Mixing
Why is the set of all positive real numbers with addition defined by $x + y = xy$ and scalar multiplication...
$begingroup$
The set of all positive real numbers, with addition defined by $x + y = x * y$ and scalar multiplication defined by $c * x = x ^ c$. How is this a vector space? Doesn't the axiom stating that $(c+d)u = cu + du$ fail?
example: $u = 2$ (since two is a positive real number, it is in the set). $c = 3$ and $d = 4$ ( these two are just scalars).
$(c+d)u = (3+4)(2) = 2^7$
$cu + du = 2^3 + 2^4$, which does not equal $2^7$, so the axiom fails.
How is this a vector space if an axiom fails? Or does this axiom somehow not fail? Any help is appreciated.
linear-algebra vector-spaces
asked Jan 11 at 14:12
James RonaldJames Ronald
1257
$endgroup$
add a comment |
$begingroup$
The set of all positive real numbers, with addition defined by $x + y = x * y$ and scalar multiplication defined by $c * x = x ^ c$. How is this a vector space? Doesn't the axiom stating that $(c+d)u = cu + du$ fail?
example: $u = 2$ (since two is a positive real number, it is in the set). $c = 3$ and $d = 4$ ( these two are just scalars).
$(c+d)u = (3+4)(2) = 2^7$
$cu + du = 2^3 + 2^4$, which does not equal $2^7$, so the axiom fails.
How is this a vector space if an axiom fails? Or does this axiom somehow not fail? Any help is appreciated.
linear-algebra vector-spaces
asked Jan 11 at 14:12
James RonaldJames Ronald
1257
$endgroup$
1
$begingroup$
check cu + du you forgot a step
$endgroup$
– T. Fo
Jan 11 at 14:13
$begingroup$
It's always a bad idea to abuse notation like this...here you, or your source, is using $x+y$ (and $xy$) in two different ways. Even the core definition seems to use $x*y$ in two different ways. Better to introduce new notation for the new operation, otherwise confusion is nearly certain.
$endgroup$
– lulu
Jan 11 at 14:15
$begingroup$
But $cu$ and $du$ are vectors and vectors sum is $x * y$. This means that $cu+du=2^3 * 2^4=2^7$.
$endgroup$
– Mauro ALLEGRANZA
Jan 11 at 14:15
add a comment |
$begingroup$
The set of all positive real numbers, with addition defined by $x + y = x * y$ and scalar multiplication defined by $c * x = x ^ c$. How is this a vector space? Doesn't the axiom stating that $(c+d)u = cu + du$ fail?
example: $u = 2$ (since two is a positive real number, it is in the set). $c = 3$ and $d = 4$ ( these two are just scalars).
$(c+d)u = (3+4)(2) = 2^7$
$cu + du = 2^3 + 2^4$, which does not equal $2^7$, so the axiom fails.
How is this a vector space if an axiom fails? Or does this axiom somehow not fail? Any help is appreciated.
linear-algebra vector-spaces
asked Jan 11 at 14:12
James RonaldJames Ronald
1257
$endgroup$
The set of all positive real numbers, with addition defined by $x + y = x * y$ and scalar multiplication defined by $c * x = x ^ c$. How is this a vector space? Doesn't the axiom stating that $(c+d)u = cu + du$ fail?
example: $u = 2$ (since two is a positive real number, it is in the set). $c = 3$ and $d = 4$ ( these two are just scalars).
$(c+d)u = (3+4)(2) = 2^7$
$cu + du = 2^3 + 2^4$, which does not equal $2^7$, so the axiom fails.
How is this a vector space if an axiom fails? Or does this axiom somehow not fail? Any help is appreciated.
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Jan 11 at 14:12
James RonaldJames Ronald
1257
asked Jan 11 at 14:12
James RonaldJames Ronald
1257
asked Jan 11 at 14:12
James RonaldJames Ronald
1257
asked Jan 11 at 14:12
James RonaldJames Ronald
1257
asked Jan 11 at 14:12
James RonaldJames Ronald
1257
1257
1
$begingroup$
check cu + du you forgot a step
$endgroup$
– T. Fo
Jan 11 at 14:13
$begingroup$
It's always a bad idea to abuse notation like this...here you, or your source, is using $x+y$ (and $xy$) in two different ways. Even the core definition seems to use $x*y$ in two different ways. Better to introduce new notation for the new operation, otherwise confusion is nearly certain.
$endgroup$
– lulu
Jan 11 at 14:15
$begingroup$
But $cu$ and $du$ are vectors and vectors sum is $x * y$. This means that $cu+du=2^3 * 2^4=2^7$.
$endgroup$
– Mauro ALLEGRANZA
Jan 11 at 14:15
add a comment |
1
$begingroup$
check cu + du you forgot a step
$endgroup$
– T. Fo
Jan 11 at 14:13
$begingroup$
It's always a bad idea to abuse notation like this...here you, or your source, is using $x+y$ (and $xy$) in two different ways. Even the core definition seems to use $x*y$ in two different ways. Better to introduce new notation for the new operation, otherwise confusion is nearly certain.
$endgroup$
– lulu
Jan 11 at 14:15
$begingroup$
But $cu$ and $du$ are vectors and vectors sum is $x * y$. This means that $cu+du=2^3 * 2^4=2^7$.
$endgroup$
– Mauro ALLEGRANZA
Jan 11 at 14:15
1
1
$begingroup$
check cu + du you forgot a step
$endgroup$
– T. Fo
Jan 11 at 14:13
$begingroup$
check cu + du you forgot a step
$endgroup$
– T. Fo
Jan 11 at 14:13
$begingroup$
It's always a bad idea to abuse notation like this...here you, or your source, is using $x+y$ (and $xy$) in two different ways. Even the core definition seems to use $x*y$ in two different ways. Better to introduce new notation for the new operation, otherwise confusion is nearly certain.
$endgroup$
– lulu
Jan 11 at 14:15
$begingroup$
It's always a bad idea to abuse notation like this...here you, or your source, is using $x+y$ (and $xy$) in two different ways. Even the core definition seems to use $x*y$ in two different ways. Better to introduce new notation for the new operation, otherwise confusion is nearly certain.
$endgroup$
– lulu
Jan 11 at 14:15
$begingroup$
But $cu$ and $du$ are vectors and vectors sum is $x * y$. This means that $cu+du=2^3 * 2^4=2^7$.
$endgroup$
– Mauro ALLEGRANZA
Jan 11 at 14:15
$begingroup$
But $cu$ and $du$ are vectors and vectors sum is $x * y$. This means that $cu+du=2^3 * 2^4=2^7$.
$endgroup$
– Mauro ALLEGRANZA
Jan 11 at 14:15
add a comment |
1 Answer
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$begingroup$
It definitely is a vector space. You seem to be getting confused about the two $+$ and two $*$ operations, as well as the fact that vectors and scalars have some overlap. It would be better to give the operations their own names:
begin{align*}
u oplus v &= uv \
lambda odot u &= u^lambda.
end{align*}
The distributivity law that you're trying to verify is as follows:
$$(lambda + mu) odot u = (lambda odot u) oplus (mu odot u).$$
Please note the scalar $+$ and the vector $oplus$, and where they belong. Don't forget that $lambda$ and $mu$ are scalars, and so they must be added by regular addition on $mathbb{R}$.
When simplifying this rule, we get,
$$u^{lambda + mu} = u^lambda cdot u^mu,$$
which is a well-known exponential law.
answered Jan 11 at 14:21
Theo BenditTheo Bendit
17.9k12152
$endgroup$
1
$begingroup$
Ohhh I see, yes I completely missed that and that makes sense. Thank you so much!
$endgroup$
– James Ronald
Jan 11 at 14:57
add a comment |
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1 Answer
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1 Answer
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$begingroup$
It definitely is a vector space. You seem to be getting confused about the two $+$ and two $*$ operations, as well as the fact that vectors and scalars have some overlap. It would be better to give the operations their own names:
begin{align*}
u oplus v &= uv \
lambda odot u &= u^lambda.
end{align*}
The distributivity law that you're trying to verify is as follows:
$$(lambda + mu) odot u = (lambda odot u) oplus (mu odot u).$$
Please note the scalar $+$ and the vector $oplus$, and where they belong. Don't forget that $lambda$ and $mu$ are scalars, and so they must be added by regular addition on $mathbb{R}$.
When simplifying this rule, we get,
$$u^{lambda + mu} = u^lambda cdot u^mu,$$
which is a well-known exponential law.
answered Jan 11 at 14:21
Theo BenditTheo Bendit
17.9k12152
$endgroup$
1
$begingroup$
Ohhh I see, yes I completely missed that and that makes sense. Thank you so much!
$endgroup$
– James Ronald
Jan 11 at 14:57
add a comment |
$begingroup$
It definitely is a vector space. You seem to be getting confused about the two $+$ and two $*$ operations, as well as the fact that vectors and scalars have some overlap. It would be better to give the operations their own names:
begin{align*}
u oplus v &= uv \
lambda odot u &= u^lambda.
end{align*}
The distributivity law that you're trying to verify is as follows:
$$(lambda + mu) odot u = (lambda odot u) oplus (mu odot u).$$
Please note the scalar $+$ and the vector $oplus$, and where they belong. Don't forget that $lambda$ and $mu$ are scalars, and so they must be added by regular addition on $mathbb{R}$.
When simplifying this rule, we get,
$$u^{lambda + mu} = u^lambda cdot u^mu,$$
which is a well-known exponential law.
answered Jan 11 at 14:21
Theo BenditTheo Bendit
17.9k12152
$endgroup$
1
$begingroup$
Ohhh I see, yes I completely missed that and that makes sense. Thank you so much!
$endgroup$
– James Ronald
Jan 11 at 14:57
add a comment |
$begingroup$
It definitely is a vector space. You seem to be getting confused about the two $+$ and two $*$ operations, as well as the fact that vectors and scalars have some overlap. It would be better to give the operations their own names:
begin{align*}
u oplus v &= uv \
lambda odot u &= u^lambda.
end{align*}
The distributivity law that you're trying to verify is as follows:
$$(lambda + mu) odot u = (lambda odot u) oplus (mu odot u).$$
Please note the scalar $+$ and the vector $oplus$, and where they belong. Don't forget that $lambda$ and $mu$ are scalars, and so they must be added by regular addition on $mathbb{R}$.
When simplifying this rule, we get,
$$u^{lambda + mu} = u^lambda cdot u^mu,$$
which is a well-known exponential law.
answered Jan 11 at 14:21
Theo BenditTheo Bendit
17.9k12152
$endgroup$
It definitely is a vector space. You seem to be getting confused about the two $+$ and two $*$ operations, as well as the fact that vectors and scalars have some overlap. It would be better to give the operations their own names:
begin{align*}
u oplus v &= uv \
lambda odot u &= u^lambda.
end{align*}
The distributivity law that you're trying to verify is as follows:
$$(lambda + mu) odot u = (lambda odot u) oplus (mu odot u).$$
Please note the scalar $+$ and the vector $oplus$, and where they belong. Don't forget that $lambda$ and $mu$ are scalars, and so they must be added by regular addition on $mathbb{R}$.
When simplifying this rule, we get,
$$u^{lambda + mu} = u^lambda cdot u^mu,$$
which is a well-known exponential law.
answered Jan 11 at 14:21
Theo BenditTheo Bendit
17.9k12152
edited Jan 11 at 15:05
answered Jan 11 at 14:21
Theo BenditTheo Bendit
17.9k12152
answered Jan 11 at 14:21
Theo BenditTheo Bendit
17.9k12152
answered Jan 11 at 14:21
Theo BenditTheo Bendit
17.9k12152
17.9k12152
1
$begingroup$
Ohhh I see, yes I completely missed that and that makes sense. Thank you so much!
$endgroup$
– James Ronald
Jan 11 at 14:57
add a comment |
1
$begingroup$
Ohhh I see, yes I completely missed that and that makes sense. Thank you so much!
$endgroup$
– James Ronald
Jan 11 at 14:57
1
1
$begingroup$
Ohhh I see, yes I completely missed that and that makes sense. Thank you so much!
$endgroup$
– James Ronald
Jan 11 at 14:57
$begingroup$
Ohhh I see, yes I completely missed that and that makes sense. Thank you so much!
$endgroup$
– James Ronald
Jan 11 at 14:57
add a comment |
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1
$begingroup$
check cu + du you forgot a step
$endgroup$
– T. Fo
Jan 11 at 14:13
$begingroup$
It's always a bad idea to abuse notation like this...here you, or your source, is using $x+y$ (and $xy$) in two different ways. Even the core definition seems to use $x*y$ in two different ways. Better to introduce new notation for the new operation, otherwise confusion is nearly certain.
$endgroup$
– lulu
Jan 11 at 14:15
$begingroup$
But $cu$ and $du$ are vectors and vectors sum is $x * y$. This means that $cu+du=2^3 * 2^4=2^7$.
$endgroup$
– Mauro ALLEGRANZA
Jan 11 at 14:15