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Mathematics LibreTexts

20.5: Area of solid rectangles

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    23714
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    Theorem \(\PageIndex{1}\)

    A solid rectangle with sides \(a\) and \(b\) has area \(a\cdot b\).

    Proof

    Suppose that \(\mathcal{R}_{a,b}\) denotes the solid rectangle with sides \(a\) and \(b\). Set

    \(s(a,b)=\text{area } \mathcal{R}_{a,b}.\)

    By definition of area, \(s(1,1)=\text{area }(\mathcal{K})=1\). That is, the first identity in the algebraic lemma holds.

    截屏2021-03-03 上午8.56.29.png

    Note that the rectangle \(\mathcal{R}_{a+b,c}\) can be subdivided into two rectangle congruent to \(\mathcal{R}_{a,c}\) and \(\mathcal{R}_{b,c}\). Therefore, by Proposition 20.4.2,

    \(\text{area }\mathcal{R}_{a+b,c}=\text{area } \mathcal{R}_{a,c}+\text{area } \mathcal{R}_{b,c}\)

    That is, the second identity in the algebraic lemma holds. The proof of the third identity is analogues.

    It remains to apply the algebraic lemma.

    Lemma \(\PageIndex{1}\) Algebraic lemma

    Assume that a function \(s\) returns a nonnegative real number \(s(a,b)\) for any pair of positive real numbers \((a,b)\) and it satisfies the following identities:

    \(\begin{aligned} s(1,1)&=1; \\ s(a,b+c)&=s(a,b)+s(a,c) \\ s(a+b,c)&=s(a,c)+s(b,c)\end{aligned}\)

    for any \(a,b,c>0\). Then

    \(s(a,b)=a\cdot b\)

    for any \(a,b>0\).

    The proof is similar to the proof of Lemma 14.4.1.

    Proof

    Note that if \(a>a'\) and \(b>b'\) then

    \[s(a,b)\ge s(a',b').\]

    Indeed, since \(s\) returns nonnegative numbers, we get that

    \(\begin{aligned} s(a,b)&=s(a',b)+s(a-a',b)\ge \\ &\ge s(a',b)= \\ &\ge s(a',b')+s(a',b-b')\ge \\ &\ge s(a',b').\end{aligned}\)

    Applying the second and third identity few times we get that

    \(\begin{aligned} s(a,m\cdot b)=s(m\cdot a,b)=m\cdot s(a,b)\end{aligned}\)

    for any positive integer \(m\). Therefore

    \(\begin{aligned} s(\tfrac kl,\tfrac mn)&=k \cdot s(\tfrac 1l,\tfrac mn)= \\ &=k\cdot m \cdot s(\tfrac 1l,\tfrac 1n)= \\ &=k\cdot m\cdot \tfrac 1l\cdot s(1, \tfrac 1n)= \\ &=k\cdot m\cdot \tfrac 1l\cdot \tfrac 1n\cdot s(1,1)= \\ &=\tfrac kl\cdot\tfrac mn\end{aligned}\)

    for any positive integers \(k\), \(l\), \(m\), and \(n\). That is, the needed identity holds for any pair of rational numbers \(a=\tfrac kl\) and \(b=\tfrac mn\).

    Arguing by contradiction, assume \(s(a,b)\ne a\cdot b\) for some pair of positive real numbers \((a,b)\). We will consider two cases: \(s(a,b)> a\cdot b\) and \(s(a,b)< a\cdot b\).

    If \(s(a,b)> a\cdot b\), we can choose a positive integer \(n\) such that

    \[s(a,b)> (a+\tfrac1n)\cdot (b+\tfrac1n).\]

    Set \(k=\lfloor a\cdot n \rfloor+1\) and \(m=\lfloor b\cdot n \rfloor+1\); equivalently, \(k\) and \(m\) are positive integers such that

    \(a< \tfrac kn\le a+\tfrac1n \quad\text{and}\quad b<\tfrac mn\le b+\tfrac1n.\)

    By 20.5.1, we get that

    \(\begin{aligned} s(a,b)&\le s(\tfrac kn,\tfrac mn)= \\ &=\tfrac kn\cdot\tfrac mn\le \\ &\le (a+\tfrac1n)\cdot(b+\tfrac1n),\end{aligned}\)

    which contradicts 20.5.2.

    The case \(s(a,b)< a\cdot b\) is similar. Fix a positive integer \(n\) such that \(a>\tfrac1n\), \(b>\tfrac1n\), and

    \[s(a,b)< (a-\tfrac1n)\cdot (b-\tfrac1n).\]

    Set \(k=\lceil a\cdot n \rceil-1\) and \(m=\lceil b\cdot n \rceil-1\); that is,

    \(a> \tfrac kn\ge a-\tfrac1n \quad\text{and}\quad b>\tfrac mn\ge b-\tfrac1n.\)

    Applying 20.5.1 again, we get that

    \(\begin{aligned} s(a,b)&\ge s(\tfrac kn,\tfrac mn)= \\ &=\tfrac kn\cdot\tfrac mn\ge \\ &\ge (a-\tfrac1n)\cdot(b-\tfrac1n),\end{aligned}\)

    which contradicts 20.5.3.