(a/b) * (c/d) = ac/bd ... is there a proof?
Tom
It is a truism that
(a/b) * (c/d) = ac/bd
But is there actually a proof for this? Thanks,
--Tom
Arturo Magidin
> To sci.math --
>
> It is a truism that
>
> (a/b) * (c/d) = ac/bd
Presumably, a, b, c, and d integers? Or are they just any old real
numbers?
> But is there actually a proof for this?
What is the definition of multiplication for rationals? Is this a
corollary of the definition, or is this *the* definition?
In many settings, one *defines* the multiplication of "fractions" this
way (for example, in abstract algebra when one defines rings of
fractions and localizations). So there is no "proof" because it is
*defined* that way.
So, one can simply *define* multiplication of rationals this way.
Another is to interpret the symbol "a/b" to mean "a multiplied by the
multiplicative inverse of b" (the multiplicative inverse of b is the
unique number x such that bx=1). If you do that, then proving that (a/
b)(c/d) = (ac/bd) comes down, using associativity and commutativity of
multiplication, to proving that the multiplicative inverse of bd is
equal to the product of the multiplicative inverses of b and d.
To prove the latter, suppose that x is the unique element such that
bx=1; and y is the unique element such that dy = 1. You want to prove
that the unique element z such that (bd)z = 1 is xy. Well:
(bd)(xy) = b(d(xy)) = b(d(yx)) = b((dy)x) = b(1x) = bx = 1.
So 1/(bd) = (1/b)*(1/d). Thus,
(a/b)(c/d) = (a* (1/b))*(c*(1/d)) = a*c*(1/b)*(1/d) = (ac)*((1/b)*(1/
d)) = (ac)(1/(bd)) = (ac)/(bd).
If you have other definitions for the product, you need other
arguments.
For example, if x/y is a rational, x and y integers, y positive, then
one can define a*(x/y) to be "the unique number z such that yz = ax."
If you do that, then to prove that (a/c)(b/d) = (ac)/(bd), you just
need to show that if you mulitply (a/c)(b/d) by bd, you'll get ac.
I'll let you do that.
--
Arturo Magidin
Bill Dubuque
>
> It is a truism that
>
> (a/b) * (c/d) = ac/bd
>
> But is there actually a proof for this?
HINT b x = a
d y = c
=> bdxy = ac
=> xy = ac/(bd)
--Bill Dubuque
Frederick Williams
>
> ... you just
> need to show that if you mulitply (a/c)(b/d) by bd, you'll get ac.
You didn't mean that. I should the last person to mention it because I
write such bollocks in newsgroups, but the OP may be a beginner and
easily confused.
Arturo Magidin
wrote:
> Arturo Magidin wrote:
>
> > ... you just
> > need to show that if you mulitply (a/c)(b/d) by bd, you'll get ac.
>
> You didn't mean that.
Thanks; meant "multiply (a/b)(c/d) by bd, you get ac."
--
Arturo Magidin
achille
You cannot proof this because this is part of
definition of multiplication between fractions.
Since two fractions a/b and a'/b' are considered to be
equal when ab' = ba', what you really need to check is
the multiplication defined this way is well defined.
ie if a/b = a'/b' AND c/d = c'/d' then ac/bd = a'c'/b'd'
Proof:
a/b = a'/b' <=> ab' = a'b
AND c/d = c'/d' <=> cd' = c'd
=> (ac)(b'd') = (ab')(cd') = (a'b)(c'd) = (a'c')(bd)
<=> ac/bd = a'c'/b'd'.
Perverse 19 mathematics
2) AOL email address
3) Fuckwit.