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Help needed to find a mistake

  • Thread starter mindauggas
  • Start date
  • #1
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Homework Statement



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

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

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


The Attempt at a Solution



My solution:

(0) [itex]\frac{M_{0}}{\sqrt{1-\frac{c^{2}}{v^{2}}}}[/itex]

(1) [itex]\frac{M_{0}}{\sqrt{\frac{c^{2}}{c^{2}}-\frac{c^{2}}{v^{2}}}}[/itex]

(2) [itex]\frac{M_{0}}{\sqrt{\frac{c^{2}-v^{2}}{c^{2}}}}[/itex]

(3) [itex]\frac{M_{0}}{\frac{\sqrt{c^{2}-v^{2}}}{c}}[/itex]

But this is the same as:

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

So it is equal to

(5) [itex]\frac{M_{0}}{\sqrt{c^{2}-v^{2}}*c}[/itex]

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

Where did I go wrong?
 
Last edited:

Answers and Replies

  • #2
1,941
238
Is it [itex] \frac {v^2} {c^2} [/itex] or [itex] \frac {c^2} {v^2} [/itex]

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

[tex]\frac{M_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/tex]


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
127
0
Sorry, i made a mistake writing in the previous post ... it should be [itex] \frac {v^2} {c^2} [/itex] in (1) and (2).

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

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

And then one has to multiply by the riciprocal [itex]\frac{1}{c}[/itex] and get (5)???
 
  • #4
Put a little emphasis on the divisor in (3). Re-written as such, (3) is really,

[tex]
\frac{M_0}{\left(\frac{\sqrt{c^2-v^2}}{c} \right)}
[/tex]

Does that help?
 
  • #5
127
0
Put a little emphasis on the divisor in (3). Re-written as such, (3) is really,

[tex]
\frac{M_0}{\left(\frac{\sqrt{c^2-v^2}}{c} \right)}
[/tex]

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 thats left to ask is WHY are the expresions not equivalent? In other words WHY is
[tex]
\frac{M_0}{\left(\frac{\sqrt{c^2-v^2}}{c} \right)}
[/tex]

the case?
 
  • #6
eumyang
Homework Helper
1,347
10
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)
[tex]\frac{\frac{a}{b}}{c} = \frac{a}{b} \cdot \frac{1}{c} = \frac{a}{bc}[/tex]
but
[tex]\frac{a}{\frac{b}{c}} = \frac{a}{1} \cdot \frac{c}{b} = \frac{ac}{b}[/tex]
(assuming that b and c are nonzero, of course)
 
  • #7
127
0
Useful realization ... thank's to everyone ...
 

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