11.11: Algebra Connections
- Page ID
- 83039
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In an advanced algebra course students are often asked to work with complicated expressions like:
\[\frac{\frac{1}{x} + 1}{\frac{3}{x}} \nonumber \]
We can make it look friendlier by using the key fraction rule, exactly the same technique we used in the chapter on “Dividing Fractions: Invert and Multiply.” In this example, let us multiply the numerator and denominator each by 
. (Do you see why this is a good choice?) We obtain:
\[\frac{(\frac{1}{x} + 1) \cdot x}{(\frac{1}{x}) \cdot x} = \frac{1 + x}{3}, \nonumber \]
and \(\frac{1+x}{3}\) is much less scary.
Notice that expressions like
\[\frac{1}{x} \nonumber \]
cannot be rewritten as a decimal. Expressions like this arise in numerous applications, so it is important for math and science students to be able to work with fractions in fraction form, without always resorting to converting to decimals.
As another example, given:
\[\frac{\frac{1}{a} - \frac{1}{b}}{ab}, \nonumber \]
one might find it helpful to multiply the numerator and the denominator each by 
and then each by 
:
\[\frac{(\frac{1}{a} - \frac{1}{b}) \cdot a \cdot b}{(ab) \cdot a \cdot b} = \frac{b - a}{a^{2} b^{2}} \ldotp \nonumber \]
For
\[\frac{\frac{1}{(w+1)^{2}} - 2}{\frac{1}{w+1)^{2}} + 5}, \nonumber \]
it might be good to multiply numerator and denominator each by \((w+1)^{2}\). (Why?)
\[\frac{(\frac{1}{(w+1)^{2}} - 2) \cdot (w+1)^{2}}{(\frac{1}{w+1)^{2}} + 5) \cdot (w+1)^{2}} = \frac{1-2(w+1)^{2}}{1+5(w+1)^{2}} \ldotp \nonumber \]
On Your Own
Can you make each of these expressions look less scary?
\[\frac{2 - \frac{1}{x}}{1 + \frac{1}{x}}, \qquad \frac{\frac{1}{x+h} + 3}{\frac{1}{x+h}}, \qquad \frac{1}{\frac{1}{a} + \frac{1}{b}}, \qquad \frac{\frac{1}{x+a} - \frac{1}{x}}{a} \ldotp \nonumber \]