Proving the area formula for a rectangle for all positive real numbers

[logicgate]

TL;DR Summary

I wanna know why the area formula for a rectangle equals length times width for all positive real numbers, not only for integers.

It's very easy to prove the area formula for a rectangle when both length and width are positive integers, but I cannot prove it when length or width or both are rational or irrational numbers. I need an intuitive proof that is as simple as possible without using very advanced math like calculus.

 

[Ibix]

If your rectangle has sides $\frac ab$ and $\frac cd$, where $a$, $b$, $c$, and $d$ are integers, draw a rectangle with sides $a$ and $c$. What's its area? How many of the original rectangles could you fit in it? So what's the area of the original?

 

[jbriggs444]

TL;DR Summary: I wanna know why the area formula for a rectangle equals length times width for all positive real numbers, not only for integers.

It's very easy to prove the area formula for a rectangle when both length and width are positive integers, but I cannot prove it when length or width or both are rational or irrational numbers. I need an intuitive proof that is as simple as possible without using very advanced math like calculus.

Are you after an intuition or a proof? For a proof, you'll need to define "area" first. For an intuition, the requirements are more relaxed. Let me try to provide an intuition.

You agree that the area of a 1 by 1 square is 1, surely?

If we change the length of the horizontal side of this 1 by 1 square to $x$ then the area of the resulting $x$ by $1$ rectangle is $x \times 1$. Would you agree?

If we change the length of the vertical side of this $x$ by 1 square to $y$ then the area of the resulting $x$ by $y$ rectangle is $x \times y$. Would you agree?

We might need to fiddle around with definitions to account for zero and negative side lengths and zero or negative areas.

 

[fresh_42]

TL;DR Summary: I wanna know why the area formula for a rectangle equals length times width for all positive real numbers, not only for integers.

It's very easy to prove the area formula for a rectangle when both length and width are positive integers, but I cannot prove it when length or width or both are rational or irrational numbers. I need an intuitive proof that is as simple as possible without using very advanced math like calculus.

If you accept it for integers, just take another folding rule. The area won't change, but the units on your meter did and you will get a different numerical value. Now instead of changing the ruler, we could equally change the area and get non-integer results.

 

[PeroK]

TL;DR Summary: I wanna know why the area formula for a rectangle equals length times width for all positive real numbers, not only for integers.

It's very easy to prove the area formula for a rectangle when both length and width are positive integers, but I cannot prove it when length or width or both are rational or irrational numbers. I need an intuitive proof that is as simple as possible without using very advanced math like calculus.

I would say that the definition of the area of a rectangle is length times width. If you want to try to prove that, then you would need a more fundamental definition of area.

 

[jbriggs444]

I would say that the definition of the area of a rectangle is length times width. If you want to try to prove that, then you would need a more fundamental definition of area.

I would guess that the underlying theory of areas that @logicgate is assuming is along the lines of an abstract measure.

You start with an axiom that the area of a 1 by 1 square is 1. [Or, more generally $k$ with $k$ depending on the choice of units for length and area]

You add an axiom that says that if you join two geometric figures so that the two overlap, if at all, only at their boundaries then the area of the combined figure is the sum of the areas of the component figures.

This puts you in a position to evaluate the area of any rectangle with sides of rational length. Either by joining finitely many identical squares to make a rectangle. Or dividing a rectangle into finitely many squares. Or both.

To make the jump to sides of irrational length, we might choose to exploit a new axiom:

"A rectangle whose perpendicular sides both have a positive length has a positive area"

This puts you in a position to sandwich the area of a rectangle with irrational sides in between the areas of sets of rectangles with rational sides. Then you use the magic of convergent sequences, Dedekind cuts or least upper bounds to argue that if the rectangle has an area at all (as it must, given the axiom above), that area must be the real number given by $\text{width} \times \text{length}$

 

[Lnewqban]

It's very easy to prove the area formula for a rectangle when both length and width are positive integers, but I cannot prove it when length or width or both are rational or irrational numbers...

If the dimension of at least one of the sides of that rectangle is irrational, shouldn't the value of its area be an irrational number as well?

 

[fresh_42]

If the dimension of at least one of the sides of that rectangle is irrational, shouldn't the value of its area be an irrational number as well?

View attachment 349412

You mean as in $\sqrt{2} \cdot \sqrt{8}$?

 

[Lnewqban]

You mean as in $\sqrt{2} \cdot \sqrt{8}$?

As shown in the above diagram.
Two of the sides of the rectangle are shared with the hypotenuse of two triangles of sides equal to 1 unit.

 

[pbuk]

As shown in the above diagram.
Two of the sides of the rectangle are shared with the hypotenuse of two triangles of sides equal to 1 unit.

@fresh_42's post was a counter-example not a request for clarification.

 

[pbuk]

I need an intuitive proof that is as simple as possible without using very advanced math like calculus.

@Ibix has provided a very elegant intuitive proof in #2.

Do you understand it? Do you understand the answers to any of your other threads?