Concept on Euler's formula
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Is there a much better way to proof and derive Euler's formula in geometrical figures? In that,F+V-2=E. For example an enclosed cube with 8 vertices, 6 faces and 12 edges. It is true that the edges, E=14-2
E=12
The idea is, where integer 2 comes in place in the equation as an abstract value.
I will appreciate anyone's contribution.
Thanks
polyhedra
edited Aug 29 at 7:28
Gerry Myerson
144k8145295
asked Aug 29 at 7:07
Makau Elijah
355
add a comment |Â
up vote
1
down vote
favorite
Is there a much better way to proof and derive Euler's formula in geometrical figures? In that,F+V-2=E. For example an enclosed cube with 8 vertices, 6 faces and 12 edges. It is true that the edges, E=14-2
E=12
The idea is, where integer 2 comes in place in the equation as an abstract value.
I will appreciate anyone's contribution.
Thanks
polyhedra
edited Aug 29 at 7:28
Gerry Myerson
144k8145295
asked Aug 29 at 7:07
Makau Elijah
355
4
You ask for a "much better way" – much better than what? Do you already know one way to prove it, but you don't like that way? How can I know whether my way is much better than the one(s) you already know, if you won't tell me which way(s) you already know?
– Gerry Myerson
Aug 29 at 7:28
4
I usually see the formula as $F-E+V = 2$. It's easier to generalize to higher (or lower) dimensions that way. As for what the $2$ has to do with anything, that's a very interesting question, and is intiricately linked to (net) curvature. Basically, in each of the 8 corners of the cube, 3 squares meet for an angle sum of $270^circ$, which is $90^circ$ away from a flat corner. $8cdot 90^circ = 720^circ$, which is two full rotations, which is the same $2$ as in your formula.
– Arthur
Aug 29 at 7:28
Are you still here, Makau? Care to engage with the comments/answer?
– Gerry Myerson
Aug 30 at 13:52
@GerryMyerson,Hi?well,the only way I know to proof is by induction method
– Makau Elijah
Aug 31 at 6:16
So, you asked for a better proof than by induction, and when John posted an induction proof, you accepted that answer. Very strange.
– Gerry Myerson
Aug 31 at 9:54
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Is there a much better way to proof and derive Euler's formula in geometrical figures? In that,F+V-2=E. For example an enclosed cube with 8 vertices, 6 faces and 12 edges. It is true that the edges, E=14-2
E=12
The idea is, where integer 2 comes in place in the equation as an abstract value.
I will appreciate anyone's contribution.
Thanks
polyhedra
edited Aug 29 at 7:28
Gerry Myerson
144k8145295
asked Aug 29 at 7:07
Makau Elijah
355
Is there a much better way to proof and derive Euler's formula in geometrical figures? In that,F+V-2=E. For example an enclosed cube with 8 vertices, 6 faces and 12 edges. It is true that the edges, E=14-2
E=12
The idea is, where integer 2 comes in place in the equation as an abstract value.
I will appreciate anyone's contribution.
Thanks
polyhedra
edited Aug 29 at 7:28
Gerry Myerson
144k8145295
asked Aug 29 at 7:07
Makau Elijah
355
edited Aug 29 at 7:28
Gerry Myerson
144k8145295
edited Aug 29 at 7:28
Gerry Myerson
144k8145295
edited Aug 29 at 7:28
Gerry Myerson
144k8145295
144k8145295
asked Aug 29 at 7:07
Makau Elijah
355
asked Aug 29 at 7:07
Makau Elijah
355
asked Aug 29 at 7:07
Makau Elijah
355
355
4
You ask for a "much better way" – much better than what? Do you already know one way to prove it, but you don't like that way? How can I know whether my way is much better than the one(s) you already know, if you won't tell me which way(s) you already know?
– Gerry Myerson
Aug 29 at 7:28
4
I usually see the formula as $F-E+V = 2$. It's easier to generalize to higher (or lower) dimensions that way. As for what the $2$ has to do with anything, that's a very interesting question, and is intiricately linked to (net) curvature. Basically, in each of the 8 corners of the cube, 3 squares meet for an angle sum of $270^circ$, which is $90^circ$ away from a flat corner. $8cdot 90^circ = 720^circ$, which is two full rotations, which is the same $2$ as in your formula.
– Arthur
Aug 29 at 7:28
Are you still here, Makau? Care to engage with the comments/answer?
– Gerry Myerson
Aug 30 at 13:52
@GerryMyerson,Hi?well,the only way I know to proof is by induction method
– Makau Elijah
Aug 31 at 6:16
So, you asked for a better proof than by induction, and when John posted an induction proof, you accepted that answer. Very strange.
– Gerry Myerson
Aug 31 at 9:54
add a comment |Â
4
You ask for a "much better way" – much better than what? Do you already know one way to prove it, but you don't like that way? How can I know whether my way is much better than the one(s) you already know, if you won't tell me which way(s) you already know?
– Gerry Myerson
Aug 29 at 7:28
4
I usually see the formula as $F-E+V = 2$. It's easier to generalize to higher (or lower) dimensions that way. As for what the $2$ has to do with anything, that's a very interesting question, and is intiricately linked to (net) curvature. Basically, in each of the 8 corners of the cube, 3 squares meet for an angle sum of $270^circ$, which is $90^circ$ away from a flat corner. $8cdot 90^circ = 720^circ$, which is two full rotations, which is the same $2$ as in your formula.
– Arthur
Aug 29 at 7:28
Are you still here, Makau? Care to engage with the comments/answer?
– Gerry Myerson
Aug 30 at 13:52
@GerryMyerson,Hi?well,the only way I know to proof is by induction method
– Makau Elijah
Aug 31 at 6:16
So, you asked for a better proof than by induction, and when John posted an induction proof, you accepted that answer. Very strange.
– Gerry Myerson
Aug 31 at 9:54
4
4
You ask for a "much better way" – much better than what? Do you already know one way to prove it, but you don't like that way? How can I know whether my way is much better than the one(s) you already know, if you won't tell me which way(s) you already know?
– Gerry Myerson
Aug 29 at 7:28
You ask for a "much better way" – much better than what? Do you already know one way to prove it, but you don't like that way? How can I know whether my way is much better than the one(s) you already know, if you won't tell me which way(s) you already know?
– Gerry Myerson
Aug 29 at 7:28
4
4
I usually see the formula as $F-E+V = 2$. It's easier to generalize to higher (or lower) dimensions that way. As for what the $2$ has to do with anything, that's a very interesting question, and is intiricately linked to (net) curvature. Basically, in each of the 8 corners of the cube, 3 squares meet for an angle sum of $270^circ$, which is $90^circ$ away from a flat corner. $8cdot 90^circ = 720^circ$, which is two full rotations, which is the same $2$ as in your formula.
– Arthur
Aug 29 at 7:28
I usually see the formula as $F-E+V = 2$. It's easier to generalize to higher (or lower) dimensions that way. As for what the $2$ has to do with anything, that's a very interesting question, and is intiricately linked to (net) curvature. Basically, in each of the 8 corners of the cube, 3 squares meet for an angle sum of $270^circ$, which is $90^circ$ away from a flat corner. $8cdot 90^circ = 720^circ$, which is two full rotations, which is the same $2$ as in your formula.
– Arthur
Aug 29 at 7:28
Are you still here, Makau? Care to engage with the comments/answer?
– Gerry Myerson
Aug 30 at 13:52
Are you still here, Makau? Care to engage with the comments/answer?
– Gerry Myerson
Aug 30 at 13:52
@GerryMyerson,Hi?well,the only way I know to proof is by induction method
– Makau Elijah
Aug 31 at 6:16
@GerryMyerson,Hi?well,the only way I know to proof is by induction method
– Makau Elijah
Aug 31 at 6:16
So, you asked for a better proof than by induction, and when John posted an induction proof, you accepted that answer. Very strange.
– Gerry Myerson
Aug 31 at 9:54
So, you asked for a better proof than by induction, and when John posted an induction proof, you accepted that answer. Very strange.
– Gerry Myerson
Aug 31 at 9:54
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
A typical proof is by induction (best done for planar graphs). Imagine you have a connected graph drawn in the plane with no edge crossings and you are redrawing the graph.
You start by drawing a single vertex. Thus, in your new drawing you've got $V = 1$, $F = 1$, and $E = 0$, so $F-E+V = 2$. So the 2 is right there from the start.
That it stays 2 comes from the following observation. You can use two moves to draw the rest of your graph (this requires a little argument, of course). Either draw an edge you haven't drawn yet and end it at a vertex you haven't drawn yet. This does not divide a face and thus if we had thus far drawn $V$ vertices, $E$ edges and $F$ faces, we have now drawn $V+1$ vertices, $E+1$ edges, and $F$ faces and therefore $$F - (E+1) + (F+1) = F - E + F$$ and thus we haven't changed the value of $V - E + F$.
Similarly, the second move is to add a single edge between two already existing drawn vertices. Because of the connectedness this always subdivides a face into two. In this case, then, we go from V, E, F in our drawing to V, E+1, F+1, but again:
$$(F+1) - (E+1) + V = F - E + V.$$
So, if you accept that all planar connected graphs can be drawn using those two moves starting from a single vertex (which I didn't really prove but is fairly intuitive if you play with a few examples), and see that both moves preserve the value of $V - E + F$, then the fact that $V - E + F = 2$ just comes from the base case of the induction.
answered Aug 30 at 13:41
John
1334
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
A typical proof is by induction (best done for planar graphs). Imagine you have a connected graph drawn in the plane with no edge crossings and you are redrawing the graph.
You start by drawing a single vertex. Thus, in your new drawing you've got $V = 1$, $F = 1$, and $E = 0$, so $F-E+V = 2$. So the 2 is right there from the start.
That it stays 2 comes from the following observation. You can use two moves to draw the rest of your graph (this requires a little argument, of course). Either draw an edge you haven't drawn yet and end it at a vertex you haven't drawn yet. This does not divide a face and thus if we had thus far drawn $V$ vertices, $E$ edges and $F$ faces, we have now drawn $V+1$ vertices, $E+1$ edges, and $F$ faces and therefore $$F - (E+1) + (F+1) = F - E + F$$ and thus we haven't changed the value of $V - E + F$.
Similarly, the second move is to add a single edge between two already existing drawn vertices. Because of the connectedness this always subdivides a face into two. In this case, then, we go from V, E, F in our drawing to V, E+1, F+1, but again:
$$(F+1) - (E+1) + V = F - E + V.$$
So, if you accept that all planar connected graphs can be drawn using those two moves starting from a single vertex (which I didn't really prove but is fairly intuitive if you play with a few examples), and see that both moves preserve the value of $V - E + F$, then the fact that $V - E + F = 2$ just comes from the base case of the induction.
answered Aug 30 at 13:41
John
1334
add a comment |Â
up vote
0
down vote
accepted
A typical proof is by induction (best done for planar graphs). Imagine you have a connected graph drawn in the plane with no edge crossings and you are redrawing the graph.
You start by drawing a single vertex. Thus, in your new drawing you've got $V = 1$, $F = 1$, and $E = 0$, so $F-E+V = 2$. So the 2 is right there from the start.
That it stays 2 comes from the following observation. You can use two moves to draw the rest of your graph (this requires a little argument, of course). Either draw an edge you haven't drawn yet and end it at a vertex you haven't drawn yet. This does not divide a face and thus if we had thus far drawn $V$ vertices, $E$ edges and $F$ faces, we have now drawn $V+1$ vertices, $E+1$ edges, and $F$ faces and therefore $$F - (E+1) + (F+1) = F - E + F$$ and thus we haven't changed the value of $V - E + F$.
Similarly, the second move is to add a single edge between two already existing drawn vertices. Because of the connectedness this always subdivides a face into two. In this case, then, we go from V, E, F in our drawing to V, E+1, F+1, but again:
$$(F+1) - (E+1) + V = F - E + V.$$
So, if you accept that all planar connected graphs can be drawn using those two moves starting from a single vertex (which I didn't really prove but is fairly intuitive if you play with a few examples), and see that both moves preserve the value of $V - E + F$, then the fact that $V - E + F = 2$ just comes from the base case of the induction.
answered Aug 30 at 13:41
John
1334
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
A typical proof is by induction (best done for planar graphs). Imagine you have a connected graph drawn in the plane with no edge crossings and you are redrawing the graph.
You start by drawing a single vertex. Thus, in your new drawing you've got $V = 1$, $F = 1$, and $E = 0$, so $F-E+V = 2$. So the 2 is right there from the start.
That it stays 2 comes from the following observation. You can use two moves to draw the rest of your graph (this requires a little argument, of course). Either draw an edge you haven't drawn yet and end it at a vertex you haven't drawn yet. This does not divide a face and thus if we had thus far drawn $V$ vertices, $E$ edges and $F$ faces, we have now drawn $V+1$ vertices, $E+1$ edges, and $F$ faces and therefore $$F - (E+1) + (F+1) = F - E + F$$ and thus we haven't changed the value of $V - E + F$.
Similarly, the second move is to add a single edge between two already existing drawn vertices. Because of the connectedness this always subdivides a face into two. In this case, then, we go from V, E, F in our drawing to V, E+1, F+1, but again:
$$(F+1) - (E+1) + V = F - E + V.$$
So, if you accept that all planar connected graphs can be drawn using those two moves starting from a single vertex (which I didn't really prove but is fairly intuitive if you play with a few examples), and see that both moves preserve the value of $V - E + F$, then the fact that $V - E + F = 2$ just comes from the base case of the induction.
answered Aug 30 at 13:41
John
1334
A typical proof is by induction (best done for planar graphs). Imagine you have a connected graph drawn in the plane with no edge crossings and you are redrawing the graph.
You start by drawing a single vertex. Thus, in your new drawing you've got $V = 1$, $F = 1$, and $E = 0$, so $F-E+V = 2$. So the 2 is right there from the start.
That it stays 2 comes from the following observation. You can use two moves to draw the rest of your graph (this requires a little argument, of course). Either draw an edge you haven't drawn yet and end it at a vertex you haven't drawn yet. This does not divide a face and thus if we had thus far drawn $V$ vertices, $E$ edges and $F$ faces, we have now drawn $V+1$ vertices, $E+1$ edges, and $F$ faces and therefore $$F - (E+1) + (F+1) = F - E + F$$ and thus we haven't changed the value of $V - E + F$.
Similarly, the second move is to add a single edge between two already existing drawn vertices. Because of the connectedness this always subdivides a face into two. In this case, then, we go from V, E, F in our drawing to V, E+1, F+1, but again:
$$(F+1) - (E+1) + V = F - E + V.$$
So, if you accept that all planar connected graphs can be drawn using those two moves starting from a single vertex (which I didn't really prove but is fairly intuitive if you play with a few examples), and see that both moves preserve the value of $V - E + F$, then the fact that $V - E + F = 2$ just comes from the base case of the induction.
answered Aug 30 at 13:41
John
1334
answered Aug 30 at 13:41
John
1334
answered Aug 30 at 13:41
John
1334
answered Aug 30 at 13:41
John
1334
1334
add a comment |Â
add a comment |Â
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4
You ask for a "much better way" – much better than what? Do you already know one way to prove it, but you don't like that way? How can I know whether my way is much better than the one(s) you already know, if you won't tell me which way(s) you already know?
– Gerry Myerson
Aug 29 at 7:28
4
I usually see the formula as $F-E+V = 2$. It's easier to generalize to higher (or lower) dimensions that way. As for what the $2$ has to do with anything, that's a very interesting question, and is intiricately linked to (net) curvature. Basically, in each of the 8 corners of the cube, 3 squares meet for an angle sum of $270^circ$, which is $90^circ$ away from a flat corner. $8cdot 90^circ = 720^circ$, which is two full rotations, which is the same $2$ as in your formula.
– Arthur
Aug 29 at 7:28
Are you still here, Makau? Care to engage with the comments/answer?
– Gerry Myerson
Aug 30 at 13:52
@GerryMyerson,Hi?well,the only way I know to proof is by induction method
– Makau Elijah
Aug 31 at 6:16
So, you asked for a better proof than by induction, and when John posted an induction proof, you accepted that answer. Very strange.
– Gerry Myerson
Aug 31 at 9:54