Sign In
Not register? Register Now!
Pages:
2 pages/β‰ˆ550 words
Sources:
Check Instructions
Style:
APA
Subject:
Mathematics & Economics
Type:
Coursework
Language:
English (U.S.)
Document:
MS Word
Date:
Total cost:
$ 10.37
Topic:

Toulmin Analysis of Mathematical Argumentation Math Coursework

Coursework Instructions:

Last Monday, we separated the ideas of statements being TRUE and WARRANTED from
each other. The goal of that discussion was to prepare you to read a paper (Weber & Alcock
) that applies a method of analyzing argumentation from
rhetoric, called Toulmin analysis, to mathematical argumentation (a.k.a. proof).
**In your reading, skip the monotone convergence proof and example analysis from the paper
- instead substitute the proof with corresponding sample Toulmin analysis that I wrote for you -
you’ll find two pdf files I sent.
While you read the paper: First focus on the ways that claims in a proof can be warranted and
on learning to apply Toulmin analysis. Second, note other interesting ideas for discussion.
When you are done reading the paper and reading through the sample proof/analysis I wrote,
revise your proof of the Pythagorean Theorem, then write up a Toulmin analysis of your mathematical proof (like my example).
Note that many of the sentences in your proof will include new conclusions (as opposed to there
being only one large conclusion for the whole argument). In particular, I’m asserting that each
small claim in a mathematical argument can be analyzed via Toulmin analysis, not just the entire
proof as a whole.
Your Reflection Assignment will have two parts: a revised proof of the Pythagorean Theorem and
a separate Toulmin analysis. Format your Toulmin analysis in three columns: data/warrants/conclusions
like my example. This assignment will be assessed for quality of engagement with the ideas from
Weber and Alcock’s paper, rather than the clarity and correctness of your proof or the Toulmin
analysis.

 

Math 120 – The Nature of Mathematics Section 6.2 Definition: The Euler Characteristic of a graph with V vertices, E edges, and R regions is V − E + R. The Euler Characteristic Theorem: Given a plane graph, if the graph is connected, then its Euler Characteristic is equal to 2. Proof. Suppose we have a connected graph in the plane (I’m not drawing it here because I don’t want to accidentally assume anything about the graph beyond that it is connected in the plane). According to our definition, our mystery graph is a collection of vertices connected by edges, so we can think of building up to our graph (whatever it looks like) from a single vertex. We can start from this single-vertex graph and carefully consider how each addition might affect the counts of V , E and R (we will show that the Euler Characteristic begins at 2 and remains 2 throughout the building-up process). First we will demonstrate that if we start from a single-vertex graph, we only need three moves to build our mystery graph: #1: add an loose edge to an existing vertex and cap it with a new vertex #2: add a loop to an existing vertex #3: connect two existing vertices with a new edge A graph is a collection of vertices connected by edges and our mystery graph is a connected graph (so any two vertices in the graph are connected by some path within the graph), which means we can start building our mystery graph from the single vertex, one vertex at a time, using move #1 until we have all of the vertices drawn (we may not have all of the edges yet...). Next we can add any loops to our graph using move #2. Finally we can add any edges we missed using move #3 and our mystery graph is drawn in the plane. Now we can examine how each of these moves affects the edge, vertex, and region counts as we built our mystery graph to see how the Euler Characteristic changes (or doesn’t change) as we build. • Move #1 increases the edge and vertex count by one, but doesn’t create or remove any regions, so the region count stays the same. • Move #2 doesn’t change the vertex count, but it does increase both the edge and region counts by one. • Move #3 doesn’t change the vertex count, but it increases the edge count by one and, because the two vertices we are connecting are already connected by another path, the new edge encloses a new region and the region count goes up by one. Finally, let’s look at our build-up process again and track the Euler Characteristic after each step: 1 • We started with a single-vertex graph. Note that for the single-vertex graph V = 1, E = 0 and R = 1, so V − E + R = 2. • Next we added each vertex one-by-one using move #1, which increases V by one AND increases E by one, so in each step we have V + 1 − (E + 1) + R = 2. • After each vertex was in place, we added all the loops we needed using move #2 and then any remaining edges using move #3, both of which increases E by one AND increases R by one, so in each step we have V − (E + 1) + (R + 1) = 2. So the mystery graph has Euler Characteristic 2. And since we assumed nothing special about our mystery graph beyond that it is a connected plane graph, the result must be true for ANY connected plane graph. Therefore, any connected plane graph has V − E + R = 2. 2

Coursework Sample Content Preview:
Toulmin Analysis of Mathematical Argumentation
The Pythagorean Theorem states that: “The square on the hypotenuse of a right-angled triangle equal the sum of the squares on the two legs.”
147955026225500The formula is simply represented as, A2 + B2 = C2
3276600156210B00B179817024283900
2406650120015C00C
325120011430A00A
Proof: #1: we will calculate the area of the square with side A + B,
The area will be, A2+ 2AB + B2……………………(i)
#2: we will calculate the area of the 4 triangles, and that of the small square,
which is given by: ½*AB*4 = 2AB…………………. (ii)
The area of the small square is C2…………. (iii)
#3 calculate the area of the bigger square by adding (ii) and (iii),
Therefore, the area of the C2+2AB……. (iv)
#4 Equate (i) and (iv)
This gives, A2+ 2AB + B2 = C2 + 2AB………(v)
#5 we will subtract 2AB from both sides of the equation (v)
If we subtract 2AB from both sides of the equation, we get:
A2 + B2 = C2
We have proved that the Pythagorean Theorem is true using three steps only, this is an examination of how each of the moves proves the Pythagorean Theorem.
Move #1 gives the area of the bigger square using the sum of A and B, which gives us equation (i)
Move #2 gives the area of 4 triangles, which is given by equation (ii), and that of the small square which is given by equation (iii)
Move #3 gives the area of the bigger square by adding equation (ii) and (iii)
Move #4 equates equation (i) and (iv), which are both areas of the big square
Move #5 removes 2AB from step (iv), giving us a proof that A2 + B2 = C2
Therefore, A2 + B...
Updated on
Get the Whole Paper!
Not exactly what you need?
Do you need a custom essay? Order right now:

πŸ‘€ Other Visitors are Viewing These APA Coursework Samples:

HIRE A WRITER FROM $11.95 / PAGE
ORDER WITH 15% DISCOUNT!