April 10th, 2018, 05:37 AM  #11 
Senior Member Joined: Oct 2009 Posts: 770 Thanks: 276  But can we though? Sure, you can axiomatically define the area of a square and a rectangle. That's it. Is that enough to say that we can define area without calculus. Shouldn't there be minimum requirements before we can say we have an area? I am thinking of the following requirements at least for $A$ to be an area function 1) $A(\emptyset) = 0$ 2) $A$ must be defined on at least all the compact sets of $\mathbb{R}^2$ 3) If $F$ and $F'$ are disjoint and $A$ is defined on them, then $A(F\cup F') = A(F) + A(F')$ 4) The area must remain stable under rigid motions (=rotations, translations, reflections) 5) If $F_1\subseteq F_2\subseteq F_3 \subseteq ...$, then $A\left(\bigcup_n F_n\right) = \sup_n A(F_n)$ 6) If $D=[a,b]\times [c,d]$ is a rectangle with sides parallel to the x and y axes, then $A(D) = (ba)(dc)$ Is an elementary definition of this possible? I would say measure theory provides us with such, but that might not fit the bill of the OP. An elementary definition of length is even harder. 
April 10th, 2018, 06:48 AM  #12 
Senior Member Joined: Jun 2015 From: England Posts: 905 Thanks: 271 
Micromass, Whilst I understand your list, is area not a functional defined on $\displaystyle R \times R \to R$ since it needs must have a two parameter input in the domain and a single real output in the codomain? 
April 10th, 2018, 06:53 AM  #13  
Senior Member Joined: Oct 2009 Posts: 770 Thanks: 276  Quote:
$$A:\mathcal{P}(\mathbb{R}^2)\rightarrow \mathbb{R}$$ with as domain the power set of $\mathbb{R}^2$. However, as is known, not all subsets of the plane have an area, so $A$ is not a function rather than a partial function that is not everywhere defined.  
April 10th, 2018, 07:07 AM  #14  
Senior Member Joined: Jun 2015 From: England Posts: 905 Thanks: 271  Quote:
Either that or there is a kernal which maps to zero for lines and such?  
April 10th, 2018, 11:50 PM  #15 
Senior Member Joined: Oct 2009 Posts: 770 Thanks: 276  

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