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Chapter 6: Problem 259

Problem 259 A tree is a connected graph, or network, consisting of vertices and edges, but with no cycles (or circuits). Prove that a tree with \(n\) vertices has exactly \(n-1\) edges. The next problem concerns spherical polyhedra. A spherical polyhedron is a polyhedral surface in 3-dimensions, which can be inflated to form a sphere (where we assume that the faces and edges can stretch as required). For example, a cube is a spherical polyhedron; but the surface of a picture frame is not. A spherical polyhedron has \- faces (flat 2-dimensional polygons, which can be stretched to take the form of a disc), \- edges (1-dimensional line segments, where exactly two faces meet), and \- vertices (0-dimensional points, where several faces and edges meet, in such a way that they form a single cycle around the vertex). Each face must clearly have at least 3 edges; and there must be at least three edges and three faces meeting at each vertex. If a spherical polyhedron has \(V\) vertices, \(E\) edges, and \(F\) faces, then the numbers \(V, E, F\) satisfy Euler's formula $$ V-E+F=2 $$ For example, a cube has \(V=8\) vertices, \(E=12\) edges, and \(F=6\) faces, and \(8-12+6=2\)

Short Answer

Expert verified
A tree with n vertices has n-1 edges. Euler's formula for spherical polyhedra is V - E + F = 2.

Step by step solution

01

Understand the Definition of a Tree

A tree is defined as a connected graph with no cycles. This means that for any two vertices in the tree, there is exactly one path connecting them, and the graph does not form any loops.
02

Initial Observation

Start with a single vertex, which forms a tree by itself with 0 edges. Notice that when we add an additional vertex, an edge must also be added to keep the graph connected and cycle-free.
03

Induction Basis

Consider the base case where there is 1 vertex: a single vertex is a tree with exactly 0 edges, and indeed, 1 vertex has 1-1=0 edges, satisfying our relation.
04

Induction Hypothesis

Assume that for a tree with k vertices, the number of edges is exactly k-1.
05

Inductive Step

Now, consider adding one more vertex to the tree with k vertices. To connect this new vertex to the existing tree, exactly 1 new edge is needed. Thus, the graph will have k+1 vertices and k edges.
06

Conclusion of Induction Proof

By our induction steps, a tree with n vertices must have exactly n-1 edges, since adding any additional vertex also adds exactly one edge to maintain the tree structure without cycles.
07

Understanding Euler's Formula for Spherical Polyhedra

Euler's formula for a spherical polyhedron states that V - E + F = 2, where V is the number of vertices, E is edges, and F is faces. This is a topological invariant for spherical polyhedra.
08

Example of Euler's Formula with a Cube

For a cube: V = 8, E = 12, F = 6. Plug these into Euler's formula: V - E + F = 8 - 12 + 6 = 2, which satisfies the formula, confirming its applicability.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trees in Graph Theory
A tree in graph theory is a special type of graph. It consists of a set of vertices connected by edges. The defining property of a tree is that it is "cycle-free", meaning it contains no loops or circuits. For any two vertices, there is exactly one unique path that connects them. This property ensures that trees are a minimal connected graph.
  • Each tree with \( n \) vertices has precisely \( n-1 \) edges.
  • If you add a vertex, you must add an edge to keep the graph a tree, without forming a cycle.
  • This gives trees their simple, branching structure, resembling a real-life tree.
In graph theory, trees are fundamental because they are seen everywhere in computer science applications like database indexing and network topology. They help solve problems efficiently using their straightforward structure.
Euler's Formula
Euler's formula is a vital link between geometry and graph theory, especially with polyhedra. The formula for any convex polyhedron is given by:\[ V - E + F = 2 \]where \( V \) is the number of vertices, \( E \) is the number of edges, and \( F \) is the number of faces.
  • This relationship is known as the Euler characteristic for polyhedra.
  • It highlights the balance between these three properties, regardless of the shape of the polyhedron.
  • For complex shapes and networks, Euler's formula provides a simple check to understand their structure.
A practical example is a cube: it has 8 vertices, 12 edges, and 6 faces. Plugging these into Euler's formula confirms \( 8 - 12 + 6 = 2 \). It holds true and exemplifies the formula's simplicity and power.
Spherical Polyhedra
Spherical polyhedra are 3D shapes that can be formed into a sphere. Imagine a shape whose surface is primarily flat faces, but can be stretched to enclose a spherical form. They include familiar objects like cubes and tetrahedrons.
  • Each face typically has at least three edges.
  • At least three edges and faces meet at each vertex.
  • The whole surface wraps around a central point, forming a complete cycle.
Spherical polyhedra respect Euler's formula. This connection is crucial in topology and geometry. Understanding these forms helps in fields like geodesy and computer graphics, where spatial representation is key.
Inductive Proof
An inductive proof is a powerful mathematical technique to demonstrate that a statement is true for all natural numbers. It involves several steps and is commonly used in graph theory to prove properties like those of trees.
  • Start with a base case: Verify that the statement holds true for an initial case, typically for the smallest number or simplest object.
  • Inductive Hypothesis: Assume the statement is true for some arbitrary case \( k \).
  • Inductive Step: Prove that if it holds for \( k \), then it must also hold for \( k+1 \).
  • Conclusion: Conclude the proof by showing it is valid for all integers beyond your base case.
When applied to show that a tree with \( n \) vertices contains \( n-1 \) edges, it simplifies the reasoning process. It lets one build from simple cases and seamlessly extend to more complex graphs. Inductive proof is widely used in computer science and mathematics for its elegance and thoroughness in establishing truths.

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Most popular questions from this chapter

(a) Prove by induction that \(n\) points on a straight line divide the line into \(n+1\) parts. (b)(i) By experimenting with small values of \(n,\) guess a formula \(R_{n}\) for the maximum number of regions which can be created in the plane by an array of \(n\) straight lines. (ii) Prove by induction that \(n\) straight lines in the plane divide the plane into at most \(R_{n}\) regions. (c)(i) By experimenting with small values of \(n,\) guess a formula \(S_{n}\) for the maximum number of regions which can be created in 3-dimensions by an array of \(n\) planes. (ii) Prove by induction that \(n\) planes in 3 -dimensions divide space into at most \(S_{n}\) regions.

(a)(i) Describe a spherical polyhedron with exactly 6 edges. (ii) Describe a spherical polyhedron with exactly 8 edges. (b) Prove that no spherical polyhedron can have exactly 7 edges. (c) Prove that for every \(n \geqslant 8\), there exists a spherical polyhedron with \(n\) edges.

(a) Are 3,7,31,211 all prime? (b) Is \(2 \times 3 \times 5 \times 7 \times 11+1\) prime? (c) Is \(2 \times 3 \times 5 \times 7 \times 11 \times 13+1\) prime? We have already met two excellent historical examples of the dangers of plausible pattern-spotting in connection with Problem 118. There you proved that: "if \(2^{n}-1\) is prime, then \(n\) must be prime." You then showed that \(2^{2}-1,2^{3}-1,2^{5}-1,2^{7}-1\) are all prime, but that \(2^{11}-1=2047=23 \times 89\) is not. This underlines the need to avoid jumping to (possibly false) conclusions, and never to confuse a statement with its converse. In the same problem you showed that: "if \(a^{b}+1\) is to be prime and \(a \neq 1,\) then \(a\) must be even, and \(b\) must be a power of \(2 . "\) You then looked at the simplest family of such candidate primes namely the sequence of Fermat numbers \(f_{n}\) : $$ f_{0}=2^{1}+1=3, f_{1}=2^{2}+1=5, f_{2}=2^{4}+1=17, f_{3}=2^{8}+1=257, f_{4}=2^{16}+1 $$ It turned out that, although \(f_{0}, f_{1}, f_{2}, f_{3}, f_{4}\) are all prime, and although Fermat (1601-1665) claimed (in a letter to Marin Mersenne \((1588-1648))\) that all Fermat numbers \(f_{n}\) are prime, we have yet to discover a sixth Fermat prime! There are times when a mathematician may appear to guess a general result on the basis of what looks like very modest evidence (such as noticing that it appears to be true in a few small cases). Such "informed guesses" are almost always rooted in other experience, or in some unnoticed feature of the particular situation, or in some striking analogy: that is, an apparent pattern strikes a chord for reasons that go way beyond the mere numbers. However those with less experience need to realise that apparent patterns or trends are often no more than numerical accidents. Pell's equation (John Pell (1611-1685)) provides some dramatic examples. \- If we evaluate the expression " \(n^{2}+1\) " for \(n=1,2,3, \ldots,\) we may notice that the outputs \(2,5,10,17,26, \ldots\) never give a perfect square. And this is to be expected, since the next square after \(n^{2}\) is $$ (n+1)^{2}=n^{2}+2 n+1 $$ and this is always greater than \(n^{2}+1\). However, if we evaluate \(" 991 n^{2}+1 "\) for \(n=1,2,3, \ldots,\) we may observe that the outputs never seem to include a perfect square. But this time there is no obvious reason why this should be so - so we may anticipate that this is simply an accident of "small" numbers. And we should hesitate to change our view, even though this accident goes on happening for a very, very, very long time: the smallest value of \(n\) for which \(991 n^{2}+1\) gives rise to a perfect square is apparently

Problem 230 Prove by induction the statement: "for each \(n \geqslant 3,\) the angles of any \(n\) -gon in the plane have sum equal to \((n-2) \pi\) radians." The formulation certainly involves a parameter \(n \geqslant 3\); so you clearly need to begin by formulating the statement \(\mathbf{P}(n)\). For the proof to have a chance of working, finding the right formulation involves a modest twist! So if you get stuck, it may be worth checking the first couple of lines of the solution. No matter how \(\mathbf{P}(n)\) is formulated, you should certainly know how to prove the statement \(\mathbf{P}(3)\) (essentially the formula for the sum of the angles in a triangle). But it is far from obvious how to prove the "induction step": "if we know that \(\mathbf{P}(k)\) is true for some particular \(k \geqslant 1,\) then \(\mathbf{P}(k+1)\) must also be true" When tackling the induction step, we certainly cannot start with \(\mathbf{P}(k) !\) The statement \(\mathbf{P}(k+1)\) says something about polygons with \(k+1\) sides: and there is no way to obtain a typical \((k+1)\) -gon by fiddling with some statement about polygons with \(k\) sides. (If you start with a \(k\) -gon, you can of course draw a triangle on one side to get a \((k+1)\) -gon; but this is a very special construction, and there is no easy way of knowing whether all \((k+1)\) -gons can be obtained in this way from some \(k\) -gon.) The whole thrust of mathematical induction is that we must always start the induction step by thinking about the hypothesis of \(\mathbf{P}(k+1)-\) that is in this case, by considering an arbitrary \((k+1)\) -gon and then finding some guaranteed way of "reducing" it in order to make use of \(\mathbf{P}(k)\) The next two problems invite you to prove some classical algebraic identities. Most of these may be familiar. The challenge here is to think carefully about the way you lay out your induction proof, to learn from the examples above, and (later) to learn from the detailed solutions provided.

When John von Neumann \((1903-1957)\) was seriously ill in hospital, a visitor tried (rather insensitively) to distract him with the following elementary mathematics problem. Have you heard the one about the two trains and the fly? Two trains are on a collision course on the same track, each travelling at \(30 \mathrm{~km} / \mathrm{h}\). A super-fly starts on Train \(A\) when the trains are 120 \(\mathrm{km}\) apart, and flies at a constant speed of \(50 \mathrm{~km} / \mathrm{h}-\) from Train \(A\) to Train \(B\), then back to Train \(A\), and so on. Eventually the two trains collide and the fly is squashed. How far did the fly travel before this sad outcome?
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