I ' m on my phone so quoting you is annoying . 4 vector proofYour velocity can be represented as x 2 y 2 z 2 t 2 c 2 Because c is constant , as you know . When x , y , z are 0 you move through time only . The moment you add to the x y or z component , you must reduce your speed through time . If x 2 y 2 z 2 is just barely under c 2 , then your passage through time is incredibly small . If you and I both use the same coordinate system , which is required to maintain consistency , then I will pass through time slower than you as my velocity increases , and thus you will pass through time faster than me . I fail to see how you are still telling me I am wrong about this when , as I said , there are experiments that show this to be entirely correct . Next , it does disprove your argument , because you ' re saying from what I can tell If we both observe each other ' s clocks going slower , then neither clock is going slower than the other which is simply not true . You must maintain an origin and measure all passages of time or distances in space relative to that origin . I have not once said that the other clock does NOT move slower relative to me . I am saying that in all cases , the clock that is at rest or moving slower relative to the other clock when viewed from the outside will always experience more time as having passed . Finally , I fail to see how invoking general relativity somehow invalidates my claim . Special relativity and general relativity can exist at the same time , and at no point did I say that my premise was based entirely in special relativity and looking at it from any other perspective wasn ' t allowed . I am not debating what is and is not possible using different ways of looking at the problem I am stating what will end up happening at the end of the problem when the two clocks are compared , which is that the relatively faster clock will have experienced less time as having passed .