Rearranging terms of a relativity equation

[Mark Goldman]

This is not a homework assignment. I'm trying to follow an example in a book and I keep getting the wrong answer. I'd appreciate your help. Thank you.

 

[Loren]

\(\displaystyle \frac{c}{\sqrt{c^2-v^2}}=\sqrt{\frac{c^2}{c^2-v^2}}=\sqrt{\frac{1}{\frac{c^2-v^2}{c^2}}}=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\)

 

[Mark Goldman]

Thank you very much. I guess I should have seen that myself, but it just didn't occur to me to square the numerator and put it under the radical. I sure wish the publishers of the book had splurged on ink and shown the 2 intermediate steps. Now I can continue reading the book. thanks again

 

[Deleted member 4993]

Thank you very much. I guess I should have seen that myself, but it just didn't occur to me to square the numerator and put it under the radical. I sure wish the publishers of the book had splurged on ink and shown the 2 intermediate steps. Now I can continue reading the book. thanks again

But I think some intermediate steps should be left out - that induces active interaction with learning. I am sure you will not forget those steps for rest of your life.

And think of it - but for the frugality of the publisher - you wouldn't have met us !!!

 

[Loren]

And sometimes, it pays to start with the right side and work back to the left side. Once you see that, you can reverse it's order to get from the left side to the right side.