TITLE: Formal Proof that area of a rectangle is $ab$
QUESTION [25 upvotes]: I tried to prove that the area of a rectangle is $ab$ given side lengths $a$ and $b$.
The best I can do is the assume the area of a $1\times1$ square is $1$. Then not the number of $1\times1$ squares that fit in an $a\times b$ rectangle is $ab$. Therefore area is $a\cdot b$. This does not seem rigorous however.

REPLY [4 votes]: One can also start from the following axioms:
$A1$: The area of a $a \times a$ square is $a^2$.
$A2$: Any two congruent shapes have the same area.
$A3$: If a shape $R$ is disjoint union of shapes $S$ and $T$, then the area of $R$ is equal to the sum of the areas of $S$ and $T$.
With these axioms at hand one readily finds the area of a rectangle with sides $a$ and $b$:

On the sides of the rectangle $ABCD$ build the squares $ABC_1D_1$ and $A_1BCD_2$. Let $D_3= (A_1D_2)\cap (C_1D_1)$.
Obviously the rectangles $ABCD$ and $A_1BC_1D_3$ are congruent. Therefore from our axioms one obtains:
$$\begin{align}
2[ABCD]&=[DD_1D_3D_2]-[ABC_1D_1]-[A_1BCD_2]\\
&=(a+b)^2-a^2-b^2=2ab.
\end{align}
$$