Just to add to the other answers, one might ask, "why the square root"?

The heart of it is the underlying geometry. When we ask "what is the diagonal of a square with sides $a$ and $b$, Euclid's rules eventually land us at the conclusion $\sqrt{a^{2} + b^{2}}$, which then comes from the fact that, in a flat Euclidean space, distances are given by $ds^{2} = dx^{2} + dy^{2} + dz^{2}$

For reasons that ultimately come down to "we need a way to differentiate space from time in order to preserve causaility, while also combining the two", it works out that the distance in spacetime is given by (the sign is an arbitrary choice that once made, must be adhered to, but is otherwise arbitrary -- most people in quantum field theory choose the top signs, most in general relativity choose the bottom signs)

$$ds^{2} = \pm(c\, dt)^{2} \mp dx^{2} \mp dy^{2} \mp dz^{2}$$

where, for this purposes, $c$ has to be some "conversion factor" with the units of velocity for this to make sense, and it works out that this has been shown to be the "speed of light". Now, choosing the top signs, and noting that $\frac{dt}{dt} =1$, we can divide both sides by $(c\,dt)^{2}$, and the RHS of the above becomes:

$${}^{4}v^{2} = 1 - \left(\frac{{}^{3}v}{c}\right)^{2}$$

where the right handed superscripts indicate three-and four-dimensional velocities. We can definitely be a lot more rigorous here, and there are great depths to work through, but if you're to ask "where does the square root and that minus sign come from?", the heart of it is that negative sign in the distance function, which can be taken as a core axiom of special relativity.

Just to add to the other answers, one might ask, "why the square root"?

The heart of it is the underlying geometry. When we ask "what is the diagonal of a square with sides $a$ and $b$, Euclid's rules eventually land us at the conclusion $\sqrt{a^{2} + b^{2}}$, which then comes from the fact that, in a flat Euclidean space, distances are given by $ds^{2} = dx^{2} + dy^{2} + dz^{2}$

For reasons that ultimately come down to "we need a way to differentiate space from time in order to preserve causaility, while also combining the two", it works out that the distance in spacetime is given by (the sign is an arbitrary choice that once made, must be adhered to, but is otherwise arbitrary -- most people in quantum field theory choose the top signs, most in general relativity choose the bottom signs)

$$ds^{2} = \pm(c\, dt)^{2} \mp dx^{2} \mp dy^{2} \mp dz^{2}$$

where, for this purposes, $c$ has to be some "conversion factor" with the units of velocity for this to make sense, and it works out that this has been shown to be the "speed of light". Now, choosing the top signs, and noting that $\frac{dt}{dt} =1$, we can divide both sides by $(c\,dt)^{2}$, and the RHS of the above becomes:

$${}^{4}v^{2} = 1 - \left(\frac{{}^{3}v}{c}\right)^{2}$$

where the left-sided superscripts indicate three-and four-dimensional velocities. We can definitely be a lot more rigorous here, and there are great depths to work through, but if you're to ask "where does the square root and that minus sign come from?", the heart of it is that negative sign in the distance function, which can be taken as a core axiom of special relativity.

Just to add to the other answers, one might ask, "why the square root"?

The heart of it is the underlying geometry. When we ask "what is the diagonal of a square with sides $a$ and $b$, Euclid's rules eventually land us at the conclusion $\sqrt{a^{2} + b^{2}}$, which then comes from the fact that, in a flat Euclidean space, distances are given by $ds^{2} = dx^{2} + dy^{2} + dz^{2}$

For reasons that ultimately come down to "we need a way to differentiate space from time in order to preserve causaility, while also combining the two", it works out that the distance in spacetime is given by

$$ds^{2} = \pm(c\, dt)^{2} \mp dx^{2} \mp dy^{2} \mp dz^{2}$$

where, for this purposes, $c$ has to be some "conversion factor" with the units of velocity for this to make sense, and it works out that this has been shown to be the "speed of light". Now, choosing the top signs, and noting that $\frac{dt}{dt} =1$, we can divide both sides by $(c\,dt)^{2}$, and the RHS of the above becomes:

$${}^{4}v^{2} = 1 - \left(\frac{{}^{3}v}{c}\right)^{2}$$

where the right handed superscripts indicate three-and four-dimensional velocities. We can definitely be a lot more rigorous here, and there are great depths to work through, but if you're to ask "where does the square root and that minus sign come from?", the heart of it is that negative sign in the distance function, which can be taken as a core axiom of special relativity.

Just to add to the other answers, one might ask, "why the square root"?

The heart of it is the underlying geometry. When we ask "what is the diagonal of a square with sides $a$ and $b$, Euclid's rules eventually land us at the conclusion $\sqrt{a^{2} + b^{2}}$, which then comes from the fact that, in a flat Euclidean space, distances are given by $ds^{2} = dx^{2} + dy^{2} + dz^{2}$

For reasons that ultimately come down to "we need a way to differentiate space from time in order to preserve causaility, while also combining the two", it works out that the distance in spacetime is given by (the sign is an arbitrary choice that once made, must be adhered to, but is otherwise arbitrary -- most people in quantum field theory choose the top signs, most in general relativity choose the bottom signs)

$$ds^{2} = \pm(c\, dt)^{2} \mp dx^{2} \mp dy^{2} \mp dz^{2}$$

where, for this purposes, $c$ has to be some "conversion factor" with the units of velocity for this to make sense, and it works out that this has been shown to be the "speed of light". Now, choosing the top signs, and noting that $\frac{dt}{dt} =1$, we can divide both sides by $(c\,dt)^{2}$, and the RHS of the above becomes:

$${}^{4}v^{2} = 1 - \left(\frac{{}^{3}v}{c}\right)^{2}$$

where the right handed superscripts indicate three-and four-dimensional velocities. We can definitely be a lot more rigorous here, and there are great depths to work through, but if you're to ask "where does the square root and that minus sign come from?", the heart of it is that negative sign in the distance function, which can be taken as a core axiom of special relativity.