Inline formulas:
\n\\(\\textbf{Example 2.}\\)Let \\(m\\), \\(n\\neq 0\\), \\(k\\), and \\(r\\neq 0\\) be whole numbers. Explain why \\(\\frac{m(\\frac{k}{r})}{n}=\\frac{mk}{nr}\\).\n<\/p>\n
DisplayMath with inserted text:
\n\\[ \\textbf{Step 1:}\\text{By Definition 1, in Module 1, }(nr)(\\frac{m(\\frac{k}{r})}{n})=r(n(\\frac{m(\\frac{k}{r})}{n}))=r({m(\\frac{k}{r})})=m({r(\\frac{k}{r})})=mk\\]\n
DisplayMath with inserted text:
\n\\[\\textbf{Step 2:}\\text{ By Definition 1, in Module 1, }(nr)(\\frac{mk}{nr})=mk\\]\nInline Formulas w\/o textbf:
\nStep 3:<\/b> From steps 1 and 2 we have, \\((nr)(\\frac{m(\\frac{k}{r})}{n})=(nr)(\\frac{mk}{nr})\\) which can only happen if \\(\\frac{m(\\frac{k}{r})}{n}=\\frac{mk}{nr}\\).\n<\/p>\n
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\\textbf{Exercise 1.}
Geometrically prove why $A(R(5,b))=A(R(1,5b))$.\\\\
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\\textbf{Exercise 2.}
Geometrically prove why $A(R(7,b))=A(R(1,7b))$.\\\\
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\\textbf{Exercise 3.}
Let $n$ be any whole number. Explain why $A(R(n,b))=A(R(1,nb))$.\\\\
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\\textbf{Example 2.}
Geometrically prove why $A(R(\\frac{1}{5},b))=A(R(1,\\frac{b}{5}))$.\\\\
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\\textbf{Step 1:} The figure below shows that $R(1,b)$ can be divided into five rectangles that are all congruent to $R(\\frac{1}{5},b))$. This implies $A(R(1,b))=5A(R(\\frac{1}{5},b))$.\\\\<\/p>\n
Geometrically prove why $$A(R(\\frac{1}{5},b))=A(R(1,\\frac{b}{5}))$$<\/p>\n
Geometrically prove why $A(R(\\frac{1}{5},b))=A(R(1,\\frac{b}{5}))$<\/p>\n","protected":false},"excerpt":{"rendered":"
Inline formulas: \\(\\textbf{Example 2.}\\)Let \\(m\\), \\(n\\neq 0\\), \\(k\\), and \\(r\\neq 0\\) be whole numbers. Explain why \\(\\frac{m(\\frac{k}{r})}{n}=\\frac{mk}{nr}\\). DisplayMath with inserted text: \\[ \\textbf{Step 1:}\\text{By Definition 1, in Module 1, }(nr)(\\frac{m(\\frac{k}{r})}{n})=r(n(\\frac{m(\\frac{k}{r})}{n}))=r({m(\\frac{k}{r})})=m({r(\\frac{k}{r})})=mk\\] DisplayMath with inserted text: \\[\\textbf{Step 2:}\\text{ By Definition 1, in … Continue reading