
# 20.5: Area of solid rectangles


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.

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}

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}