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

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

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.