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September 5th, 2014, 11:40 AM   #1
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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.
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September 10th, 2014, 11:02 PM   #2
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Quote:
Originally Posted by Mathbound View Post
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".
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