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This is a exercise that I came across while studying a geometry textbook.

Use the method of induction to prove the area formula of a rectangle whose two sides are natural numbers.

I can derive the simple formula of the area of a rectangle by using a method like this:

enter image description here

And so we can derive that $A = xy$ by using $x^2 + y^2 + 2A = (x + y)^2$.

However, I'm having some trouble using induction to prove this. How should I start this problem? Any tips are welcome, thanks.

Sean
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I believe what is assumed is that

  1. You show that if $x,y=1$, then $A(x,y)=1$. (base)
  2. You show that if $A(x,1)=1x=x$, then $A(x+1,1)=1(x+1)$. (induction on $x$)
  3. You show that if $A(x,y)=xy$, then $A(x,y+1)=x(y+1)$. (induction on $y$)
Vasily Mitch
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    It's the base step that is the problem: why should the area of a $1\times 1$ rectangle be $1$? That is not something that you can show; it has to be an axiom (at least in this context). See this question for an example. – TonyK Mar 13 '20 at 12:30
  • The book that I'm using says that if we assume that the area of a rectangle with sides $1$ is $1$ (i.e. If $\mathbf{x} = (1, 0)$ and $\mathbf{y} = (0, 1)$ then $A(\mathbf{x}, \mathbf{y}) = 1$ and also that if $D = D_1 \cup D_2$ then $A(D) = A(D_1) + A(D_2)$. And we can use these two assumptions to say that the area of a rectangle that has natural numbers $a$ and $b$ as its sides has area $ab$. Is this true?? I can't help but think that it's insufficient and very hand-wavy. – Sean Mar 18 '20 at 03:09