Riemann Hypothesis Based on the complex numbers system. A complex number is in the form a+bi
The real numbers are a sub group consisting of (Reals)+0i. Of course 0i=0. So for convenience we leave of the i part when dealing with the Reals.
The intersection of i to Reals is a perpendicular line. Simply put, a 90 degree intersection.
For the Riemann zeta function, it is known that all the trivial zeros fall on every negative even integer starting at -2. This can be shown to correspond to the perpendicular intersection by adding 2 to negative 2 which equals zero. Thus the relation of an even spacing holds throughout the line of Reals. Note of course 0 is not a 0 output of the function. It has the value of negative 1/2.
Since we added to get to zero, we simply subtract this value from 0 or 0-(-1/2)=1/2 or ½ +0i.
Any non-trivial zero of the Riemann zeta function will be on the line of ½+bi, or there will exist a trivial zero that is not at a negative even real number. That of course is absurd.
As trivial as it is, from this one can easily understand that the Riemann Hypothesis is true.
For those that desire to find a more eloquent solution to the proof of RH, study the relation of the trivial zeros, and then the wave forms of the imaginary part to the real part of the non-trivial zeros.
Watch the polar version of the output, and you will easily see that the phase difference is such that for all known non-trivial zeros that any were off the line of ½+bi the phase difference will not allow a non-trivial zero to exist.
I have no doubt some will eventually be able to show the phasing of the wave forms off the line of ½+bi will not allow a non-trivial zero to exist.
To put this in a form easier to understand both the real and imaginary parts are mirrored. One mirror position is in the form of the line of the Reals. Fold along the real line, the upper and lower sections, the real part will be mirrored. The other is a line through a point in the Reals on a z axis. The imaginary part is folded up and over in the z axis direction, the point on the real line and it will match.
Again all this does is again show the perpendicular input is matched to the output,as I have stated above.