For this question I created another account because I'm soooooo embarrassed to ask: Is there a proof of why the method of multiplying fractions (a/b * c/d = ac/bd) actually works all time? [edited by - peter_walsh_ on March 20, 2004 10:25:57 AM]

That's a perfectly reasonable question and one that you'd only really look at in a University course. It's easy once you write it in a different form viewing division as multiplication of the multiplicative inverse. Then using the commutative/associative properties of multiplication to rearrange it.

If you're interested in this sort of maths I'd recommend Cohn's Classic Algebra. It's pretty rigorous and I have spent 2 hours working and only got through 5 pages before, but it's a lot of fun and really requires little knowledge of school maths to read if that's a problem. However it is heavy going and not to everyone's tastes.

[edited by - higherspeed on March 20, 2004 10:48:41 AM]

EDIT: Had it here but took it out due to the homework policy here. Anyways: The answer is "yes" and the way proposed by higherspeed is quite straightforward

[edited by - Atheist on March 20, 2004 10:49:14 AM]

Thanks for you comments.

In case your want to know why I asked such a question:

I'm teaching myself mathematics using the Wiley Self Teaching Guides, because I want to study computer science next year.
I finished "Practical Algebra" and right now I'm studying "Geometry and Trigonometry for Calculus". These books are very well written, with only one caveat: they sometimes omit the "why" (which is very important to me)




[edited by - peter_walsh_ on March 20, 2004 12:48:22 PM]

Well, if you still want to work it out yourself (it´s quite straighforward using higherspeeds proposal, as I already mentioned), then there´s one result you will probably need:

quote:
from: 1 = 1*1 = [b*(1/b)]*[d*(1/d)] = b*(1/b)*d*(1/d) = [b*d]*[(1/b)*(1/d)]
=> (1/b)*(1/d) = 1/(b*d)

The reason I gave you this is that this part of the proof uses a technique called "factorisation of the one" which you probably do not know if you haven´t studied math at university level.

note: 1/x is the multiplicative inverse of x, which is defined by (1/x)*x = x*(1/x) = 1.

[edited by - Atheist on March 20, 2004 1:08:17 PM]

Why open another account - is that allowed on Gamedev, having aliases?

quote:Original post by peter_walsh_
For this question I created another account because I''m soooooo
embarrassed to ask:

Is there a proof of why the method of multiplying fractions
(a/b * c/d = ac/bd) actually works all time?

[edited by - peter_walsh_ on March 20, 2004 10:25:57 AM]

just for fun, without any pretense...

if b != 0 and d != 0
then

a/b == a*b^(-1) and c/d == c*d^(-1) by definition

then a/b * c/d is equivalent to

a*b^(-1) * c*d^(-1)

which is also equivalent to

a*b/c*d...


So its always true if b and d are != 0, if not then you are trying to mutiply 2 infitite and thats indeterminate

Id guess what we really want to prove is that :

(1/c)*(1/d) == (1/(c*d)) which is a good way to prove this for SCALARS. Becuase if youre using non associative algebras thaen this proof wont hold, and you can see why :

(1/c)*(1/d) == (1/(c*d))
Multiply both sides by "c"
c*(1/c)*(1/d) == c*((1/(c*d))
since for scalars c^-1 is defined as (1/c) we have
c*(c^-1)*(1/d) == c*((1/(c*d))
and c times its multiplicative inverse is 1 so we have
1*(1/d) == c*((1/c*d))
now we multiply again in the same way by d
d*1*(1/d) == d*c*((1/(c*d))
since the identity scalar "1" times any other scalar is the scalar
d*(1/d)
again we use the inverse identity
1 == d*c*(1/(c*d))

and we get

1== (d*c)/(c*d)

Now since we are dealing with scalars (regular numbers, not matrices)

d*c == c*d
so (d*c)/(c*d) == (c*d)/(c*d)
and since a scalar times a scalar (c*d) still belongs to the same "space", the inverse property still holds true.
(c*d) * (1/(c*d)) = a scalar times its inverse which == 1

so 1 ==1 and this is pretty much a real proof. Not just saying "this equates this" without providing why.

As you can see, it doesnt apply to ALL types of algebras. Becuase c*d does not always equal d*c.


[edited by - Healeyx76 on March 23, 2004 3:50:47 PM]