Prove that in any convex polyhedron, the number of faces that have an odd number of edges is even.
$begingroup$
Prove that in any convex polyhedron, the number of faces that have an odd number of edges is even.
I attempted to prove this by contradiction but didn't make any progress.
graph-theory polyhedra
asked Nov 27 '18 at 3:31
thetravellerthetraveller
1515
$endgroup$
add a comment |
$begingroup$
Prove that in any convex polyhedron, the number of faces that have an odd number of edges is even.
I attempted to prove this by contradiction but didn't make any progress.
graph-theory polyhedra
asked Nov 27 '18 at 3:31
thetravellerthetraveller
1515
$endgroup$
$begingroup$
Each edge borders exactly two faces.
$endgroup$
– Lord Shark the Unknown
Nov 27 '18 at 3:35
add a comment |
$begingroup$
Prove that in any convex polyhedron, the number of faces that have an odd number of edges is even.
I attempted to prove this by contradiction but didn't make any progress.
graph-theory polyhedra
asked Nov 27 '18 at 3:31
thetravellerthetraveller
1515
$endgroup$
Prove that in any convex polyhedron, the number of faces that have an odd number of edges is even.
I attempted to prove this by contradiction but didn't make any progress.
graph-theory polyhedra
graph-theory polyhedra
asked Nov 27 '18 at 3:31
thetravellerthetraveller
1515
asked Nov 27 '18 at 3:31
thetravellerthetraveller
1515
asked Nov 27 '18 at 3:31
thetravellerthetraveller
1515
asked Nov 27 '18 at 3:31
thetravellerthetraveller
1515
asked Nov 27 '18 at 3:31
thetravellerthetraveller
1515
1515
$begingroup$
Each edge borders exactly two faces.
$endgroup$
– Lord Shark the Unknown
Nov 27 '18 at 3:35
add a comment |
$begingroup$
Each edge borders exactly two faces.
$endgroup$
– Lord Shark the Unknown
Nov 27 '18 at 3:35
$begingroup$
Each edge borders exactly two faces.
$endgroup$
– Lord Shark the Unknown
Nov 27 '18 at 3:35
$begingroup$
Each edge borders exactly two faces.
$endgroup$
– Lord Shark the Unknown
Nov 27 '18 at 3:35
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Create a graph in the following way:
- Vertices are faces of the polyhedron
- Two vertices are connected if their corresponding faces share an edge
Observe that the odd-edged faces are exactly the vertices of odd degree in our graph. We know that the number of vertices of odd degree in any graph $G$ is even.
answered Nov 28 '18 at 3:47
Santana AftonSantana Afton
2,6192629
$endgroup$
add a comment |
$begingroup$
Let the faces be $F_1,ldots,F_k$. Consider the sum $S=n_1+cdots+n_k$
where face $F_i$ has $n_i$ edges.
This sum counts each edge twice: $S$ is even. So the number of $j$ for
which $n_j$ is odd must be even.
answered Nov 27 '18 at 4:42
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
$begingroup$
well @Lord, this argument alone only holds for 1 type of faces, or for 1 odd-polygonal face type, while all others are even-polygonal. But why this rules out say 7 triangles plus 3 pentagons?
$endgroup$
– Dr. Richard Klitzing
Nov 27 '18 at 19:47
add a comment |
$begingroup$
I'll elaborate a bit more, which'll helpfully help anybody interested get off on the right foot.
Let $G$ be a graph depicting the planar representation of a convex polyhedron. Let $F in Bbb Z^{+}$ denote the total number of faces in $G$ and $f_{1},f_{2},...,f_{F}in Bbb Z^{+}$ denote the respective number of edges of each of the $F$ faces of $G$. Recall $$sum_{i=1}^F f_{i} =2e,$$ where $ein Bbb Z^{+}$ denotes the total of number of edges of $G$. Hence $$f_{1}+f_{2}+...+f_{F}=2e.$$
There we have it. You just use some basic properties of number theory/ arithmetic/ algebra to deduce that there must be an even number of odd $f$'s so that ultimately their sum is an even number. For example, note that $2+3+5=10$ is an even number, but $2+3+5+7=17$ is not.
answered Nov 28 '18 at 3:38
greycatbirdgreycatbird
1377
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3015296%2fprove-that-in-any-convex-polyhedron-the-number-of-faces-that-have-an-odd-number%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Create a graph in the following way:
- Vertices are faces of the polyhedron
- Two vertices are connected if their corresponding faces share an edge
Observe that the odd-edged faces are exactly the vertices of odd degree in our graph. We know that the number of vertices of odd degree in any graph $G$ is even.
answered Nov 28 '18 at 3:47
Santana AftonSantana Afton
2,6192629
$endgroup$
add a comment |
$begingroup$
Create a graph in the following way:
- Vertices are faces of the polyhedron
- Two vertices are connected if their corresponding faces share an edge
Observe that the odd-edged faces are exactly the vertices of odd degree in our graph. We know that the number of vertices of odd degree in any graph $G$ is even.
answered Nov 28 '18 at 3:47
Santana AftonSantana Afton
2,6192629
$endgroup$
add a comment |
$begingroup$
Create a graph in the following way:
- Vertices are faces of the polyhedron
- Two vertices are connected if their corresponding faces share an edge
Observe that the odd-edged faces are exactly the vertices of odd degree in our graph. We know that the number of vertices of odd degree in any graph $G$ is even.
answered Nov 28 '18 at 3:47
Santana AftonSantana Afton
2,6192629
$endgroup$
Create a graph in the following way:
- Vertices are faces of the polyhedron
- Two vertices are connected if their corresponding faces share an edge
Observe that the odd-edged faces are exactly the vertices of odd degree in our graph. We know that the number of vertices of odd degree in any graph $G$ is even.
answered Nov 28 '18 at 3:47
Santana AftonSantana Afton
2,6192629
answered Nov 28 '18 at 3:47
Santana AftonSantana Afton
2,6192629
answered Nov 28 '18 at 3:47
Santana AftonSantana Afton
2,6192629
answered Nov 28 '18 at 3:47
Santana AftonSantana Afton
2,6192629
2,6192629
add a comment |
add a comment |
$begingroup$
Let the faces be $F_1,ldots,F_k$. Consider the sum $S=n_1+cdots+n_k$
where face $F_i$ has $n_i$ edges.
This sum counts each edge twice: $S$ is even. So the number of $j$ for
which $n_j$ is odd must be even.
answered Nov 27 '18 at 4:42
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
$begingroup$
well @Lord, this argument alone only holds for 1 type of faces, or for 1 odd-polygonal face type, while all others are even-polygonal. But why this rules out say 7 triangles plus 3 pentagons?
$endgroup$
– Dr. Richard Klitzing
Nov 27 '18 at 19:47
add a comment |
$begingroup$
Let the faces be $F_1,ldots,F_k$. Consider the sum $S=n_1+cdots+n_k$
where face $F_i$ has $n_i$ edges.
This sum counts each edge twice: $S$ is even. So the number of $j$ for
which $n_j$ is odd must be even.
answered Nov 27 '18 at 4:42
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
$begingroup$
well @Lord, this argument alone only holds for 1 type of faces, or for 1 odd-polygonal face type, while all others are even-polygonal. But why this rules out say 7 triangles plus 3 pentagons?
$endgroup$
– Dr. Richard Klitzing
Nov 27 '18 at 19:47
add a comment |
$begingroup$
Let the faces be $F_1,ldots,F_k$. Consider the sum $S=n_1+cdots+n_k$
where face $F_i$ has $n_i$ edges.
This sum counts each edge twice: $S$ is even. So the number of $j$ for
which $n_j$ is odd must be even.
answered Nov 27 '18 at 4:42
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
Let the faces be $F_1,ldots,F_k$. Consider the sum $S=n_1+cdots+n_k$
where face $F_i$ has $n_i$ edges.
This sum counts each edge twice: $S$ is even. So the number of $j$ for
which $n_j$ is odd must be even.
answered Nov 27 '18 at 4:42
Lord Shark the UnknownLord Shark the Unknown
103k1160132
answered Nov 27 '18 at 4:42
Lord Shark the UnknownLord Shark the Unknown
103k1160132
answered Nov 27 '18 at 4:42
Lord Shark the UnknownLord Shark the Unknown
103k1160132
answered Nov 27 '18 at 4:42
Lord Shark the UnknownLord Shark the Unknown
103k1160132
103k1160132
$begingroup$
well @Lord, this argument alone only holds for 1 type of faces, or for 1 odd-polygonal face type, while all others are even-polygonal. But why this rules out say 7 triangles plus 3 pentagons?
$endgroup$
– Dr. Richard Klitzing
Nov 27 '18 at 19:47
add a comment |
$begingroup$
well @Lord, this argument alone only holds for 1 type of faces, or for 1 odd-polygonal face type, while all others are even-polygonal. But why this rules out say 7 triangles plus 3 pentagons?
$endgroup$
– Dr. Richard Klitzing
Nov 27 '18 at 19:47
$begingroup$
well @Lord, this argument alone only holds for 1 type of faces, or for 1 odd-polygonal face type, while all others are even-polygonal. But why this rules out say 7 triangles plus 3 pentagons?
$endgroup$
– Dr. Richard Klitzing
Nov 27 '18 at 19:47
$begingroup$
well @Lord, this argument alone only holds for 1 type of faces, or for 1 odd-polygonal face type, while all others are even-polygonal. But why this rules out say 7 triangles plus 3 pentagons?
$endgroup$
– Dr. Richard Klitzing
Nov 27 '18 at 19:47
add a comment |
$begingroup$
I'll elaborate a bit more, which'll helpfully help anybody interested get off on the right foot.
Let $G$ be a graph depicting the planar representation of a convex polyhedron. Let $F in Bbb Z^{+}$ denote the total number of faces in $G$ and $f_{1},f_{2},...,f_{F}in Bbb Z^{+}$ denote the respective number of edges of each of the $F$ faces of $G$. Recall $$sum_{i=1}^F f_{i} =2e,$$ where $ein Bbb Z^{+}$ denotes the total of number of edges of $G$. Hence $$f_{1}+f_{2}+...+f_{F}=2e.$$
There we have it. You just use some basic properties of number theory/ arithmetic/ algebra to deduce that there must be an even number of odd $f$'s so that ultimately their sum is an even number. For example, note that $2+3+5=10$ is an even number, but $2+3+5+7=17$ is not.
answered Nov 28 '18 at 3:38
greycatbirdgreycatbird
1377
$endgroup$
add a comment |
$begingroup$
I'll elaborate a bit more, which'll helpfully help anybody interested get off on the right foot.
Let $G$ be a graph depicting the planar representation of a convex polyhedron. Let $F in Bbb Z^{+}$ denote the total number of faces in $G$ and $f_{1},f_{2},...,f_{F}in Bbb Z^{+}$ denote the respective number of edges of each of the $F$ faces of $G$. Recall $$sum_{i=1}^F f_{i} =2e,$$ where $ein Bbb Z^{+}$ denotes the total of number of edges of $G$. Hence $$f_{1}+f_{2}+...+f_{F}=2e.$$
There we have it. You just use some basic properties of number theory/ arithmetic/ algebra to deduce that there must be an even number of odd $f$'s so that ultimately their sum is an even number. For example, note that $2+3+5=10$ is an even number, but $2+3+5+7=17$ is not.
answered Nov 28 '18 at 3:38
greycatbirdgreycatbird
1377
$endgroup$
add a comment |
$begingroup$
I'll elaborate a bit more, which'll helpfully help anybody interested get off on the right foot.
Let $G$ be a graph depicting the planar representation of a convex polyhedron. Let $F in Bbb Z^{+}$ denote the total number of faces in $G$ and $f_{1},f_{2},...,f_{F}in Bbb Z^{+}$ denote the respective number of edges of each of the $F$ faces of $G$. Recall $$sum_{i=1}^F f_{i} =2e,$$ where $ein Bbb Z^{+}$ denotes the total of number of edges of $G$. Hence $$f_{1}+f_{2}+...+f_{F}=2e.$$
There we have it. You just use some basic properties of number theory/ arithmetic/ algebra to deduce that there must be an even number of odd $f$'s so that ultimately their sum is an even number. For example, note that $2+3+5=10$ is an even number, but $2+3+5+7=17$ is not.
answered Nov 28 '18 at 3:38
greycatbirdgreycatbird
1377
$endgroup$
I'll elaborate a bit more, which'll helpfully help anybody interested get off on the right foot.
Let $G$ be a graph depicting the planar representation of a convex polyhedron. Let $F in Bbb Z^{+}$ denote the total number of faces in $G$ and $f_{1},f_{2},...,f_{F}in Bbb Z^{+}$ denote the respective number of edges of each of the $F$ faces of $G$. Recall $$sum_{i=1}^F f_{i} =2e,$$ where $ein Bbb Z^{+}$ denotes the total of number of edges of $G$. Hence $$f_{1}+f_{2}+...+f_{F}=2e.$$
There we have it. You just use some basic properties of number theory/ arithmetic/ algebra to deduce that there must be an even number of odd $f$'s so that ultimately their sum is an even number. For example, note that $2+3+5=10$ is an even number, but $2+3+5+7=17$ is not.
answered Nov 28 '18 at 3:38
greycatbirdgreycatbird
1377
answered Nov 28 '18 at 3:38
greycatbirdgreycatbird
1377
answered Nov 28 '18 at 3:38
greycatbirdgreycatbird
1377
answered Nov 28 '18 at 3:38
greycatbirdgreycatbird
1377
1377
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3015296%2fprove-that-in-any-convex-polyhedron-the-number-of-faces-that-have-an-odd-number%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Each edge borders exactly two faces.
$endgroup$
– Lord Shark the Unknown
Nov 27 '18 at 3:35