How do you prove that an equation is dimensionally correct?

Question

anonymous · Answer

By performing the mathematics on the units. For example, consider$$d=v_0t+\frac{ 1 }{ 2 }at^2$$Forgetting about any values and just looking at the units gives$$m=\left( \frac{ m }{ s } \right)\left( s \right) + \left( \frac{ m }{ s^2 } \right)\left( s \right)^2$$$$m = m$$Hence, the equation is dimensionally correct.

unklerhaukus · Answer

Do-not confuse units with dimensions.

$$\sf L=\left( \frac{ \sf L }{ \sf T } \right)\left( \sf T \right) + \left( \frac{ \sf L }{ \sf T^2 } \right)\left( \sf T \right)^2\
\sf L = \sf L+\sf L\
\sf L = \sf L$$

anonymous · Answer

Right you are @UnkleRhaukas. I'm getting lazy in my old age. However, as long as you are consistent in the use of base units in a single measurement system (SI, Imperial, cgs, etc.) the two approaches are equivalent. Thanks!