Simplifying square root

[lioric]

How is it equal to v in the end?
I'm sorry for asking such questions. But I'm just trying to understand

 

[fresh_42]

Because addition is associative and commutative and $c^2≠0$. For short: just calculate the left hand side.

 

[DrClaude]

Can you simplify the denominator?

 

[lioric]

Can you simplify the denominator?

Dear god I must be blind
v^2-v^2 = 0
c^2 and c^2 cancels
And all that's left is a rooted v^2 which cancels itself

But look at this

This is how the same solution is illustrated in another book.
(1-u^2/c^2) cancels each other
The u^2 subtracts each other in the denominator like in the previous sollution
Which leaves a rooted u^2 / c^2
How does that simplify?

 

[DrClaude]

Which leaves a rooted u^2 / c^2
How does that simplify?

Look to the left of the square root sign...

 

[lioric]

Look to the left of the square root sign...

I can see the c root What does it mean?

 

[DrClaude]

I can see the c root What does it mean?

No, it's c times the root.

 

[lioric]

No, it's c times the root.

So are you saying that the large root will cancel the squares of u^2 / c^2 making it like u / c * c/1
and c and c cancels?

 

[DrClaude]

So are you saying that the large root will cancel the squares of u^2 / c^2 making it like u / c * c/1
and c and c cancels?

Yes. Note that the author there takes only the positive root, while in the OP the two roots are kept.

 

[lioric]

Yes. Note that the author there takes only the positive root, while in the OP the two roots are kept.

Thank you very much

 

[Mark44]

View attachment 96291
How is it equal to v in the end?
I'm sorry for asking such questions. But I'm just trying to understand

This is actually incorrect. On the right side it should be |v|, not $\pm v$. In other words, the square root evaluates to a single nonnegative number, not two numbers.