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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:
$\dfrac{a}{b} = ab^{1}$, 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 rewrite $\dfrac{a}{b}$ as $ab^{1}$ and $\dfrac{c}{d} = cd^{1}$. Then, letting $x = \dfrac{a}{b}$ and $y = \dfrac{c}{d}$, we rewrite: $\dfrac{x}{y} = xy^{1} = (ab^{1})(cd)^{1})^{1}$. 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 differentlooking expressions in mathematics often turn out to be the "same thing".  

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