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Prove that every right triangular region is measurable and its area is 1/2bh

  1. Jul 7, 2011 #1
    1. The problem statement, all variables and given/known data
    Prove that every right triangular region is measurable because it can be obtained as the intersection of two rectangles. Prove that every triangular region is measurable and its area is one half the product of its base and altitude. (Apostol's Calculus Vol1.- 1.7 Exercises)


    2. Relevant equations

    We assume that there exists a class M of measurable sets in the plane and a set function a whose domain is M with the following properties:
    1) A(S) >= 0 for each set S in M.
    2) if S and T are two sets in M their intersection and union is also in M and we have:
    A(S U T) = A(S) + A(T) - A(S ∩ T)
    3)If S and T are in M with S ⊆ T then T − S is in M and a(T − S) = a(T) − a(S).
    4) If a set S is in M and S is congruent to T then T is also in M and a(S) = a(T).
    5) Every rectangle R is in M. If the rectangle has length h and breadth k then a(R) = hk.
    6) Let Q be a set enclosed between two step regions S and T. A step region is formed from a finite union of adjacent rectangles resting on a common base, i.e. S⊆ Q ⊆ T. If there is a unique number c such that a(S) <= c <= a(T) for all such step regions S and T, then a(Q) = c.

    3. The attempt at a solution

    Well, I first tried to define a right triangle as the intersection of two rectangles as Apostol suggests but I failed to do that. I think It's impossible to define a right triangle as the intersection of two rectangles unless we rotate one of them which makes it hard to describe the triangle by an algebraic equation as simple as the one we have for rectangles.
    I then tried to define a right triangle by writing down three equations and then intersect them. I wrote down 0 <= x <= k, 0 <= y <= h and y<= -h/k x + h where k and h are two positive numbers. the intersection of those three equations gives us the set that consists of all pairs like (x,y) s.t 0 <= x <= k & 0 <= y <= -h/k x + h. so let's take T to be the set defined in the following way:

    T = {(x,y)| (0 <= x <= k) & (0 <= ky <= -hx + kh)}. This is supposed to be a triangle that looks like |_\.

    Now assume S to be the intersection of the three equations: 0<= x <= k & 0<= y < h & y >= -h/k x + h. so we define the set S to be:

    S = {(x,y)| (0 <= x <= k) & (-hx + kh <= ky <= kh)}. This one is supposed to be a triangle that looks like \-|.

    It it easy to show that the union of these two algebraic equations is the set S U T:

    S U T = {(x,y)| 0<= x <= k & 0 <= y =< h} which is indeed a rectangle.

    Using the 2nd Axiom we have:
    A(S U T) = A(S) + A(T) - A(S ∩ T)

    It's easy to show that the intersection of S & T is the line hx + ky = hk. I have previously proved that the area of a line is 0. so if I show that S and T can be congruent by defining a one-to-one correspondence between them that keeps the distance between two arbitrary points the same then using the 4th axiom I can conclude that A(S) = A(T) and I'm done.

    Well, to prove that the two sets T and S are congruent I define a reflection on the points in T across the point (k/2,h/2) that lies on the center of the line hx + ky = hk.
    if I define r: T -> S by r(t) = 2p - t then r is a reflection of t across the point p. any reflection in the Euclidean plane is isometric, so it definitely keeps the distance between any two points unchanged. as I mentioned earlier, we let p be (k/2,h/2) and let t=(x,y) be any point in T. then r(x,y) = 2(k/2,h/2) - (x,y) = (k,h) - (x,y) = (k-x,h-y). The function r is bijective because for any t in T: ror(t) = t. that means r has an inverse therefore it is bijective. to show that T and S are congruent we show that r which is a bijective isometric function puts the points in T into one-to-one correspondence with the points in S.

    for any (x,y)∈T : 0<= x <= k & 0 <= ky <= -hx + kh if we apply r on (x,y) we'll have:
    0<= k-x <= k --> -k <= -x <= 0 --> 0 <= x =< k.
    0<= k(h-y) <= -h(k-x) + kh --> 0 <= kh - ky <= -hk +hx + kh --> 0 <= kh - ky <= +hx --> 0 <= -ky =< +hx - kh --> -hx + kh <= ky =< kh which proves that r(x,y)∈S which proves that S and T are congruent. therefore A(S) = A(T) using the 4th axiom.

    Now, as stated earlier: A(S U T) = A(S) + A(T) - A(S ∩ T).
    A(S U T) = A(S) + A(T) - A(S ∩ T) => A(S U T) = A(S) + A(T) - 0 => A(S U T) = 2A(S) => hk = 2 A(S) => A(S) = hk/2.

    This proves that the area of a right triangle is one half the product of its base and its height. but to prove that a right triangle is a measurable set seems to be impossible or very hard by my approach so I think Apostol's suggestion works well to prove that a right triangle is a measurable set ( the 2nd axiom proves that a right triangle is a measurable set because It's the intersection of two rectangles and the intersection of two measurable sets is a measurable set).

    Now to generalize this result to any arbitrary triangle one just need to split the triangle into two disjoint right triangles and then use the 2nd property.

    Well, It seems that the problem has been solved but the way that I've proved it is a hard way of proving this because although It's rigorous but It's very lengthy. I'm here to ask if my proof is right and if yes, does anyone here know a simpler way of proving this using the six axioms given above?
     
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  3. Jul 8, 2011 #2

    micromass

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    Hi AdrianZ! :smile:

    Here are some comments on your proof:

    It is possible to describe a right triangle as the intersection of two rectangles, but it is indeed true that you need to rotate at least one of the rectangles. I don't quite understand why you bother with algebraic equations here, they have nothing to do with the problem?

    This will only describe the triangles who's rectangular sides are the x-axis or y-axis. (or at best the triangles who's rectangular sides are parallel to the x-axis or y-axis). This is not good enough, you'll need all rectangular triangles!!!!

    Same as above.

    This is good.

    Yes, you'll still need to prove that it's measurable. So you'll have to apply Apostol's approach anyway. So why not apply it for the entire argument?

    Like I said, you'll need to prove that the result holds for arbitrary triangles first, not just triangles who's sides are the x-axis and the y-axis!

    I must say that the proof is quite ingenious, but it has some holes. So you can fill up the holes, or you can try Apostol's method. Both will work really :smile:
     
  4. Jul 8, 2011 #3
    Apostol has defined rectangle as the set {(x,y)| 0<=x<=h & 0<=y<=k}. so I tried to show that the union of two right triangles is a rectangle by the definition of rectangle that has been given to us. surely there are zillions of ways to define a rectangle but I tried to use the definition that has been given to us and rely only on that definition. Can I do this without algebraic equations?

    True. but It's possible to show that any other right triangular region is congruent to a right triangular region like this by a proper translation and rotation. can't I?

    I don't know Apostol's approach. I mean I understand the idea behind it but I don't know how he wants to show that a right triangular region is the intersection of two properly rotated rectangles using only his definition of rectangle and the six axioms of area function. I think if I want to use his approach first I need to show that any rectangular region in R*R can be written as the set {(x,y)| 0<=x<=h & 0<=y<=k} then I show that the intersection of two nice rectangles is a right triangular region. but then I'll still face hardship to find the height and base of the right triangle by their equations, It won't be impossible but it'll definitely be hard.

    I have to show that any two right triangular regions with the base b and the height h are congruent which sounds tricky. right? because first I need to say what I mean by a triangular region in general and then show that any two such sets are congruent by proper translations and rotations. This will complete the proof and make it rigorous. Now what is Apostol's approach to this problem if not using algebraic equations?
     
  5. Jul 8, 2011 #4

    micromass

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    Well, rotated rectangles are congruent to the rectangles that Apostol defines. So you can use rotated rectangles. I don't think there's a need for algebraic equations.


    Here's my method: For a triangle ABC with rectangular sides AB and AC, you can define a rectangle X with sides AB and AC. And you can define a rectangle Y with side BC. The intersection of these two rectangles X and Y will be the triangle. This is what Apostol wants you to do.

    Your method isn't wrong! But I guess it can be shortened.

    True, but you need to show that somehow. I don't think it's trivial.

    It isn't hard, it's actually quite easy. I wish I could make drawings here, so I could point out what Apostol means. But I hope my discussion above clarifies it.

    Indeed, showing that any two right triangular regions with the same dimensions are congruent is the tricky part. It is possible, but Apostol did not intended you to do this.

    If my method above is not clear, then I'll try to clear it up a bit...
     
  6. Jul 8, 2011 #5
    well, then first we need to show that any two rectangular regions are congruent. well, It's easy though, but the difficult part is to define a general rectangular region. well, I can define a general rectangular region as the region that is between 4 lines that are parallel to each other two by two and the set of points that is enclosed between them is bounded. It's easy to write down their equations as well. then I need to define a one-to-one correspondence between this general rectangular region and Apostol's defined rectangle and show that the bijective function I've defined is isometric. that would be complete the proof. right?

    Yea, your method is quite understandable and convincing.

    well, I understood your method.

    It's clear. Thanks for the help.
     
  7. Jul 8, 2011 #6

    micromass

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    Right!
     
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