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# Most stupid question ever

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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]

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communitive property of multiplication/division i believe? dont hold me to it

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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]

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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]

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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]

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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]

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

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

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

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quote:
a*b^(-1) * c*d^(-1)
which is also equivalent to
a*b/c*d...

Excuse me, but... wasn''t this "equivalence" the thing that was meant to be proven?

Besides, atheist already provided a good proof...

Victor Nicollet, INT13 game programmer

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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]

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I''m pretty sure that we learned this stuff in school already, though the fancy wording might''ve only officially made its way into varsity.
but yes

a/b = (a/1) * (1/b)

so
a/b*c/d
= (a/1) * (1/b) * (c/1) * (d/1)
= (a*c) * 1/(b*d)
= (a*c) / (b*d)
which looks nicer as
= (a.c)/(b.d)

since ac/bd can be misread if you feel evil enough as
a*c/b*d = (a*c*d)/b

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Haeleyx76:
I think you mean commutative rather than associative. Fortunately matrix multiplication is associative, otherwise things would get really, really nasty.

Also your use of the word scalar is a little misleading, I know what you mean, you''re talking about a field, for which commuativity is given.

a/b * c/d = ac/bd holds for any field, but not necessarily for a group, which does not have to be commutative. In fact it holds for any commutative group.

I''m beginning to think this post is a little pointless, but if you try to view the problem from a pure perspective, taking it into a more general case, where it doesn''t have to be on the real numbers, you might as well go the whole way.

I''m pretty sure the most general system where it holds is an abelian group though, as the only things given are identity, closure, associativity, inverses and commutativity.

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