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Arrange the steps in the correct order to prove that a simple graph T is a tree if and only if it is connected, but the deletion of any of its edges produces a graph that is not connected. (First, prove that if a simple graph T is a tree, then it is connected and the deletion of any of its edges produces a graph that is not connected, and then prove the converse of it.) Rank the options below. Suppose that a simple connected graph 7 satisfies the condition that the removal of any edge will disconnect it. Suppose that T is a tree; then, by definition, it is connected. If T is not a tree, then it has a simple circuit, say, x1, x2, ..., xr, x1. 1 4 LQ 5 Now, T with (x, y) deleted has no path from x to y, since there was only one simple path from x to y in T, and the edge itself was it. Therefore, the graph with (x, y) deleted is not connected. 2 Let (x, y) be an edge of T such that x * y. This is a contradiction to the condition. Therefore, our assumption was wrong, and T is a tree. If we delete the edge (x, x1) from 7, then the graph will remain connected, since wherever the deleted edge was used in forming paths between vertices we can instead use the rest of the circuit x1, x2, ..., xr or its reverse. 6 ☑ × × 3 × 7 For which values of n do these graphs have an Euler circuit? 1 Wn 2 Kn 3 Сп 4 Qn Match each of the options above to the items below. n≥ 3 and n is odd. n≥ 3 No Euler circuit is possible for any value of n. n> 0 and n is even. 2 1 × × 4 × 3

Arrange the steps in the correct order to prove that a simple graph T is a tree if and only if it is connected, but the deletion of any of its edges produces a graph that is not connected. (First, prove that if a simple graph T is a tree, then it is connected and the deletion of any of its edges produces a graph that is not connected, and then prove the converse of it.) Rank the options below. Suppose that a simple connected graph 7 satisfies the condition that the removal of any edge will disconnect it. Suppose that T is a tree; then, by definition, it is connected. If T is not a tree, then it has a simple circuit, say, x1, x2, ..., xr, x1. 1 4 LQ 5 Now, T with (x, y) deleted has no path from x to y, since there was only one simple path from x to y in T, and the edge itself was it. Therefore, the graph with (x, y) deleted is not connected. 2 Let (x, y) be an edge of T such that x * y. This is a contradiction to the condition. Therefore, our assumption was wrong, and T is a tree. If we delete the edge (x, x1) from 7, then the graph will remain connected, since wherever the deleted edge was used in forming paths between vertices we can instead use the rest of the circuit x1, x2, ..., xr or its reverse. 6 ☑ × × 3 × 7 For which values of n do these graphs have an Euler circuit? 1 Wn 2 Kn 3 Сп 4 Qn Match each of the options above to the items below. n≥ 3 and n is odd. n≥ 3 No Euler circuit is possible for any value of n. n> 0 and n is even. 2 1 × × 4 × 3

Algebra: Structure And Method, Book 1
Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter4: Polynomials
Section4.1: Exponents
Problem 1CE
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Arrange the steps in the correct order to prove that a simple graph T is a tree if and only if it is connected, but the deletion of any of its edges produces a graph that is not connected. (First, prove that if a simple graph T is a tree, then it is connected and the deletion of any of its edges produces a graph that is not connected, and then prove the converse of it.) Rank the options below. Suppose that a simple connected graph 7 satisfies the condition that the removal of any edge will disconnect it. Suppose that T is a tree; then, by definition, it is connected. If T is not a tree, then it has a simple circuit, say, x1, x2, ..., xr, x1. 1 4 LQ 5 Now, T with (x, y) deleted has no path from x to y, since there was only one simple path from x to y in T, and the edge itself was it. Therefore, the graph with (x, y) deleted is not connected. 2 Let (x, y) be an edge of T such that x * y. This is a contradiction to the condition. Therefore, our assumption was wrong, and T is a tree. If we delete the edge (x, x1) from 7, then the graph will remain connected, since wherever the deleted edge was used in forming paths between vertices we can instead use the rest of the circuit x1, x2, ..., xr or its reverse. 6 ☑ × × 3 × 7
Transcribed Image Text:Arrange the steps in the correct order to prove that a simple graph T is a tree if and only if it is connected, but the deletion of any of its edges produces a graph that is not connected. (First, prove that if a simple graph T is a tree, then it is connected and the deletion of any of its edges produces a graph that is not connected, and then prove the converse of it.) Rank the options below. Suppose that a simple connected graph 7 satisfies the condition that the removal of any edge will disconnect it. Suppose that T is a tree; then, by definition, it is connected. If T is not a tree, then it has a simple circuit, say, x1, x2, ..., xr, x1. 1 4 LQ 5 Now, T with (x, y) deleted has no path from x to y, since there was only one simple path from x to y in T, and the edge itself was it. Therefore, the graph with (x, y) deleted is not connected. 2 Let (x, y) be an edge of T such that x * y. This is a contradiction to the condition. Therefore, our assumption was wrong, and T is a tree. If we delete the edge (x, x1) from 7, then the graph will remain connected, since wherever the deleted edge was used in forming paths between vertices we can instead use the rest of the circuit x1, x2, ..., xr or its reverse. 6 ☑ × × 3 × 7
For which values of n do these graphs have an Euler circuit? 1 Wn 2 Kn 3 Сп 4 Qn Match each of the options above to the items below. n≥ 3 and n is odd. n≥ 3 No Euler circuit is possible for any value of n. n> 0 and n is even. 2 1 × × 4 × 3
Transcribed Image Text:For which values of n do these graphs have an Euler circuit? 1 Wn 2 Kn 3 Сп 4 Qn Match each of the options above to the items below. n≥ 3 and n is odd. n≥ 3 No Euler circuit is possible for any value of n. n> 0 and n is even. 2 1 × × 4 × 3
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