Finding and Fixing Mistakes in Simplifying Algebraic Expressions

[mindauggas]

Homework Statement

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.

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?

 

[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)

 

[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)?

 

[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?

 

[mindauggas]

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?

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

Now the only thing that's 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?

 

[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)

 

[mindauggas]

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