# Help needed to find a mistake

1. Nov 16, 2011

### mindauggas

1. The problem statement, all variables and given/known data

It is asked to write $\frac{M_{0}}{\sqrt{1-\frac{c^{2}}{v^{2}}}}$ in lowest terms.

The answer that's given is: $\frac{M_{0}*c\sqrt{c^{2}-v^{2}}}{c^{2}-v^{2}}$

The problem is that I don't get the correct answer.

3. The attempt at a solution

My solution:

(0) $\frac{M_{0}}{\sqrt{1-\frac{c^{2}}{v^{2}}}}$

(1) $\frac{M_{0}}{\sqrt{\frac{c^{2}}{c^{2}}-\frac{c^{2}}{v^{2}}}}$

(2) $\frac{M_{0}}{\sqrt{\frac{c^{2}-v^{2}}{c^{2}}}}$

(3) $\frac{M_{0}}{\frac{\sqrt{c^{2}-v^{2}}}{c}}$

But this is the same as:

(4) $\frac{M_{0}}{\sqrt{c^{2}-v^{2}}}$ / c

So it is equal to

(5) $\frac{M_{0}}{\sqrt{c^{2}-v^{2}}*c}$

And I will not get rid of c in the denominator ...

Where did I go wrong?

Last edited: Nov 16, 2011
2. Nov 16, 2011

### willem2

Is it $\frac {v^2} {c^2}$ or $\frac {c^2} {v^2}$

your step (2) doesn't follow from (1) , but if the initial expression contains $\frac {v^2} {c^2}$ it would be valid. The answer you gave also needs the initial expression to be

$$\frac{M_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$$

from step (3) to step (4) you make a mistake.

you have to multiply the divisor and the dividend with c, and then you won't end up with (4)

3. Nov 17, 2011

### mindauggas

Sorry, i made a mistake writing in the previous post ... it should be $\frac {v^2} {c^2}$ in (1) and (2).

What botther me actually is the (4) and (5).

Isn't (4) $\frac{M_{0}}{\sqrt{c^{2}-v^{2}}}$ / $\frac{c}{1}$ ???

And then one has to multiply by the riciprocal $\frac{1}{c}$ and get (5)???

4. Nov 17, 2011

### TheoMcCloskey

Put a little emphasis on the divisor in (3). Re-written as such, (3) is really,

$$\frac{M_0}{\left(\frac{\sqrt{c^2-v^2}}{c} \right)}$$

Does that help?

5. Nov 17, 2011

### mindauggas

Well of course it helps, I tried it myself and if this is the case then everything turns out right.

Now the only thing thats left to ask is WHY are the expresions not equivalent? In other words WHY is
$$\frac{M_0}{\left(\frac{\sqrt{c^2-v^2}}{c} \right)}$$

the case?

6. Nov 17, 2011

### eumyang

No, (3) does not equal (4). Order of operations come into play. You're interpreting (3) as a/(b/c) and (4) as (a/b)/c, but
(a/b)/c ≠ a/(b/c)
$$\frac{\frac{a}{b}}{c} = \frac{a}{b} \cdot \frac{1}{c} = \frac{a}{bc}$$
but
$$\frac{a}{\frac{b}{c}} = \frac{a}{1} \cdot \frac{c}{b} = \frac{ac}{b}$$
(assuming that b and c are nonzero, of course)

7. Nov 17, 2011

### mindauggas

Useful realization ... thank's to everyone ...