Mixing
Verification of a vector space
$begingroup$
Suppose we have a vector space $
langle V, oplus, otimes rangle$ where $V = mathbb{C}$. Define $z oplus w := |zw| quad forall z,w in V$ and $lambda otimes z := lambda cdot z quad forall lambda in mathbb{C} quad forall z in V$. Verify whether each axiom for vector spaces holds.
I can verify whether most of the axioms hold/fail except I run into the following logical dilemma:
To verify whether there exists a $overrightarrow{mathbf{0}}$ for all $z in V$ such that $z oplus overrightarrow{mathbf{0}} = z quad forall z in V$, note that $|zw|$ is always a nonnegative real number (by definition of modulus of a complex number). We take $z = 1 + i$ as a counterexample, and take an arbitrary $x$ where $x$ is our choice of the zero element (identity element under the operation $oplus$). Then $z oplus x neq z$ since we get that $z oplus x = |(1+i)x| geq 0$ i.e. $z oplus x in mathbb{R}$ for any $x$ we choose to be the zero vector. Hence such $mathbf{overrightarrow{0}}$ does not exist.
Next, to verify whether $forall z in V$, $exists z^* in V$ such that $z oplus z^* = mathbf{overrightarrow{0}}$ (inverse element under the operation $oplus$): We have proven that $mathbf{overrightarrow{0}}$ does not exist in $V$.We can take $z = 1+i$ as a counterexample, and clearly, no $z^*$ exists such that $z oplus z^* = mathbf{overrightarrow{0}}$ as $mathbf{overrightarrow{0}}$ does not even exist in $V$ and since the zero vector of any vector space $langle V, oplus, otimes rangle$ must be unique, we conclude that the axiom fails i.e. for all $z$, there cannot exist a $z^*$ such that $z + z^* = mathbf{overrightarrow{0}}$. No inverse element under $oplus$ exists.
However, since $mathbf{overrightarrow{0}}$ does not even exist in $V$, shouldn't I be able to conclude that the second aforementioned axiom holds by vacuous truth? Then we would have a logical contradiction. Is my approach given above correct?
proof-verification vector-spaces
asked Feb 1 at 4:51
uznamuznam
437
$endgroup$
add a comment |
$begingroup$
Suppose we have a vector space $
langle V, oplus, otimes rangle$ where $V = mathbb{C}$. Define $z oplus w := |zw| quad forall z,w in V$ and $lambda otimes z := lambda cdot z quad forall lambda in mathbb{C} quad forall z in V$. Verify whether each axiom for vector spaces holds.
I can verify whether most of the axioms hold/fail except I run into the following logical dilemma:
To verify whether there exists a $overrightarrow{mathbf{0}}$ for all $z in V$ such that $z oplus overrightarrow{mathbf{0}} = z quad forall z in V$, note that $|zw|$ is always a nonnegative real number (by definition of modulus of a complex number). We take $z = 1 + i$ as a counterexample, and take an arbitrary $x$ where $x$ is our choice of the zero element (identity element under the operation $oplus$). Then $z oplus x neq z$ since we get that $z oplus x = |(1+i)x| geq 0$ i.e. $z oplus x in mathbb{R}$ for any $x$ we choose to be the zero vector. Hence such $mathbf{overrightarrow{0}}$ does not exist.
Next, to verify whether $forall z in V$, $exists z^* in V$ such that $z oplus z^* = mathbf{overrightarrow{0}}$ (inverse element under the operation $oplus$): We have proven that $mathbf{overrightarrow{0}}$ does not exist in $V$.We can take $z = 1+i$ as a counterexample, and clearly, no $z^*$ exists such that $z oplus z^* = mathbf{overrightarrow{0}}$ as $mathbf{overrightarrow{0}}$ does not even exist in $V$ and since the zero vector of any vector space $langle V, oplus, otimes rangle$ must be unique, we conclude that the axiom fails i.e. for all $z$, there cannot exist a $z^*$ such that $z + z^* = mathbf{overrightarrow{0}}$. No inverse element under $oplus$ exists.
However, since $mathbf{overrightarrow{0}}$ does not even exist in $V$, shouldn't I be able to conclude that the second aforementioned axiom holds by vacuous truth? Then we would have a logical contradiction. Is my approach given above correct?
proof-verification vector-spaces
asked Feb 1 at 4:51
uznamuznam
437
$endgroup$
$begingroup$
The inversion axiom must assume a zero first, or it makes no sense. None of this matters, since it takes only one failed axiom to kill the whole thing. It hardly matters if multiple axioms fail.
$endgroup$
– Randall
Feb 1 at 4:53
$begingroup$
@Randall Your response is much appreciated! I am aware that it only takes one failed axiom for us to conclude that $V$ is not a vector space, but for conceptual (logical?) understanding I would just like to identify individually which axiom holds and why. So in this case, I cannot conclude via vacuous truth that the inversion axiom holds?
$endgroup$
– uznam
Feb 1 at 4:58
$begingroup$
@uznam I think the second axiom is vacuously true, but it has no meaning. In mathematical logic, the statement "If I can fly to the Moon for myself, then I can breath on the surface of the Moon." is vacuously true. However, it has no meaning in real life. Like this, the vacuously true axiom (or statement) has no effect in mathematical world. All of them are just my personal thoughts.
$endgroup$
– Doyun Nam
Feb 1 at 5:36
add a comment |
$begingroup$
Suppose we have a vector space $
langle V, oplus, otimes rangle$ where $V = mathbb{C}$. Define $z oplus w := |zw| quad forall z,w in V$ and $lambda otimes z := lambda cdot z quad forall lambda in mathbb{C} quad forall z in V$. Verify whether each axiom for vector spaces holds.
I can verify whether most of the axioms hold/fail except I run into the following logical dilemma:
To verify whether there exists a $overrightarrow{mathbf{0}}$ for all $z in V$ such that $z oplus overrightarrow{mathbf{0}} = z quad forall z in V$, note that $|zw|$ is always a nonnegative real number (by definition of modulus of a complex number). We take $z = 1 + i$ as a counterexample, and take an arbitrary $x$ where $x$ is our choice of the zero element (identity element under the operation $oplus$). Then $z oplus x neq z$ since we get that $z oplus x = |(1+i)x| geq 0$ i.e. $z oplus x in mathbb{R}$ for any $x$ we choose to be the zero vector. Hence such $mathbf{overrightarrow{0}}$ does not exist.
Next, to verify whether $forall z in V$, $exists z^* in V$ such that $z oplus z^* = mathbf{overrightarrow{0}}$ (inverse element under the operation $oplus$): We have proven that $mathbf{overrightarrow{0}}$ does not exist in $V$.We can take $z = 1+i$ as a counterexample, and clearly, no $z^*$ exists such that $z oplus z^* = mathbf{overrightarrow{0}}$ as $mathbf{overrightarrow{0}}$ does not even exist in $V$ and since the zero vector of any vector space $langle V, oplus, otimes rangle$ must be unique, we conclude that the axiom fails i.e. for all $z$, there cannot exist a $z^*$ such that $z + z^* = mathbf{overrightarrow{0}}$. No inverse element under $oplus$ exists.
However, since $mathbf{overrightarrow{0}}$ does not even exist in $V$, shouldn't I be able to conclude that the second aforementioned axiom holds by vacuous truth? Then we would have a logical contradiction. Is my approach given above correct?
proof-verification vector-spaces
asked Feb 1 at 4:51
uznamuznam
437
$endgroup$
Suppose we have a vector space $
langle V, oplus, otimes rangle$ where $V = mathbb{C}$. Define $z oplus w := |zw| quad forall z,w in V$ and $lambda otimes z := lambda cdot z quad forall lambda in mathbb{C} quad forall z in V$. Verify whether each axiom for vector spaces holds.
I can verify whether most of the axioms hold/fail except I run into the following logical dilemma:
To verify whether there exists a $overrightarrow{mathbf{0}}$ for all $z in V$ such that $z oplus overrightarrow{mathbf{0}} = z quad forall z in V$, note that $|zw|$ is always a nonnegative real number (by definition of modulus of a complex number). We take $z = 1 + i$ as a counterexample, and take an arbitrary $x$ where $x$ is our choice of the zero element (identity element under the operation $oplus$). Then $z oplus x neq z$ since we get that $z oplus x = |(1+i)x| geq 0$ i.e. $z oplus x in mathbb{R}$ for any $x$ we choose to be the zero vector. Hence such $mathbf{overrightarrow{0}}$ does not exist.
Next, to verify whether $forall z in V$, $exists z^* in V$ such that $z oplus z^* = mathbf{overrightarrow{0}}$ (inverse element under the operation $oplus$): We have proven that $mathbf{overrightarrow{0}}$ does not exist in $V$.We can take $z = 1+i$ as a counterexample, and clearly, no $z^*$ exists such that $z oplus z^* = mathbf{overrightarrow{0}}$ as $mathbf{overrightarrow{0}}$ does not even exist in $V$ and since the zero vector of any vector space $langle V, oplus, otimes rangle$ must be unique, we conclude that the axiom fails i.e. for all $z$, there cannot exist a $z^*$ such that $z + z^* = mathbf{overrightarrow{0}}$. No inverse element under $oplus$ exists.
However, since $mathbf{overrightarrow{0}}$ does not even exist in $V$, shouldn't I be able to conclude that the second aforementioned axiom holds by vacuous truth? Then we would have a logical contradiction. Is my approach given above correct?
proof-verification vector-spaces
proof-verification vector-spaces
asked Feb 1 at 4:51
uznamuznam
437
asked Feb 1 at 4:51
uznamuznam
437
edited Feb 1 at 5:12
uznam
asked Feb 1 at 4:51
uznamuznam
437
asked Feb 1 at 4:51
uznamuznam
437
asked Feb 1 at 4:51
uznamuznam
437
437
$begingroup$
The inversion axiom must assume a zero first, or it makes no sense. None of this matters, since it takes only one failed axiom to kill the whole thing. It hardly matters if multiple axioms fail.
$endgroup$
– Randall
Feb 1 at 4:53
$begingroup$
@Randall Your response is much appreciated! I am aware that it only takes one failed axiom for us to conclude that $V$ is not a vector space, but for conceptual (logical?) understanding I would just like to identify individually which axiom holds and why. So in this case, I cannot conclude via vacuous truth that the inversion axiom holds?
$endgroup$
– uznam
Feb 1 at 4:58
$begingroup$
@uznam I think the second axiom is vacuously true, but it has no meaning. In mathematical logic, the statement "If I can fly to the Moon for myself, then I can breath on the surface of the Moon." is vacuously true. However, it has no meaning in real life. Like this, the vacuously true axiom (or statement) has no effect in mathematical world. All of them are just my personal thoughts.
$endgroup$
– Doyun Nam
Feb 1 at 5:36
add a comment |
$begingroup$
The inversion axiom must assume a zero first, or it makes no sense. None of this matters, since it takes only one failed axiom to kill the whole thing. It hardly matters if multiple axioms fail.
$endgroup$
– Randall
Feb 1 at 4:53
$begingroup$
@Randall Your response is much appreciated! I am aware that it only takes one failed axiom for us to conclude that $V$ is not a vector space, but for conceptual (logical?) understanding I would just like to identify individually which axiom holds and why. So in this case, I cannot conclude via vacuous truth that the inversion axiom holds?
$endgroup$
– uznam
Feb 1 at 4:58
$begingroup$
@uznam I think the second axiom is vacuously true, but it has no meaning. In mathematical logic, the statement "If I can fly to the Moon for myself, then I can breath on the surface of the Moon." is vacuously true. However, it has no meaning in real life. Like this, the vacuously true axiom (or statement) has no effect in mathematical world. All of them are just my personal thoughts.
$endgroup$
– Doyun Nam
Feb 1 at 5:36
$begingroup$
The inversion axiom must assume a zero first, or it makes no sense. None of this matters, since it takes only one failed axiom to kill the whole thing. It hardly matters if multiple axioms fail.
$endgroup$
– Randall
Feb 1 at 4:53
$begingroup$
The inversion axiom must assume a zero first, or it makes no sense. None of this matters, since it takes only one failed axiom to kill the whole thing. It hardly matters if multiple axioms fail.
$endgroup$
– Randall
Feb 1 at 4:53
$begingroup$
@Randall Your response is much appreciated! I am aware that it only takes one failed axiom for us to conclude that $V$ is not a vector space, but for conceptual (logical?) understanding I would just like to identify individually which axiom holds and why. So in this case, I cannot conclude via vacuous truth that the inversion axiom holds?
$endgroup$
– uznam
Feb 1 at 4:58
$begingroup$
@Randall Your response is much appreciated! I am aware that it only takes one failed axiom for us to conclude that $V$ is not a vector space, but for conceptual (logical?) understanding I would just like to identify individually which axiom holds and why. So in this case, I cannot conclude via vacuous truth that the inversion axiom holds?
$endgroup$
– uznam
Feb 1 at 4:58
$begingroup$
@uznam I think the second axiom is vacuously true, but it has no meaning. In mathematical logic, the statement "If I can fly to the Moon for myself, then I can breath on the surface of the Moon." is vacuously true. However, it has no meaning in real life. Like this, the vacuously true axiom (or statement) has no effect in mathematical world. All of them are just my personal thoughts.
$endgroup$
– Doyun Nam
Feb 1 at 5:36
$begingroup$
@uznam I think the second axiom is vacuously true, but it has no meaning. In mathematical logic, the statement "If I can fly to the Moon for myself, then I can breath on the surface of the Moon." is vacuously true. However, it has no meaning in real life. Like this, the vacuously true axiom (or statement) has no effect in mathematical world. All of them are just my personal thoughts.
$endgroup$
– Doyun Nam
Feb 1 at 5:36
add a comment |
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$begingroup$
The inversion axiom must assume a zero first, or it makes no sense. None of this matters, since it takes only one failed axiom to kill the whole thing. It hardly matters if multiple axioms fail.
$endgroup$
– Randall
Feb 1 at 4:53
$begingroup$
@Randall Your response is much appreciated! I am aware that it only takes one failed axiom for us to conclude that $V$ is not a vector space, but for conceptual (logical?) understanding I would just like to identify individually which axiom holds and why. So in this case, I cannot conclude via vacuous truth that the inversion axiom holds?
$endgroup$
– uznam
Feb 1 at 4:58
$begingroup$
@uznam I think the second axiom is vacuously true, but it has no meaning. In mathematical logic, the statement "If I can fly to the Moon for myself, then I can breath on the surface of the Moon." is vacuously true. However, it has no meaning in real life. Like this, the vacuously true axiom (or statement) has no effect in mathematical world. All of them are just my personal thoughts.
$endgroup$
– Doyun Nam
Feb 1 at 5:36