Question about this unit conversion principle (multiplying by "1")

[member 731016]

Homework Statement

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Relevant Equations

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Suppose I want to convert 100,000 metric Tonnes to kilograms, then I would perform, a unit cancellation:

Given that 1 t = 1000 kg

$100,000 t \times \frac{1000~kg}{1 t} = 1 \times 10^8 kg$, however, why are we allowed to multiply the 100,000 by that?

My reasoning is,
$1 t = 1000 kg$
$1 = \frac{1000 kg}{1 t}$ therefore multiplying by $100,000 t$ by $\frac{1000 kg}{1 t}$ is the same as multiplying by 1

Does someone please know whether my reasoning is correct?

Many thanks!

 

[Frabjous]

Yes

 

[member 731016]

Yes

Thank you for your reply @Frabjous!

 

[hmmm27]

Suppose I want to convert 100,000 metric Tonnes to kilograms, then I would perform, a unit cancellation:

Given that 1 t = 1000 kg

$100,000 t \times \frac{1000~kg}{1 t} = 1 \times 10^8 kg$, however, why are we allowed to multiply the 100,000 by that?

My reasoning is,
$1 t = 1000 kg$
$1 = \frac{1000 kg}{1 t}$ therefore multiplying by $100,000 t$ by $\frac{1000 kg}{1 t}$ is the same as multiplying by 1

Does someone please know whether my reasoning is correct?

No, it is not. Treat the dimensions in the equation as you would unknown variables.

 

[Frabjous]

No, it is not. Treat the dimensions in the equation as you would unknown variables.

You need to expand on that.

 

[hmmm27]

$100,000t \times 1 = 100,000t$ which is incorrect not the answer to the question.

$\frac t t=1$ as a next step yields the correct answer.

 

[DaveE]

No, it is not. Treat the dimensions in the equation as you would unknown variables.

How is that different?

I guess you mean something like this?
$ax = by \Rightarrow x = \frac{b}{a} y$
then substitute to get $cx = c (\frac{b}{a} y) = (c \frac{b}{a}) y$
where $a,b,c$ are constants and $x, y$ are units (variables?)

This is instead of
$ax = by \Rightarrow 1 = \frac{by}{ax}$
$cx = cx (1) = cx (\frac{by}{ax})= (c \frac{b}{a}) y$ as others suggested.

I feel like I'm missing your point here.

 

[phinds]

You can multiply 100,000t by 1 and you do indeed get 100,000t. So what? That not what is sought. What is sought is to get a final answer in kg, not in t. To do that you need to multiply 100,000t by 1000Kg/t to get 10E8Kg. Units matter.

 

[hmmm27]

My point is the OP used "reasoning" which, while useful in creating or validating a conversion factor, doesn't solve the equation by itself whereas, after including it in the calculation, simple cancellation does :
$100,000 \cancel t \times \frac{1,000kg}{\cancel t} = 100,000,000kg$

 

[Lnewqban]

$1 = \frac{1000 kg}{1 t}$

I believe that such equation is not correct.
$1000~kg/t$ is simply a rate, a proportion.
It is used only because it is useful in mathematical operations, where it can be cancelled to obtain the desired units.

IMHO, it is not different from $3600~s/h$ or $1000~km/m$

 

[Frabjous]

I believe that such equation is not correct.
$1000~kg/t$ is simply a rate, a proportion.
It is used only because it is useful in mathematical operations, where it can be cancelled to obtain the desired units.

IMHO, it is not different from $3600~s/h$ or $1000~km/m$

Is there a situation where assuming the equation is correct will get you in trouble?
I agee that it is not different from 3600s/h or 1000m/km.

 

[hmmm27]

Is there a situation where assuming the equation is correct will get you in trouble?
I agee that it is not different from 3600s/h or 1000m/km.

It is not incorrect if used as a conversion factor ; but the OP specifically asked "Are we allowed to multiply like that"... then proceeded to not bother, or at least not show the bother.

 

[Frabjous]

It is not incorrect if used as a conversion factor ; but the OP specifically asked "Are we allowed to multiply like that"... then proceeded to not bother, or at least not show the bother.

I am looking for the substantive reason that you believe it is sometimes wrong.

 

[hmmm27]

I am looking for the substantive reason that you believe it is wrong.

That I believe what is wrong ? That $100,000t = 100,000t$ is not a useful answer for "convert from tonnes to kg's" ?

 

[Lnewqban]

Is there a situation where assuming the equation is correct will get you in trouble?
I agree that it is not different from 3600s/h or 1000m/km.

No, unless it is taken out of context.
It may be mathematically correct, but I don't know enough to see any value in something like
$1=1~kilo-banana/1000~bananas$.

 

[Frabjous]

No, unless it is taken out of context.
It may be mathematically correct, but I don't know enough to see any value in something like
$1=1~kilo-banana/1000~bananas$.

Obviously you have never studied economics on the Planet of the Apes. Low utility situations do not invalidate a concept.

There are known pitfalls when dimensionality comes into play. For example, the addition of non-dimensional numbers or the dimensional equivalence of torque and energy. We struggle through.

 

[member 731016]

Thank you for your replies @hmmm27, @Frabjous , @DaveE, @phinds and @Lnewqban !

I understand now :)