPLEMENT. Two double identities.
>> Classic: 1/n = 1/(2n) + 1/(2n) = 1/(n+1) + 1/(n(n+1)) for n > 0.
>> Relativistic: 1/n = 1/(2n-1) '+' 1/(2n+1) = 1/(n+1) '+' 1/(n(n+1)-1) for
>> n > 1.
>>
>> T. Ordowski
>>
>> pt., 24 wrz 2021 o 13:15 Tomasz Ordowski <tomaszordowski at gmail.com>
>> napisał(a):
>>
>>> P.S. Two examples.
>>> Classic: 1/3 + 1/6 = 1/2.
>>> Relativistic: 1/3 '+' 1/5 = 1/2.
>>> I recommend this new arithmetic!
>>> Relativistically, only proper fractions can be added.
>>> Then their sum will make physical sense (SR).
>>> However, the mathematical sense is wide.
>>>
>>> T. Ordowski
>>>
>>> czw., 23 wrz 2021 o 10:39 Tomasz Ordowski <tomaszordowski at gmail.com>
>>> napisał(a):
>>>
>>>>
>>>> Here is the well-known relativistic formula for sum of velocities:
>>>> w = (u+v)/(1+uv/c^2). Let the constant c=1 (Planck units).
>>>> Note that F(u,v) = (u+v)/(uv+1) = F(1/u,1/v).
>>>> Let F(1/n,1/m) = 1/k (unit fractions),
>>>> so (nm+1)/(n+m) is an integer.
>>>>
>>>> Let x > 0 be an integer.
>>>> I noticed that for a fixed natural n > 1,
>>>> the quotient (nx+1)/(n+x) is an integer y
>>>> if and only if (n-y)|(ny-1) with 0 < y < n.
>>>> The number of solutions y
>>>> is d(n^2-1)/2 for n > 1.
>>>> Cf. A129296 - OEIS <https://oeis.org/A129296>
>>>> (see FORMULA).
>>>>
>>>> Classic (non-relativistic) version.
>>>> Let x >= 0 be an integer.
>>>> Theorem: for a fixed natural n > 0,
>>>> the quotient (nx)/(n+x) is an integer y
>>>> if and only if (n-y)|(ny) with 0 <= y < n.
>>>> The number of solutions y
>>>> is (d(n^2)+1)/2 for n > 0.
>>>> See the first comment
>>>> on A018892 - OEIS <https://oeis.org/A018892>
>>>> (see the NAME).
>>>>
>>>> Best regards,
>>>>
>>>> Thomas Ordowski
>>>> _________________________
>>>> The relativistic sum of fractions
>>>> n/N '+' m/M = (Mn+Nm)/(NM+nm).
>>>> In the school formula, nm disappears,
>>>> but young Einstein added relativistically...
>>>> Alternate arithmetic and new number theory!
>>>>
>>>>

```