1. The problem statement, all variables and given/known data This is the problem as stated in my textbook: Prove that for a timelike interval, two events can never be considered to occur simultaneously. I thought about two different ways to go about this; both of these ways make sense to me, but I'm not sure if they both make sense to all of you/if one is significantly better/different from the other 2. Relevant equations Δs'2=∆s'2 (∆s is invariant) s^2=x^2 - (ct)^2 s'^2=x'^2 - (ct')^2 For a timelike interval, it is given that: ∆s^2<0 3. The attempt at a solution The first way I thought of doing the problem: At first I thought that I should be showing that if two events occur at (x1, t1) and (x2, t2) in system K, then there is NO frame K' such that these two events could occur at the same time (simultaneously). If this is a misunderstanding or seems tautological, then please correct me. I went about proving it as follows: ∆s^2=∆x^2 - (c∆t)^2 = ∆x'^2 - (c∆t')^2 = ∆s'^2 Where, in K', we are assuming that t2'-t1'=0 such that: ∆s^2=∆x^2 - (c∆t)^2 = ∆x'^2 = ∆s'^2 From here, it follows that as ∆s^2=∆x^2 - (c∆t)^2<0 for it to be timelike separation, this entire quantity must be negative in sign. However, the right side of the equation, ∆x'^2, must necessarily be positive. Because it is given that s^2 is invariant, it must also be true that ∆s^2 is invariant from reference frame to reference frame, and so this must be impossible because ∆s^2≠∆s'^2 The second way I tried to solve the problem: At this point I had doubts about whether my first solution made sense, and so I tried to solve it all in the context of ONE inertial frame: ∆s^2=∆x^2 - (c∆t)^2 For the interval to be timelike, ∆s^2=∆x^2 - (c∆t)^2<0; thus, if we are assuming that t1=t2, then: ∆s^2=∆x^2 - 0 CANNOT be less than 0, as it MUST be a positive number or it must be zero, neither of which allow for a timelike interval. What do all of you have to say? Are both wrong? One right one wrong? One better?