My Math Forum  

Go Back   My Math Forum > High School Math Forum > Geometry

Geometry Geometry Math Forum


Reply
 
LinkBack Thread Tools Display Modes
April 10th, 2018, 05:37 AM   #11
Senior Member
 
Joined: Oct 2009

Posts: 436
Thanks: 147

Quote:
Originally Posted by v8archie View Post
We can define area perfectly well without calculus.
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 $A(D) = (b-a)(d-c)$

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.
Micrm@ss is offline  
 
April 10th, 2018, 06:48 AM   #12
Senior Member
 
Joined: Jun 2015
From: England

Posts: 844
Thanks: 252

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 co-domain?
studiot is offline  
April 10th, 2018, 06:53 AM   #13
Senior Member
 
Joined: Oct 2009

Posts: 436
Thanks: 147

Quote:
Originally Posted by studiot View Post
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 co-domain?
My area is a function taking as input a subset of $\mathbb{R}^2$ and outputs a number. Thus it would be a function
$$A:\mathcal{P}(\mathbb{R}^2)\rightarrow \mathbb{R}$$
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.
Micrm@ss is offline  
April 10th, 2018, 07:07 AM   #14
Senior Member
 
Joined: Jun 2015
From: England

Posts: 844
Thanks: 252

Quote:
$\displaystyle R \times R \to R$

Either that or there is a kernal which maps to zero for lines and such?
studiot is offline  
April 10th, 2018, 11:50 PM   #15
Senior Member
 
Joined: Oct 2009

Posts: 436
Thanks: 147

Quote:
Originally Posted by studiot View Post
Either that or there is a kernal which maps to zero for lines and such?
Right, ok, that works too!
Micrm@ss is offline  
Reply

  My Math Forum > High School Math Forum > Geometry

Tags
area, circle



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Area of circle ma78s Trigonometry 1 March 21st, 2017 12:55 AM
Finding the area of a circle inside a square only given the area of the square skimanner Math 12 July 8th, 2016 07:55 AM
Area enclosed by a semi-circle and a quarter circle yeoky Algebra 4 May 3rd, 2014 01:06 AM
Area of square with convex parts - area of circle gus Algebra 1 April 17th, 2011 04:25 PM
Area of a Sector of a Circle Julie Algebra 2 May 18th, 2009 11:41 AM





Copyright © 2018 My Math Forum. All rights reserved.