SOLUTION: from special relativity, I can get most of the way apart from the very last step how do you simplify {{{ c/sqrt(c^2 - v^2) }}}, the answer should be {{{ 1/sqrt(1 - v^2/c^2) }}}

Algebra ->  Square-cubic-other-roots -> SOLUTION: from special relativity, I can get most of the way apart from the very last step how do you simplify {{{ c/sqrt(c^2 - v^2) }}}, the answer should be {{{ 1/sqrt(1 - v^2/c^2) }}}       Log On


   

Question 317101: from special relativity, I can get most of the way apart from the very last step
how do you simplify +c%2Fsqrt%28c%5E2+-+v%5E2%29+,
the answer should be +1%2Fsqrt%281+-+v%5E2%2Fc%5E2%29+
but I don't see how it's done
thanks
John

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
It's not much of a simplification (in my opinion) as the expression doesn't get any simpler, but here it goes...


+c%2Fsqrt%28c%5E2+-+v%5E2%29+ Start with the given expression.


+%28c%2Fc%29%2F%28sqrt%28c%5E2+-+v%5E2%29%2Fc%29+ Divide both the numerator and the denominator by 'c'


+1%2F%28sqrt%28c%5E2+-+v%5E2%29%2Fc%29+ Reduce c%2Fc to get 1.


+1%2F%28sqrt%28c%5E2+-+v%5E2%29%2Fsqrt%28c%5E2%29%29+ Rewrite c as sqrt%28c%5E2%29. Note: this implies that 'c' is a non-negative number, which it is.


+1%2F%28sqrt%28%28c%5E2+-+v%5E2%29%2F%28c%5E2%29%29%29+ Combine the lower square roots using the identity sqrt%28x%29%2Fsqrt%28y%29=sqrt%28x%2Fy%29


+1%2F%28sqrt%28%28c%5E2%29%2F%28c%5E2%29+-+%28v%5E2%29%2F%28c%5E2%29%29%29+ Break up the lower inner fraction.


+1%2F%28sqrt%281+-+%28v%5E2%29%2F%28c%5E2%29%29%29+ Reduce %28c%5E2%29%2F%28c%5E2%29 to get 1.


So +c%2Fsqrt%28c%5E2+-+v%5E2%29+=+1%2F%28sqrt%281+-+%28v%5E2%29%2F%28c%5E2%29%29%29+