Mixing
Prove or disprove. The vertices of every regular tetrahedron in 3-space have at least two irrational coordinates.
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Prove or disprove. All four vertices of every regular tetrahedron in $ mathbbR^3$ have at least two irrational coordinates.
This question arose from my inability to construct a tetrahedron in $mathbbR^3$ with all the coordinates
$$
M_x,M_y,M_z,N_x,N_y,N_z,P_x,P_y,P_z,Q_x,Q_y ;mathrm
and; Q_z,
$$
of its vertices $M$, $N$, $P$ and $Q$ being rational numbers.
geometry analytic-geometry simplex
edited Jul 22 at 14:41
Parcly Taxel
33.5k136588
asked Jul 22 at 14:37
MathOverview
7,95242962
add a comment |Â
up vote
0
down vote
favorite
Prove or disprove. All four vertices of every regular tetrahedron in $ mathbbR^3$ have at least two irrational coordinates.
This question arose from my inability to construct a tetrahedron in $mathbbR^3$ with all the coordinates
$$
M_x,M_y,M_z,N_x,N_y,N_z,P_x,P_y,P_z,Q_x,Q_y ;mathrm
and; Q_z,
$$
of its vertices $M$, $N$, $P$ and $Q$ being rational numbers.
geometry analytic-geometry simplex
edited Jul 22 at 14:41
Parcly Taxel
33.5k136588
asked Jul 22 at 14:37
MathOverview
7,95242962
2
Think about the tetrahedra in the cube.
– Lord Shark the Unknown
Jul 22 at 14:45
2
This question answers yours. It's also Lord Shark's comment and @ParclyTaxel 's deleted answer. math.stackexchange.com/questions/1423014/…
– Ethan Bolker
Jul 22 at 14:51
You may want to think more carefully about the order of quantifiers in your question. As stated, the obvious interpretation is "For every tetrahedron $T subset mathbb R^3$, and for ever vertex $V = (V_x,V_y,V_z)$ of $T$, at least two of the three coordinates $V_x,V_y,V_z$ are irrational", and it is way too easy to construct a counterexample to that statement as the answer of @RossMillikan shows.
– Lee Mosher
Jul 22 at 16:48
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Prove or disprove. All four vertices of every regular tetrahedron in $ mathbbR^3$ have at least two irrational coordinates.
This question arose from my inability to construct a tetrahedron in $mathbbR^3$ with all the coordinates
$$
M_x,M_y,M_z,N_x,N_y,N_z,P_x,P_y,P_z,Q_x,Q_y ;mathrm
and; Q_z,
$$
of its vertices $M$, $N$, $P$ and $Q$ being rational numbers.
geometry analytic-geometry simplex
edited Jul 22 at 14:41
Parcly Taxel
33.5k136588
asked Jul 22 at 14:37
MathOverview
7,95242962
Prove or disprove. All four vertices of every regular tetrahedron in $ mathbbR^3$ have at least two irrational coordinates.
This question arose from my inability to construct a tetrahedron in $mathbbR^3$ with all the coordinates
$$
M_x,M_y,M_z,N_x,N_y,N_z,P_x,P_y,P_z,Q_x,Q_y ;mathrm
and; Q_z,
$$
of its vertices $M$, $N$, $P$ and $Q$ being rational numbers.
geometry analytic-geometry simplex
edited Jul 22 at 14:41
Parcly Taxel
33.5k136588
asked Jul 22 at 14:37
MathOverview
7,95242962
edited Jul 22 at 14:41
Parcly Taxel
33.5k136588
edited Jul 22 at 14:41
Parcly Taxel
33.5k136588
edited Jul 22 at 14:41
Parcly Taxel
33.5k136588
33.5k136588
asked Jul 22 at 14:37
MathOverview
7,95242962
asked Jul 22 at 14:37
MathOverview
7,95242962
asked Jul 22 at 14:37
MathOverview
7,95242962
7,95242962
2
Think about the tetrahedra in the cube.
– Lord Shark the Unknown
Jul 22 at 14:45
2
This question answers yours. It's also Lord Shark's comment and @ParclyTaxel 's deleted answer. math.stackexchange.com/questions/1423014/…
– Ethan Bolker
Jul 22 at 14:51
You may want to think more carefully about the order of quantifiers in your question. As stated, the obvious interpretation is "For every tetrahedron $T subset mathbb R^3$, and for ever vertex $V = (V_x,V_y,V_z)$ of $T$, at least two of the three coordinates $V_x,V_y,V_z$ are irrational", and it is way too easy to construct a counterexample to that statement as the answer of @RossMillikan shows.
– Lee Mosher
Jul 22 at 16:48
add a comment |Â
2
Think about the tetrahedra in the cube.
– Lord Shark the Unknown
Jul 22 at 14:45
2
This question answers yours. It's also Lord Shark's comment and @ParclyTaxel 's deleted answer. math.stackexchange.com/questions/1423014/…
– Ethan Bolker
Jul 22 at 14:51
You may want to think more carefully about the order of quantifiers in your question. As stated, the obvious interpretation is "For every tetrahedron $T subset mathbb R^3$, and for ever vertex $V = (V_x,V_y,V_z)$ of $T$, at least two of the three coordinates $V_x,V_y,V_z$ are irrational", and it is way too easy to construct a counterexample to that statement as the answer of @RossMillikan shows.
– Lee Mosher
Jul 22 at 16:48
2
2
Think about the tetrahedra in the cube.
– Lord Shark the Unknown
Jul 22 at 14:45
Think about the tetrahedra in the cube.
– Lord Shark the Unknown
Jul 22 at 14:45
2
2
This question answers yours. It's also Lord Shark's comment and @ParclyTaxel 's deleted answer. math.stackexchange.com/questions/1423014/…
– Ethan Bolker
Jul 22 at 14:51
This question answers yours. It's also Lord Shark's comment and @ParclyTaxel 's deleted answer. math.stackexchange.com/questions/1423014/…
– Ethan Bolker
Jul 22 at 14:51
You may want to think more carefully about the order of quantifiers in your question. As stated, the obvious interpretation is "For every tetrahedron $T subset mathbb R^3$, and for ever vertex $V = (V_x,V_y,V_z)$ of $T$, at least two of the three coordinates $V_x,V_y,V_z$ are irrational", and it is way too easy to construct a counterexample to that statement as the answer of @RossMillikan shows.
– Lee Mosher
Jul 22 at 16:48
You may want to think more carefully about the order of quantifiers in your question. As stated, the obvious interpretation is "For every tetrahedron $T subset mathbb R^3$, and for ever vertex $V = (V_x,V_y,V_z)$ of $T$, at least two of the three coordinates $V_x,V_y,V_z$ are irrational", and it is way too easy to construct a counterexample to that statement as the answer of @RossMillikan shows.
– Lee Mosher
Jul 22 at 16:48
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
Choose four suitable vertices of the unit cube $[0,1]^3$, and you have a regular tetrahedron.
answered Jul 22 at 16:28
Christian Blatter
163k7107306
add a comment |Â
up vote
1
down vote
You can place one of the vertices at the origin. A second one can be at $(1,0,0)$
answered Jul 22 at 14:45
Ross Millikan
276k21186352
I wouldn't put my second vertex there....
– Lord Shark the Unknown
Jul 22 at 14:48
1
@LordSharktheUnknown: The question claimed that every point had two irrational coordinates. I have two points with no irrational coordinates which disproves the assertion. I saw your comment indicating that we can do it with no irrational coordinates.
– Ross Millikan
Jul 22 at 14:50
1
Also, the first sentence alone suffices, I suppose?
– user202729
Jul 22 at 14:56
add a comment |Â
up vote
0
down vote
I think you want to prove that there doesn't exist a regular tetrahedron such that all its vertices have only rational coordinates. Assume the contrary, let $ABCD$ be such a tetrahedron. Consider the plane $(ABC)$ as a complex plane with origin in $A$ and let the points $B$ and $C$ have associated complex numbers $a+bi$ and $c+di$. By our assumption, we would have that $a,b,c,dinmathbbQ$. We know that the points associated to complex numbers $z_1,z_2,z_3$ form an equilateral triangle if and only if $z_1+varepsilon z_2+varepsilon^2z_3=0$ where $varepsilon=-frac12+fracsqrt32i$ is the root of order 3 of the unity. By plugging our numbers, we obtain
$$a+bi+varepsilon(c+di)+varepsilon^2cdot 0=0Rightarrow$$
$$Rightarrow a+bi+left(-frac12+fracsqrt32iright)(c+di)=0Rightarrow$$
$$Rightarrow a+bi-fracc2-fracd2i+fracsqrt3c2i-fracsqrt3d2=0Rightarrow$$
$$Rightarrow a-fracc2-fracsqrt3d2=0Rightarrow d=0$$
also
$$b-fracd2+fracsqrt3c2=0Rightarrow c=0$$
thus $A=C$, so we don't actually have a triangle $ABC$ hence the conclusion.
answered Jul 22 at 15:02
Andrei Cataron
1178
1
This is a correct proof about equilateral triangles in the plane. But in space you can build an equilateral triangle with all rational coordinates - the normal to its plane will have some irrationality. See my comment on the question.
– Ethan Bolker
Jul 22 at 15:10
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Choose four suitable vertices of the unit cube $[0,1]^3$, and you have a regular tetrahedron.
answered Jul 22 at 16:28
Christian Blatter
163k7107306
add a comment |Â
up vote
3
down vote
Choose four suitable vertices of the unit cube $[0,1]^3$, and you have a regular tetrahedron.
answered Jul 22 at 16:28
Christian Blatter
163k7107306
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Choose four suitable vertices of the unit cube $[0,1]^3$, and you have a regular tetrahedron.
answered Jul 22 at 16:28
Christian Blatter
163k7107306
Choose four suitable vertices of the unit cube $[0,1]^3$, and you have a regular tetrahedron.
answered Jul 22 at 16:28
Christian Blatter
163k7107306
answered Jul 22 at 16:28
Christian Blatter
163k7107306
answered Jul 22 at 16:28
Christian Blatter
163k7107306
answered Jul 22 at 16:28
Christian Blatter
163k7107306
163k7107306
add a comment |Â
add a comment |Â
up vote
1
down vote
You can place one of the vertices at the origin. A second one can be at $(1,0,0)$
answered Jul 22 at 14:45
Ross Millikan
276k21186352
I wouldn't put my second vertex there....
– Lord Shark the Unknown
Jul 22 at 14:48
1
@LordSharktheUnknown: The question claimed that every point had two irrational coordinates. I have two points with no irrational coordinates which disproves the assertion. I saw your comment indicating that we can do it with no irrational coordinates.
– Ross Millikan
Jul 22 at 14:50
1
Also, the first sentence alone suffices, I suppose?
– user202729
Jul 22 at 14:56
add a comment |Â
up vote
1
down vote
You can place one of the vertices at the origin. A second one can be at $(1,0,0)$
answered Jul 22 at 14:45
Ross Millikan
276k21186352
I wouldn't put my second vertex there....
– Lord Shark the Unknown
Jul 22 at 14:48
1
@LordSharktheUnknown: The question claimed that every point had two irrational coordinates. I have two points with no irrational coordinates which disproves the assertion. I saw your comment indicating that we can do it with no irrational coordinates.
– Ross Millikan
Jul 22 at 14:50
1
Also, the first sentence alone suffices, I suppose?
– user202729
Jul 22 at 14:56
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You can place one of the vertices at the origin. A second one can be at $(1,0,0)$
answered Jul 22 at 14:45
Ross Millikan
276k21186352
You can place one of the vertices at the origin. A second one can be at $(1,0,0)$
answered Jul 22 at 14:45
Ross Millikan
276k21186352
answered Jul 22 at 14:45
Ross Millikan
276k21186352
answered Jul 22 at 14:45
Ross Millikan
276k21186352
answered Jul 22 at 14:45
Ross Millikan
276k21186352
276k21186352
I wouldn't put my second vertex there....
– Lord Shark the Unknown
Jul 22 at 14:48
1
@LordSharktheUnknown: The question claimed that every point had two irrational coordinates. I have two points with no irrational coordinates which disproves the assertion. I saw your comment indicating that we can do it with no irrational coordinates.
– Ross Millikan
Jul 22 at 14:50
1
Also, the first sentence alone suffices, I suppose?
– user202729
Jul 22 at 14:56
add a comment |Â
I wouldn't put my second vertex there....
– Lord Shark the Unknown
Jul 22 at 14:48
1
@LordSharktheUnknown: The question claimed that every point had two irrational coordinates. I have two points with no irrational coordinates which disproves the assertion. I saw your comment indicating that we can do it with no irrational coordinates.
– Ross Millikan
Jul 22 at 14:50
1
Also, the first sentence alone suffices, I suppose?
– user202729
Jul 22 at 14:56
I wouldn't put my second vertex there....
– Lord Shark the Unknown
Jul 22 at 14:48
I wouldn't put my second vertex there....
– Lord Shark the Unknown
Jul 22 at 14:48
1
1
@LordSharktheUnknown: The question claimed that every point had two irrational coordinates. I have two points with no irrational coordinates which disproves the assertion. I saw your comment indicating that we can do it with no irrational coordinates.
– Ross Millikan
Jul 22 at 14:50
@LordSharktheUnknown: The question claimed that every point had two irrational coordinates. I have two points with no irrational coordinates which disproves the assertion. I saw your comment indicating that we can do it with no irrational coordinates.
– Ross Millikan
Jul 22 at 14:50
1
1
Also, the first sentence alone suffices, I suppose?
– user202729
Jul 22 at 14:56
Also, the first sentence alone suffices, I suppose?
– user202729
Jul 22 at 14:56
add a comment |Â
up vote
0
down vote
I think you want to prove that there doesn't exist a regular tetrahedron such that all its vertices have only rational coordinates. Assume the contrary, let $ABCD$ be such a tetrahedron. Consider the plane $(ABC)$ as a complex plane with origin in $A$ and let the points $B$ and $C$ have associated complex numbers $a+bi$ and $c+di$. By our assumption, we would have that $a,b,c,dinmathbbQ$. We know that the points associated to complex numbers $z_1,z_2,z_3$ form an equilateral triangle if and only if $z_1+varepsilon z_2+varepsilon^2z_3=0$ where $varepsilon=-frac12+fracsqrt32i$ is the root of order 3 of the unity. By plugging our numbers, we obtain
$$a+bi+varepsilon(c+di)+varepsilon^2cdot 0=0Rightarrow$$
$$Rightarrow a+bi+left(-frac12+fracsqrt32iright)(c+di)=0Rightarrow$$
$$Rightarrow a+bi-fracc2-fracd2i+fracsqrt3c2i-fracsqrt3d2=0Rightarrow$$
$$Rightarrow a-fracc2-fracsqrt3d2=0Rightarrow d=0$$
also
$$b-fracd2+fracsqrt3c2=0Rightarrow c=0$$
thus $A=C$, so we don't actually have a triangle $ABC$ hence the conclusion.
answered Jul 22 at 15:02
Andrei Cataron
1178
1
This is a correct proof about equilateral triangles in the plane. But in space you can build an equilateral triangle with all rational coordinates - the normal to its plane will have some irrationality. See my comment on the question.
– Ethan Bolker
Jul 22 at 15:10
add a comment |Â
up vote
0
down vote
I think you want to prove that there doesn't exist a regular tetrahedron such that all its vertices have only rational coordinates. Assume the contrary, let $ABCD$ be such a tetrahedron. Consider the plane $(ABC)$ as a complex plane with origin in $A$ and let the points $B$ and $C$ have associated complex numbers $a+bi$ and $c+di$. By our assumption, we would have that $a,b,c,dinmathbbQ$. We know that the points associated to complex numbers $z_1,z_2,z_3$ form an equilateral triangle if and only if $z_1+varepsilon z_2+varepsilon^2z_3=0$ where $varepsilon=-frac12+fracsqrt32i$ is the root of order 3 of the unity. By plugging our numbers, we obtain
$$a+bi+varepsilon(c+di)+varepsilon^2cdot 0=0Rightarrow$$
$$Rightarrow a+bi+left(-frac12+fracsqrt32iright)(c+di)=0Rightarrow$$
$$Rightarrow a+bi-fracc2-fracd2i+fracsqrt3c2i-fracsqrt3d2=0Rightarrow$$
$$Rightarrow a-fracc2-fracsqrt3d2=0Rightarrow d=0$$
also
$$b-fracd2+fracsqrt3c2=0Rightarrow c=0$$
thus $A=C$, so we don't actually have a triangle $ABC$ hence the conclusion.
answered Jul 22 at 15:02
Andrei Cataron
1178
1
This is a correct proof about equilateral triangles in the plane. But in space you can build an equilateral triangle with all rational coordinates - the normal to its plane will have some irrationality. See my comment on the question.
– Ethan Bolker
Jul 22 at 15:10
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I think you want to prove that there doesn't exist a regular tetrahedron such that all its vertices have only rational coordinates. Assume the contrary, let $ABCD$ be such a tetrahedron. Consider the plane $(ABC)$ as a complex plane with origin in $A$ and let the points $B$ and $C$ have associated complex numbers $a+bi$ and $c+di$. By our assumption, we would have that $a,b,c,dinmathbbQ$. We know that the points associated to complex numbers $z_1,z_2,z_3$ form an equilateral triangle if and only if $z_1+varepsilon z_2+varepsilon^2z_3=0$ where $varepsilon=-frac12+fracsqrt32i$ is the root of order 3 of the unity. By plugging our numbers, we obtain
$$a+bi+varepsilon(c+di)+varepsilon^2cdot 0=0Rightarrow$$
$$Rightarrow a+bi+left(-frac12+fracsqrt32iright)(c+di)=0Rightarrow$$
$$Rightarrow a+bi-fracc2-fracd2i+fracsqrt3c2i-fracsqrt3d2=0Rightarrow$$
$$Rightarrow a-fracc2-fracsqrt3d2=0Rightarrow d=0$$
also
$$b-fracd2+fracsqrt3c2=0Rightarrow c=0$$
thus $A=C$, so we don't actually have a triangle $ABC$ hence the conclusion.
answered Jul 22 at 15:02
Andrei Cataron
1178
I think you want to prove that there doesn't exist a regular tetrahedron such that all its vertices have only rational coordinates. Assume the contrary, let $ABCD$ be such a tetrahedron. Consider the plane $(ABC)$ as a complex plane with origin in $A$ and let the points $B$ and $C$ have associated complex numbers $a+bi$ and $c+di$. By our assumption, we would have that $a,b,c,dinmathbbQ$. We know that the points associated to complex numbers $z_1,z_2,z_3$ form an equilateral triangle if and only if $z_1+varepsilon z_2+varepsilon^2z_3=0$ where $varepsilon=-frac12+fracsqrt32i$ is the root of order 3 of the unity. By plugging our numbers, we obtain
$$a+bi+varepsilon(c+di)+varepsilon^2cdot 0=0Rightarrow$$
$$Rightarrow a+bi+left(-frac12+fracsqrt32iright)(c+di)=0Rightarrow$$
$$Rightarrow a+bi-fracc2-fracd2i+fracsqrt3c2i-fracsqrt3d2=0Rightarrow$$
$$Rightarrow a-fracc2-fracsqrt3d2=0Rightarrow d=0$$
also
$$b-fracd2+fracsqrt3c2=0Rightarrow c=0$$
thus $A=C$, so we don't actually have a triangle $ABC$ hence the conclusion.
answered Jul 22 at 15:02
Andrei Cataron
1178
answered Jul 22 at 15:02
Andrei Cataron
1178
answered Jul 22 at 15:02
Andrei Cataron
1178
answered Jul 22 at 15:02
Andrei Cataron
1178
1178
1
This is a correct proof about equilateral triangles in the plane. But in space you can build an equilateral triangle with all rational coordinates - the normal to its plane will have some irrationality. See my comment on the question.
– Ethan Bolker
Jul 22 at 15:10
add a comment |Â
1
This is a correct proof about equilateral triangles in the plane. But in space you can build an equilateral triangle with all rational coordinates - the normal to its plane will have some irrationality. See my comment on the question.
– Ethan Bolker
Jul 22 at 15:10
1
1
This is a correct proof about equilateral triangles in the plane. But in space you can build an equilateral triangle with all rational coordinates - the normal to its plane will have some irrationality. See my comment on the question.
– Ethan Bolker
Jul 22 at 15:10
This is a correct proof about equilateral triangles in the plane. But in space you can build an equilateral triangle with all rational coordinates - the normal to its plane will have some irrationality. See my comment on the question.
– Ethan Bolker
Jul 22 at 15:10
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2859463%2fprove-or-disprove-the-vertices-of-every-regular-tetrahedron-in-3-space-have-at%23new-answer', 'question_page');
);
Post as a guest
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
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
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
2
Think about the tetrahedra in the cube.
– Lord Shark the Unknown
Jul 22 at 14:45
2
This question answers yours. It's also Lord Shark's comment and @ParclyTaxel 's deleted answer. math.stackexchange.com/questions/1423014/…
– Ethan Bolker
Jul 22 at 14:51
You may want to think more carefully about the order of quantifiers in your question. As stated, the obvious interpretation is "For every tetrahedron $T subset mathbb R^3$, and for ever vertex $V = (V_x,V_y,V_z)$ of $T$, at least two of the three coordinates $V_x,V_y,V_z$ are irrational", and it is way too easy to construct a counterexample to that statement as the answer of @RossMillikan shows.
– Lee Mosher
Jul 22 at 16:48