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Non-empty, disjoint subsets of $Bbb R^2,$ both isometric to their union?
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This question was posed to me by a friend who tells me the answer is yes, but I cannot see why.
Does there exist two nonempty, disjoint sets $A, B subset mathbbR^2$ such that $A$ and $B$ are both isometric to their union?
I cannot for the life of me come up with a way of constructing both sets - I've noticed that the measure of both has to be $0$, but that's about it.
real-analysis metric-spaces
edited Jul 15 at 15:26
Cameron Buie
83.5k771153
asked Jul 15 at 15:22
NMister
333110
add a comment |Â
up vote
7
down vote
favorite
This question was posed to me by a friend who tells me the answer is yes, but I cannot see why.
Does there exist two nonempty, disjoint sets $A, B subset mathbbR^2$ such that $A$ and $B$ are both isometric to their union?
I cannot for the life of me come up with a way of constructing both sets - I've noticed that the measure of both has to be $0$, but that's about it.
real-analysis metric-spaces
edited Jul 15 at 15:26
Cameron Buie
83.5k771153
asked Jul 15 at 15:22
NMister
333110
The measure could also be $infty$. Or else they could be non-measurable.
– user357151
Jul 15 at 15:37
Perhaps A and B are of the same range and 'content' but of opposite sign.
– poetasis
Jul 15 at 15:39
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
This question was posed to me by a friend who tells me the answer is yes, but I cannot see why.
Does there exist two nonempty, disjoint sets $A, B subset mathbbR^2$ such that $A$ and $B$ are both isometric to their union?
I cannot for the life of me come up with a way of constructing both sets - I've noticed that the measure of both has to be $0$, but that's about it.
real-analysis metric-spaces
edited Jul 15 at 15:26
Cameron Buie
83.5k771153
asked Jul 15 at 15:22
NMister
333110
This question was posed to me by a friend who tells me the answer is yes, but I cannot see why.
Does there exist two nonempty, disjoint sets $A, B subset mathbbR^2$ such that $A$ and $B$ are both isometric to their union?
I cannot for the life of me come up with a way of constructing both sets - I've noticed that the measure of both has to be $0$, but that's about it.
real-analysis metric-spaces
edited Jul 15 at 15:26
Cameron Buie
83.5k771153
asked Jul 15 at 15:22
NMister
333110
edited Jul 15 at 15:26
Cameron Buie
83.5k771153
edited Jul 15 at 15:26
Cameron Buie
83.5k771153
edited Jul 15 at 15:26
Cameron Buie
83.5k771153
83.5k771153
asked Jul 15 at 15:22
NMister
333110
asked Jul 15 at 15:22
NMister
333110
asked Jul 15 at 15:22
NMister
333110
333110
The measure could also be $infty$. Or else they could be non-measurable.
– user357151
Jul 15 at 15:37
Perhaps A and B are of the same range and 'content' but of opposite sign.
– poetasis
Jul 15 at 15:39
add a comment |Â
The measure could also be $infty$. Or else they could be non-measurable.
– user357151
Jul 15 at 15:37
Perhaps A and B are of the same range and 'content' but of opposite sign.
– poetasis
Jul 15 at 15:39
The measure could also be $infty$. Or else they could be non-measurable.
– user357151
Jul 15 at 15:37
The measure could also be $infty$. Or else they could be non-measurable.
– user357151
Jul 15 at 15:37
Perhaps A and B are of the same range and 'content' but of opposite sign.
– poetasis
Jul 15 at 15:39
Perhaps A and B are of the same range and 'content' but of opposite sign.
– poetasis
Jul 15 at 15:39
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
8
down vote
accepted
Let the group with presentation $langle x,y mid x^2=e rangle$ (that is, the free product $C_2 * mathbb Z$) act on the plane by mapping $x$ to a rotation by $180^circ$ about $(1,0)$ and $y$ to a rotation about the origin by some angle $theta$ whose cosine is transcendental.
(Originally I claimed that this action is faithful, but actually it isn't -- the group element $y^-2xyxyxy^-2xyxyx$ has the identity action no matter what $theta$ is. Fortunately less can do for this particular purpose).
Let $K$ be the set of group elements that can be written on the form
$$ y^n_mxy^n_m-1xcdots xy^n_1 x y^n_0 $$
where $mge 0$, and each $n_jge 0$ except that $n_0$ may be negative.
This set is not a subgroup, but it has two useful properties:
$K$ is the disjoint union of $ykmid kin K$ and $xykmid kin K$.
Different elements of $K$ map the point $p_0 = (3,0)$ to different points. (This can be seen by going to the complex plane where $y$ is multiplication by $e^itheta$ and $x$ is the map $zmapsto 2-z$. Then each element of $K$ maps $3$ to a different Laurent polynomial in $e^itheta$ with integer coefficients, and those all have distinct values because $e^itheta$ is transcendental. Phew!)
Now set
$$ A = ykp_0 mid kin K qquad B = xykp_0 mid kin K $$
Then $A$, $B$, and $Acup B$ are related by rigid motions of the plane, namely:
$$ y^-1A = (xy)^-1B = kp_0 mid kin K = A cup B $$
so they are isometric.
This construction is inspired by the initial step of the proof for the Banach-Tarski paradox. It doesn't need the axiom of choice because it doesn't need to select a representative of each orbit, because it is not required that $Acup B$ is an entire pre-existing shape.
It is natural to ask how $A$ looks -- but unfortunately it can't really be seen: it is dense in $mathbb R^2$.
answered Jul 15 at 16:17
Henning Makholm
226k16291520
Thank you, amazing answer! Really basic question, I know, but is the group you described abelian? Also, how do you know $A$ is dense in the plane?
– NMister
Jul 16 at 19:51
@NMister: No, it is quite non-abelian -- it needs to be; otherwise $yxp_0 in A$ and $xyp_0 in B$ would be the same point and the construction wouldn't work. Nontrivial free products are never abelian.
– Henning Makholm
Jul 16 at 20:27
@NMister: A is dense in the plane because the image of $(3,0)$ under iterations of $y$ is dense in a circle around the origin with radius $3$. Doing an $x$ now moves the circle away from the origin -- so now the distances from the origin are dense in the interval $[1,5]$. Then doing more $y$s will smear the moved circle out to points that lie densely in an annulus with radii from $1$ to $5$, and now the next $x$ will move that annulus again ... as we have more and more $x$s in the word we can get arbitrarily far from the origin.
– Henning Makholm
Jul 16 at 20:33
Thanks! One more question - How do we know we can't cancel the $e^itheta$s?
– NMister
Jul 19 at 17:17
@NMister: The Laurent polynomials must be different for different $kin K$ because (a) the cofficients are always integers, (b) the ones that are not $0$ or $pm 3$ encode $x$s in the word $k$, and the distances between them count the number of $y$s between successive $x$s, (c) exactly one of the coefficients is odd, and its position relative to rightmost "not $0$ or $pm 3$" encodes the power of $y$ after the last $x$ (d) the exponent corresponding to the odd coefficient is the total power of $y$s in $k$.
– Henning Makholm
Jul 19 at 18:03
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
Let the group with presentation $langle x,y mid x^2=e rangle$ (that is, the free product $C_2 * mathbb Z$) act on the plane by mapping $x$ to a rotation by $180^circ$ about $(1,0)$ and $y$ to a rotation about the origin by some angle $theta$ whose cosine is transcendental.
(Originally I claimed that this action is faithful, but actually it isn't -- the group element $y^-2xyxyxy^-2xyxyx$ has the identity action no matter what $theta$ is. Fortunately less can do for this particular purpose).
Let $K$ be the set of group elements that can be written on the form
$$ y^n_mxy^n_m-1xcdots xy^n_1 x y^n_0 $$
where $mge 0$, and each $n_jge 0$ except that $n_0$ may be negative.
This set is not a subgroup, but it has two useful properties:
$K$ is the disjoint union of $ykmid kin K$ and $xykmid kin K$.
Different elements of $K$ map the point $p_0 = (3,0)$ to different points. (This can be seen by going to the complex plane where $y$ is multiplication by $e^itheta$ and $x$ is the map $zmapsto 2-z$. Then each element of $K$ maps $3$ to a different Laurent polynomial in $e^itheta$ with integer coefficients, and those all have distinct values because $e^itheta$ is transcendental. Phew!)
Now set
$$ A = ykp_0 mid kin K qquad B = xykp_0 mid kin K $$
Then $A$, $B$, and $Acup B$ are related by rigid motions of the plane, namely:
$$ y^-1A = (xy)^-1B = kp_0 mid kin K = A cup B $$
so they are isometric.
This construction is inspired by the initial step of the proof for the Banach-Tarski paradox. It doesn't need the axiom of choice because it doesn't need to select a representative of each orbit, because it is not required that $Acup B$ is an entire pre-existing shape.
It is natural to ask how $A$ looks -- but unfortunately it can't really be seen: it is dense in $mathbb R^2$.
answered Jul 15 at 16:17
Henning Makholm
226k16291520
Thank you, amazing answer! Really basic question, I know, but is the group you described abelian? Also, how do you know $A$ is dense in the plane?
– NMister
Jul 16 at 19:51
@NMister: No, it is quite non-abelian -- it needs to be; otherwise $yxp_0 in A$ and $xyp_0 in B$ would be the same point and the construction wouldn't work. Nontrivial free products are never abelian.
– Henning Makholm
Jul 16 at 20:27
@NMister: A is dense in the plane because the image of $(3,0)$ under iterations of $y$ is dense in a circle around the origin with radius $3$. Doing an $x$ now moves the circle away from the origin -- so now the distances from the origin are dense in the interval $[1,5]$. Then doing more $y$s will smear the moved circle out to points that lie densely in an annulus with radii from $1$ to $5$, and now the next $x$ will move that annulus again ... as we have more and more $x$s in the word we can get arbitrarily far from the origin.
– Henning Makholm
Jul 16 at 20:33
Thanks! One more question - How do we know we can't cancel the $e^itheta$s?
– NMister
Jul 19 at 17:17
@NMister: The Laurent polynomials must be different for different $kin K$ because (a) the cofficients are always integers, (b) the ones that are not $0$ or $pm 3$ encode $x$s in the word $k$, and the distances between them count the number of $y$s between successive $x$s, (c) exactly one of the coefficients is odd, and its position relative to rightmost "not $0$ or $pm 3$" encodes the power of $y$ after the last $x$ (d) the exponent corresponding to the odd coefficient is the total power of $y$s in $k$.
– Henning Makholm
Jul 19 at 18:03
add a comment |Â
up vote
8
down vote
accepted
Let the group with presentation $langle x,y mid x^2=e rangle$ (that is, the free product $C_2 * mathbb Z$) act on the plane by mapping $x$ to a rotation by $180^circ$ about $(1,0)$ and $y$ to a rotation about the origin by some angle $theta$ whose cosine is transcendental.
(Originally I claimed that this action is faithful, but actually it isn't -- the group element $y^-2xyxyxy^-2xyxyx$ has the identity action no matter what $theta$ is. Fortunately less can do for this particular purpose).
Let $K$ be the set of group elements that can be written on the form
$$ y^n_mxy^n_m-1xcdots xy^n_1 x y^n_0 $$
where $mge 0$, and each $n_jge 0$ except that $n_0$ may be negative.
This set is not a subgroup, but it has two useful properties:
$K$ is the disjoint union of $ykmid kin K$ and $xykmid kin K$.
Different elements of $K$ map the point $p_0 = (3,0)$ to different points. (This can be seen by going to the complex plane where $y$ is multiplication by $e^itheta$ and $x$ is the map $zmapsto 2-z$. Then each element of $K$ maps $3$ to a different Laurent polynomial in $e^itheta$ with integer coefficients, and those all have distinct values because $e^itheta$ is transcendental. Phew!)
Now set
$$ A = ykp_0 mid kin K qquad B = xykp_0 mid kin K $$
Then $A$, $B$, and $Acup B$ are related by rigid motions of the plane, namely:
$$ y^-1A = (xy)^-1B = kp_0 mid kin K = A cup B $$
so they are isometric.
This construction is inspired by the initial step of the proof for the Banach-Tarski paradox. It doesn't need the axiom of choice because it doesn't need to select a representative of each orbit, because it is not required that $Acup B$ is an entire pre-existing shape.
It is natural to ask how $A$ looks -- but unfortunately it can't really be seen: it is dense in $mathbb R^2$.
answered Jul 15 at 16:17
Henning Makholm
226k16291520
Thank you, amazing answer! Really basic question, I know, but is the group you described abelian? Also, how do you know $A$ is dense in the plane?
– NMister
Jul 16 at 19:51
@NMister: No, it is quite non-abelian -- it needs to be; otherwise $yxp_0 in A$ and $xyp_0 in B$ would be the same point and the construction wouldn't work. Nontrivial free products are never abelian.
– Henning Makholm
Jul 16 at 20:27
@NMister: A is dense in the plane because the image of $(3,0)$ under iterations of $y$ is dense in a circle around the origin with radius $3$. Doing an $x$ now moves the circle away from the origin -- so now the distances from the origin are dense in the interval $[1,5]$. Then doing more $y$s will smear the moved circle out to points that lie densely in an annulus with radii from $1$ to $5$, and now the next $x$ will move that annulus again ... as we have more and more $x$s in the word we can get arbitrarily far from the origin.
– Henning Makholm
Jul 16 at 20:33
Thanks! One more question - How do we know we can't cancel the $e^itheta$s?
– NMister
Jul 19 at 17:17
@NMister: The Laurent polynomials must be different for different $kin K$ because (a) the cofficients are always integers, (b) the ones that are not $0$ or $pm 3$ encode $x$s in the word $k$, and the distances between them count the number of $y$s between successive $x$s, (c) exactly one of the coefficients is odd, and its position relative to rightmost "not $0$ or $pm 3$" encodes the power of $y$ after the last $x$ (d) the exponent corresponding to the odd coefficient is the total power of $y$s in $k$.
– Henning Makholm
Jul 19 at 18:03
add a comment |Â
up vote
8
down vote
accepted
up vote
8
down vote
accepted
Let the group with presentation $langle x,y mid x^2=e rangle$ (that is, the free product $C_2 * mathbb Z$) act on the plane by mapping $x$ to a rotation by $180^circ$ about $(1,0)$ and $y$ to a rotation about the origin by some angle $theta$ whose cosine is transcendental.
(Originally I claimed that this action is faithful, but actually it isn't -- the group element $y^-2xyxyxy^-2xyxyx$ has the identity action no matter what $theta$ is. Fortunately less can do for this particular purpose).
Let $K$ be the set of group elements that can be written on the form
$$ y^n_mxy^n_m-1xcdots xy^n_1 x y^n_0 $$
where $mge 0$, and each $n_jge 0$ except that $n_0$ may be negative.
This set is not a subgroup, but it has two useful properties:
$K$ is the disjoint union of $ykmid kin K$ and $xykmid kin K$.
Different elements of $K$ map the point $p_0 = (3,0)$ to different points. (This can be seen by going to the complex plane where $y$ is multiplication by $e^itheta$ and $x$ is the map $zmapsto 2-z$. Then each element of $K$ maps $3$ to a different Laurent polynomial in $e^itheta$ with integer coefficients, and those all have distinct values because $e^itheta$ is transcendental. Phew!)
Now set
$$ A = ykp_0 mid kin K qquad B = xykp_0 mid kin K $$
Then $A$, $B$, and $Acup B$ are related by rigid motions of the plane, namely:
$$ y^-1A = (xy)^-1B = kp_0 mid kin K = A cup B $$
so they are isometric.
This construction is inspired by the initial step of the proof for the Banach-Tarski paradox. It doesn't need the axiom of choice because it doesn't need to select a representative of each orbit, because it is not required that $Acup B$ is an entire pre-existing shape.
It is natural to ask how $A$ looks -- but unfortunately it can't really be seen: it is dense in $mathbb R^2$.
answered Jul 15 at 16:17
Henning Makholm
226k16291520
Let the group with presentation $langle x,y mid x^2=e rangle$ (that is, the free product $C_2 * mathbb Z$) act on the plane by mapping $x$ to a rotation by $180^circ$ about $(1,0)$ and $y$ to a rotation about the origin by some angle $theta$ whose cosine is transcendental.
(Originally I claimed that this action is faithful, but actually it isn't -- the group element $y^-2xyxyxy^-2xyxyx$ has the identity action no matter what $theta$ is. Fortunately less can do for this particular purpose).
Let $K$ be the set of group elements that can be written on the form
$$ y^n_mxy^n_m-1xcdots xy^n_1 x y^n_0 $$
where $mge 0$, and each $n_jge 0$ except that $n_0$ may be negative.
This set is not a subgroup, but it has two useful properties:
$K$ is the disjoint union of $ykmid kin K$ and $xykmid kin K$.
Different elements of $K$ map the point $p_0 = (3,0)$ to different points. (This can be seen by going to the complex plane where $y$ is multiplication by $e^itheta$ and $x$ is the map $zmapsto 2-z$. Then each element of $K$ maps $3$ to a different Laurent polynomial in $e^itheta$ with integer coefficients, and those all have distinct values because $e^itheta$ is transcendental. Phew!)
Now set
$$ A = ykp_0 mid kin K qquad B = xykp_0 mid kin K $$
Then $A$, $B$, and $Acup B$ are related by rigid motions of the plane, namely:
$$ y^-1A = (xy)^-1B = kp_0 mid kin K = A cup B $$
so they are isometric.
This construction is inspired by the initial step of the proof for the Banach-Tarski paradox. It doesn't need the axiom of choice because it doesn't need to select a representative of each orbit, because it is not required that $Acup B$ is an entire pre-existing shape.
It is natural to ask how $A$ looks -- but unfortunately it can't really be seen: it is dense in $mathbb R^2$.
answered Jul 15 at 16:17
Henning Makholm
226k16291520
edited Jul 19 at 17:58
answered Jul 15 at 16:17
Henning Makholm
226k16291520
answered Jul 15 at 16:17
Henning Makholm
226k16291520
answered Jul 15 at 16:17
Henning Makholm
226k16291520
226k16291520
Thank you, amazing answer! Really basic question, I know, but is the group you described abelian? Also, how do you know $A$ is dense in the plane?
– NMister
Jul 16 at 19:51
@NMister: No, it is quite non-abelian -- it needs to be; otherwise $yxp_0 in A$ and $xyp_0 in B$ would be the same point and the construction wouldn't work. Nontrivial free products are never abelian.
– Henning Makholm
Jul 16 at 20:27
@NMister: A is dense in the plane because the image of $(3,0)$ under iterations of $y$ is dense in a circle around the origin with radius $3$. Doing an $x$ now moves the circle away from the origin -- so now the distances from the origin are dense in the interval $[1,5]$. Then doing more $y$s will smear the moved circle out to points that lie densely in an annulus with radii from $1$ to $5$, and now the next $x$ will move that annulus again ... as we have more and more $x$s in the word we can get arbitrarily far from the origin.
– Henning Makholm
Jul 16 at 20:33
Thanks! One more question - How do we know we can't cancel the $e^itheta$s?
– NMister
Jul 19 at 17:17
@NMister: The Laurent polynomials must be different for different $kin K$ because (a) the cofficients are always integers, (b) the ones that are not $0$ or $pm 3$ encode $x$s in the word $k$, and the distances between them count the number of $y$s between successive $x$s, (c) exactly one of the coefficients is odd, and its position relative to rightmost "not $0$ or $pm 3$" encodes the power of $y$ after the last $x$ (d) the exponent corresponding to the odd coefficient is the total power of $y$s in $k$.
– Henning Makholm
Jul 19 at 18:03
add a comment |Â
Thank you, amazing answer! Really basic question, I know, but is the group you described abelian? Also, how do you know $A$ is dense in the plane?
– NMister
Jul 16 at 19:51
@NMister: No, it is quite non-abelian -- it needs to be; otherwise $yxp_0 in A$ and $xyp_0 in B$ would be the same point and the construction wouldn't work. Nontrivial free products are never abelian.
– Henning Makholm
Jul 16 at 20:27
@NMister: A is dense in the plane because the image of $(3,0)$ under iterations of $y$ is dense in a circle around the origin with radius $3$. Doing an $x$ now moves the circle away from the origin -- so now the distances from the origin are dense in the interval $[1,5]$. Then doing more $y$s will smear the moved circle out to points that lie densely in an annulus with radii from $1$ to $5$, and now the next $x$ will move that annulus again ... as we have more and more $x$s in the word we can get arbitrarily far from the origin.
– Henning Makholm
Jul 16 at 20:33
Thanks! One more question - How do we know we can't cancel the $e^itheta$s?
– NMister
Jul 19 at 17:17
@NMister: The Laurent polynomials must be different for different $kin K$ because (a) the cofficients are always integers, (b) the ones that are not $0$ or $pm 3$ encode $x$s in the word $k$, and the distances between them count the number of $y$s between successive $x$s, (c) exactly one of the coefficients is odd, and its position relative to rightmost "not $0$ or $pm 3$" encodes the power of $y$ after the last $x$ (d) the exponent corresponding to the odd coefficient is the total power of $y$s in $k$.
– Henning Makholm
Jul 19 at 18:03
Thank you, amazing answer! Really basic question, I know, but is the group you described abelian? Also, how do you know $A$ is dense in the plane?
– NMister
Jul 16 at 19:51
Thank you, amazing answer! Really basic question, I know, but is the group you described abelian? Also, how do you know $A$ is dense in the plane?
– NMister
Jul 16 at 19:51
@NMister: No, it is quite non-abelian -- it needs to be; otherwise $yxp_0 in A$ and $xyp_0 in B$ would be the same point and the construction wouldn't work. Nontrivial free products are never abelian.
– Henning Makholm
Jul 16 at 20:27
@NMister: No, it is quite non-abelian -- it needs to be; otherwise $yxp_0 in A$ and $xyp_0 in B$ would be the same point and the construction wouldn't work. Nontrivial free products are never abelian.
– Henning Makholm
Jul 16 at 20:27
@NMister: A is dense in the plane because the image of $(3,0)$ under iterations of $y$ is dense in a circle around the origin with radius $3$. Doing an $x$ now moves the circle away from the origin -- so now the distances from the origin are dense in the interval $[1,5]$. Then doing more $y$s will smear the moved circle out to points that lie densely in an annulus with radii from $1$ to $5$, and now the next $x$ will move that annulus again ... as we have more and more $x$s in the word we can get arbitrarily far from the origin.
– Henning Makholm
Jul 16 at 20:33
@NMister: A is dense in the plane because the image of $(3,0)$ under iterations of $y$ is dense in a circle around the origin with radius $3$. Doing an $x$ now moves the circle away from the origin -- so now the distances from the origin are dense in the interval $[1,5]$. Then doing more $y$s will smear the moved circle out to points that lie densely in an annulus with radii from $1$ to $5$, and now the next $x$ will move that annulus again ... as we have more and more $x$s in the word we can get arbitrarily far from the origin.
– Henning Makholm
Jul 16 at 20:33
Thanks! One more question - How do we know we can't cancel the $e^itheta$s?
– NMister
Jul 19 at 17:17
Thanks! One more question - How do we know we can't cancel the $e^itheta$s?
– NMister
Jul 19 at 17:17
@NMister: The Laurent polynomials must be different for different $kin K$ because (a) the cofficients are always integers, (b) the ones that are not $0$ or $pm 3$ encode $x$s in the word $k$, and the distances between them count the number of $y$s between successive $x$s, (c) exactly one of the coefficients is odd, and its position relative to rightmost "not $0$ or $pm 3$" encodes the power of $y$ after the last $x$ (d) the exponent corresponding to the odd coefficient is the total power of $y$s in $k$.
– Henning Makholm
Jul 19 at 18:03
@NMister: The Laurent polynomials must be different for different $kin K$ because (a) the cofficients are always integers, (b) the ones that are not $0$ or $pm 3$ encode $x$s in the word $k$, and the distances between them count the number of $y$s between successive $x$s, (c) exactly one of the coefficients is odd, and its position relative to rightmost "not $0$ or $pm 3$" encodes the power of $y$ after the last $x$ (d) the exponent corresponding to the odd coefficient is the total power of $y$s in $k$.
– Henning Makholm
Jul 19 at 18:03
add a comment |Â
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The measure could also be $infty$. Or else they could be non-measurable.
– user357151
Jul 15 at 15:37
Perhaps A and B are of the same range and 'content' but of opposite sign.
– poetasis
Jul 15 at 15:39