User Name Remember Me? Password

 Abstract Algebra Abstract Algebra Math Forum

 September 5th, 2014, 11:40 AM #1 Member   Joined: Jun 2014 From: Alberta Posts: 56 Thanks: 2 Revision of "Basic Algebraic Proof" My last thread must not have been clear, so I downloaded a picture. Is there anything wrong with this? Math Pic.jpg The textbook has, Math pic 2.JPG . Why does b get brought up as b^(-1) to the numerator position in the beginning of the proof and then put back to b in the denominator position. It seems unnecessary, but I would like to sure. September 10th, 2014, 11:02 PM   #2
Senior Member

Joined: Mar 2012

Posts: 294
Thanks: 88

Quote:
 Originally Posted by Mathbound My last thread must not have been clear, so I downloaded a picture. Is there anything wrong with this? Attachment 5314 The textbook has, Attachment 5315 . Why does b get brought up as b^(-1) to the numerator position in the beginning of the proof and then put back to b in the denominator position. It seems unnecessary, but I would like to sure.
By definition:

$\dfrac{a}{b} = ab^{-1}$, provided $b \neq 0$

Therefore:

$\dfrac{\left(\dfrac{a}{b}\right)}{\left(\dfrac{c} {d}\right)} = \left(\dfrac{a}{b}\right)\left(\dfrac{c}{d}\right) ^{-1}$

To continue, we need to know what:

$\left(\dfrac{c}{d}\right)^{-1}$ is, that is, we need to solve:

$\dfrac{c}{d}\cdot x = 1$ for $x$.

If:

$\dfrac{c}{d}\cdot x = 1$, then:

$d\cdot\dfrac{c}{d}\cdot x = d$, and since:

$d\cdot\dfrac{c}{d} = \dfrac{d}{1}\cdot\dfrac{c}{d} = \dfrac{cd}{d} = \dfrac{c}{1} = c$

(since $cd\cdot 1 = cd$)

We have $cx = d$, and so:

$\dfrac{1}{c}\cdot cx = \dfrac{d}{c}$ (as long as $c \neq 0$).

But:

$\dfrac{1}{c}\cdot cx = \dfrac{cx}{c} = x$, so we finally have:

$x = \left(\dfrac{c}{d}\right)^{-1} = \dfrac{d}{c}$

Note we need $d \neq 0$ for $\dfrac{c}{d}$ to EXIST, and we need $c \neq 0$ for it to be INVERTIBLE.

So, back to where we left off:

$\left(\dfrac{a}{b}\right)\left(\dfrac{c}{d}\right )^{-1} =\left(\dfrac{a}{b}\right)\left(\dfrac{d}{c}\right ) = \dfrac{ad}{bc}$

To answer your question: if we tag something with an inversion sign ($^{-1}$), we are turning DIVISION into MULTIPLICATION. The reason for this, is that multiplication is ASSOCIATIVE, but division is NOT:

$\dfrac{a}{\left(\dfrac{b}{c}\right)} \neq \dfrac{\left(\dfrac{a}{b}\right)}{c}$

Turning this into multiplication makes this clearer:

$\dfrac{a}{\left(\dfrac{b}{c}\right)} = a(bc^{-1})^{-1} = ab^{-1}(c^{-1})^{-1} = acb^{-1} = \dfrac{ac}{b}$

while:

$\dfrac{\left(\dfrac{a}{b}\right)}{c} = (ab^{-1})c^{-1} = ab^{-1}c^{-1} = a(bc)^{-1} = \dfrac{a}{bc}$

To see that these aren't always them same, consider $a = b = 1,c = 2$.

Let's look at the book's proof in greater detail:

first we re-write $\dfrac{a}{b}$ as $ab^{-1}$ and $\dfrac{c}{d} = cd^{-1}$.

Then, letting $x = \dfrac{a}{b}$ and $y = \dfrac{c}{d}$, we re-write:

$\dfrac{x}{y} = xy^{-1} = (ab^{-1})(cd)^{-1})^{-1}$.

Let's focus on the second term.

The "missing link" that your book glosses over, is that:

$(st)^{-1} = s^{-1}t^{-1}$.

To see this, note that:

$(st)(s^{-1}t^{-1}) = s(ts^{-1})t^{-1}$ (by associativity)

$= s(s^{-1}t)t^{-1}$ (by commutativity, applied to the two factors "in the middle")

$= (ss^{-1})(t^{-1}t) = 1\cdot 1 = 1$ (associativity, and definition of inverse).

So $(cd^{-1}) = c^{-1}(d^{-1})^{-1}$.

Finally, we note that $(d^{-1})^{-1} = d$, since $d^{-1}d = 1$.

So our expression is now:

$ab^{-1}c^{-1}d = adb^{-1}c^{-1} = ad((bc)^{-1}) = \dfrac{ad}{bc}$.

To answer your first question, there's nothing wrong with what you did. You need to be accustomed to the fact that different-looking expressions in mathematics often turn out to be the "same thing". Tags basic algebraic proof, revision Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post honzik Math Books 3 June 11th, 2012 12:49 PM SedaKhold Calculus 0 February 13th, 2012 11:45 AM The Chaz Calculus 1 August 5th, 2011 09:03 PM omoplata Applied Math 4 June 3rd, 2011 02:53 PM katie0127 Advanced Statistics 0 December 3rd, 2008 01:54 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      