Vector Space with unusual addition?
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I'm studying before my class starts in a few weeks and I encountered this question in one of the practice problems:
The addition it has given me is defined as,
$(a,b)+(c,d)= (ac,bd)$
It's asking me if this is a vector of space and I am stuck after proving this,
There is an element $0$ in $V$ so that $v + 0 = v$ for all $v$ in $V$.
I did this -> $(a,b)+(1,1) = (1a,1b) = (a,b)$
Stuck right here,
For each $v$ in $V$ there is an element $-v$ in $V$ so that $v+(-v) = 0$.
$(a,b)+(0,0) = (0a,0b) = (0,0)$
Is $(0,0)$ $a$ $-v$ when there's no such thing as '$-0$'?
Do I stop proving right at the step?
So this is not a vector of space?
Thank you for your time.
Edit: Thank you everyone! The question is stated exactly like so,
Show that the set of ordered pairs of positive real numbers is a vector space under the addition and scalar multiplication.
$$(a,b)+(c,d) = (ac,bd),$$
$$c(a,b) = (a^c, b^c).$$
So the additive inverse is an element that, when added to $(a,b)$, will give me the additive identity, which in this case is $(1,1)$?
linear-algebra vector-spaces
asked Aug 28 at 19:49
Eddie Dylan
335
add a comment |Â
up vote
6
down vote
favorite
I'm studying before my class starts in a few weeks and I encountered this question in one of the practice problems:
The addition it has given me is defined as,
$(a,b)+(c,d)= (ac,bd)$
It's asking me if this is a vector of space and I am stuck after proving this,
There is an element $0$ in $V$ so that $v + 0 = v$ for all $v$ in $V$.
I did this -> $(a,b)+(1,1) = (1a,1b) = (a,b)$
Stuck right here,
For each $v$ in $V$ there is an element $-v$ in $V$ so that $v+(-v) = 0$.
$(a,b)+(0,0) = (0a,0b) = (0,0)$
Is $(0,0)$ $a$ $-v$ when there's no such thing as '$-0$'?
Do I stop proving right at the step?
So this is not a vector of space?
Thank you for your time.
Edit: Thank you everyone! The question is stated exactly like so,
Show that the set of ordered pairs of positive real numbers is a vector space under the addition and scalar multiplication.
$$(a,b)+(c,d) = (ac,bd),$$
$$c(a,b) = (a^c, b^c).$$
So the additive inverse is an element that, when added to $(a,b)$, will give me the additive identity, which in this case is $(1,1)$?
linear-algebra vector-spaces
asked Aug 28 at 19:49
Eddie Dylan
335
3
What is $V$? Is it $mathbbR^2$?
– zzuussee
Aug 28 at 19:55
2
To define a vector space, you need a set, a field, an internal law on the set (the addition) and a scalar multiplication. You haven’t told us what are the set, the field and the scalar multiplication. It would be good to provide us with those information!
– mathcounterexamples.net
Aug 28 at 20:02
1
After verifying you do indeed get a real vector space, note that $mathbb R^2to V$ given by $(x,y) mapsto (exp x, exp y)$ is an isomorphism of real vector spaces, where $mathbb R^2$ here carries the usual structure.
– Christoph
Aug 28 at 20:42
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I'm studying before my class starts in a few weeks and I encountered this question in one of the practice problems:
The addition it has given me is defined as,
$(a,b)+(c,d)= (ac,bd)$
It's asking me if this is a vector of space and I am stuck after proving this,
There is an element $0$ in $V$ so that $v + 0 = v$ for all $v$ in $V$.
I did this -> $(a,b)+(1,1) = (1a,1b) = (a,b)$
Stuck right here,
For each $v$ in $V$ there is an element $-v$ in $V$ so that $v+(-v) = 0$.
$(a,b)+(0,0) = (0a,0b) = (0,0)$
Is $(0,0)$ $a$ $-v$ when there's no such thing as '$-0$'?
Do I stop proving right at the step?
So this is not a vector of space?
Thank you for your time.
Edit: Thank you everyone! The question is stated exactly like so,
Show that the set of ordered pairs of positive real numbers is a vector space under the addition and scalar multiplication.
$$(a,b)+(c,d) = (ac,bd),$$
$$c(a,b) = (a^c, b^c).$$
So the additive inverse is an element that, when added to $(a,b)$, will give me the additive identity, which in this case is $(1,1)$?
linear-algebra vector-spaces
asked Aug 28 at 19:49
Eddie Dylan
335
I'm studying before my class starts in a few weeks and I encountered this question in one of the practice problems:
The addition it has given me is defined as,
$(a,b)+(c,d)= (ac,bd)$
It's asking me if this is a vector of space and I am stuck after proving this,
There is an element $0$ in $V$ so that $v + 0 = v$ for all $v$ in $V$.
I did this -> $(a,b)+(1,1) = (1a,1b) = (a,b)$
Stuck right here,
For each $v$ in $V$ there is an element $-v$ in $V$ so that $v+(-v) = 0$.
$(a,b)+(0,0) = (0a,0b) = (0,0)$
Is $(0,0)$ $a$ $-v$ when there's no such thing as '$-0$'?
Do I stop proving right at the step?
So this is not a vector of space?
Thank you for your time.
Edit: Thank you everyone! The question is stated exactly like so,
Show that the set of ordered pairs of positive real numbers is a vector space under the addition and scalar multiplication.
$$(a,b)+(c,d) = (ac,bd),$$
$$c(a,b) = (a^c, b^c).$$
So the additive inverse is an element that, when added to $(a,b)$, will give me the additive identity, which in this case is $(1,1)$?
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Aug 28 at 19:49
Eddie Dylan
335
asked Aug 28 at 19:49
Eddie Dylan
335
edited Aug 28 at 20:25
user99914
asked Aug 28 at 19:49
Eddie Dylan
335
asked Aug 28 at 19:49
Eddie Dylan
335
asked Aug 28 at 19:49
Eddie Dylan
335
335
3
What is $V$? Is it $mathbbR^2$?
– zzuussee
Aug 28 at 19:55
2
To define a vector space, you need a set, a field, an internal law on the set (the addition) and a scalar multiplication. You haven’t told us what are the set, the field and the scalar multiplication. It would be good to provide us with those information!
– mathcounterexamples.net
Aug 28 at 20:02
1
After verifying you do indeed get a real vector space, note that $mathbb R^2to V$ given by $(x,y) mapsto (exp x, exp y)$ is an isomorphism of real vector spaces, where $mathbb R^2$ here carries the usual structure.
– Christoph
Aug 28 at 20:42
add a comment |Â
3
What is $V$? Is it $mathbbR^2$?
– zzuussee
Aug 28 at 19:55
2
To define a vector space, you need a set, a field, an internal law on the set (the addition) and a scalar multiplication. You haven’t told us what are the set, the field and the scalar multiplication. It would be good to provide us with those information!
– mathcounterexamples.net
Aug 28 at 20:02
1
After verifying you do indeed get a real vector space, note that $mathbb R^2to V$ given by $(x,y) mapsto (exp x, exp y)$ is an isomorphism of real vector spaces, where $mathbb R^2$ here carries the usual structure.
– Christoph
Aug 28 at 20:42
3
3
What is $V$? Is it $mathbbR^2$?
– zzuussee
Aug 28 at 19:55
What is $V$? Is it $mathbbR^2$?
– zzuussee
Aug 28 at 19:55
2
2
To define a vector space, you need a set, a field, an internal law on the set (the addition) and a scalar multiplication. You haven’t told us what are the set, the field and the scalar multiplication. It would be good to provide us with those information!
– mathcounterexamples.net
Aug 28 at 20:02
To define a vector space, you need a set, a field, an internal law on the set (the addition) and a scalar multiplication. You haven’t told us what are the set, the field and the scalar multiplication. It would be good to provide us with those information!
– mathcounterexamples.net
Aug 28 at 20:02
1
1
After verifying you do indeed get a real vector space, note that $mathbb R^2to V$ given by $(x,y) mapsto (exp x, exp y)$ is an isomorphism of real vector spaces, where $mathbb R^2$ here carries the usual structure.
– Christoph
Aug 28 at 20:42
After verifying you do indeed get a real vector space, note that $mathbb R^2to V$ given by $(x,y) mapsto (exp x, exp y)$ is an isomorphism of real vector spaces, where $mathbb R^2$ here carries the usual structure.
– Christoph
Aug 28 at 20:42
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
5
down vote
As you have the neutral element $o=(1,1)$ you need to make sure your inverses are relative to that. Assuming $V=(a,b): a,binmathbbR, a,b>0$ or something of that kind you could use $(a,b)+(frac1a,frac1b)=(1,1)$.
What you still need is to tell us how your base field acts on $V$.
answered Aug 28 at 19:59
Kusma
3,440219
I have added the question word for word from my textbook. Thank you!
– Eddie Dylan
Aug 28 at 20:16
add a comment |Â
up vote
5
down vote
First of all $(0,0)$ is not the "zero vector $vec 0$", from what you did $vec 0 = (1,1)$. So finding
$$ v+ (0,0) = (0,0)$$
does not mean that $(0,0)$ is the negative of $v$. You are looking for $(c,d)$ so that
$$ (a, b) + (c, d) = vec 0 = (1,1).$$
Edit (as per the new edit of the OP): The set $V$ is the set of ordered pair of positive real number. So $(0,0)$ is just not an element of $V$.
Oh! That is the relationship between additive inverse and identity. I need a better textbook that has better explanation than the one I have on hand then. Thank you!
– Eddie Dylan
Aug 28 at 20:14
I've made an edit @EddieDylan
– user99914
Aug 28 at 20:32
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
As you have the neutral element $o=(1,1)$ you need to make sure your inverses are relative to that. Assuming $V=(a,b): a,binmathbbR, a,b>0$ or something of that kind you could use $(a,b)+(frac1a,frac1b)=(1,1)$.
What you still need is to tell us how your base field acts on $V$.
answered Aug 28 at 19:59
Kusma
3,440219
I have added the question word for word from my textbook. Thank you!
– Eddie Dylan
Aug 28 at 20:16
add a comment |Â
up vote
5
down vote
As you have the neutral element $o=(1,1)$ you need to make sure your inverses are relative to that. Assuming $V=(a,b): a,binmathbbR, a,b>0$ or something of that kind you could use $(a,b)+(frac1a,frac1b)=(1,1)$.
What you still need is to tell us how your base field acts on $V$.
answered Aug 28 at 19:59
Kusma
3,440219
I have added the question word for word from my textbook. Thank you!
– Eddie Dylan
Aug 28 at 20:16
add a comment |Â
up vote
5
down vote
up vote
5
down vote
As you have the neutral element $o=(1,1)$ you need to make sure your inverses are relative to that. Assuming $V=(a,b): a,binmathbbR, a,b>0$ or something of that kind you could use $(a,b)+(frac1a,frac1b)=(1,1)$.
What you still need is to tell us how your base field acts on $V$.
answered Aug 28 at 19:59
Kusma
3,440219
As you have the neutral element $o=(1,1)$ you need to make sure your inverses are relative to that. Assuming $V=(a,b): a,binmathbbR, a,b>0$ or something of that kind you could use $(a,b)+(frac1a,frac1b)=(1,1)$.
What you still need is to tell us how your base field acts on $V$.
answered Aug 28 at 19:59
Kusma
3,440219
answered Aug 28 at 19:59
Kusma
3,440219
answered Aug 28 at 19:59
Kusma
3,440219
answered Aug 28 at 19:59
Kusma
3,440219
3,440219
I have added the question word for word from my textbook. Thank you!
– Eddie Dylan
Aug 28 at 20:16
add a comment |Â
I have added the question word for word from my textbook. Thank you!
– Eddie Dylan
Aug 28 at 20:16
I have added the question word for word from my textbook. Thank you!
– Eddie Dylan
Aug 28 at 20:16
I have added the question word for word from my textbook. Thank you!
– Eddie Dylan
Aug 28 at 20:16
add a comment |Â
up vote
5
down vote
First of all $(0,0)$ is not the "zero vector $vec 0$", from what you did $vec 0 = (1,1)$. So finding
$$ v+ (0,0) = (0,0)$$
does not mean that $(0,0)$ is the negative of $v$. You are looking for $(c,d)$ so that
$$ (a, b) + (c, d) = vec 0 = (1,1).$$
Edit (as per the new edit of the OP): The set $V$ is the set of ordered pair of positive real number. So $(0,0)$ is just not an element of $V$.
Oh! That is the relationship between additive inverse and identity. I need a better textbook that has better explanation than the one I have on hand then. Thank you!
– Eddie Dylan
Aug 28 at 20:14
I've made an edit @EddieDylan
– user99914
Aug 28 at 20:32
add a comment |Â
up vote
5
down vote
First of all $(0,0)$ is not the "zero vector $vec 0$", from what you did $vec 0 = (1,1)$. So finding
$$ v+ (0,0) = (0,0)$$
does not mean that $(0,0)$ is the negative of $v$. You are looking for $(c,d)$ so that
$$ (a, b) + (c, d) = vec 0 = (1,1).$$
Edit (as per the new edit of the OP): The set $V$ is the set of ordered pair of positive real number. So $(0,0)$ is just not an element of $V$.
Oh! That is the relationship between additive inverse and identity. I need a better textbook that has better explanation than the one I have on hand then. Thank you!
– Eddie Dylan
Aug 28 at 20:14
I've made an edit @EddieDylan
– user99914
Aug 28 at 20:32
add a comment |Â
up vote
5
down vote
up vote
5
down vote
First of all $(0,0)$ is not the "zero vector $vec 0$", from what you did $vec 0 = (1,1)$. So finding
$$ v+ (0,0) = (0,0)$$
does not mean that $(0,0)$ is the negative of $v$. You are looking for $(c,d)$ so that
$$ (a, b) + (c, d) = vec 0 = (1,1).$$
Edit (as per the new edit of the OP): The set $V$ is the set of ordered pair of positive real number. So $(0,0)$ is just not an element of $V$.
First of all $(0,0)$ is not the "zero vector $vec 0$", from what you did $vec 0 = (1,1)$. So finding
$$ v+ (0,0) = (0,0)$$
does not mean that $(0,0)$ is the negative of $v$. You are looking for $(c,d)$ so that
$$ (a, b) + (c, d) = vec 0 = (1,1).$$
Edit (as per the new edit of the OP): The set $V$ is the set of ordered pair of positive real number. So $(0,0)$ is just not an element of $V$.
edited Aug 28 at 20:32
answered Aug 28 at 19:57
user99914
Oh! That is the relationship between additive inverse and identity. I need a better textbook that has better explanation than the one I have on hand then. Thank you!
– Eddie Dylan
Aug 28 at 20:14
I've made an edit @EddieDylan
– user99914
Aug 28 at 20:32
add a comment |Â
Oh! That is the relationship between additive inverse and identity. I need a better textbook that has better explanation than the one I have on hand then. Thank you!
– Eddie Dylan
Aug 28 at 20:14
I've made an edit @EddieDylan
– user99914
Aug 28 at 20:32
Oh! That is the relationship between additive inverse and identity. I need a better textbook that has better explanation than the one I have on hand then. Thank you!
– Eddie Dylan
Aug 28 at 20:14
Oh! That is the relationship between additive inverse and identity. I need a better textbook that has better explanation than the one I have on hand then. Thank you!
– Eddie Dylan
Aug 28 at 20:14
I've made an edit @EddieDylan
– user99914
Aug 28 at 20:32
I've made an edit @EddieDylan
– user99914
Aug 28 at 20:32
add a comment |Â
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3
What is $V$? Is it $mathbbR^2$?
– zzuussee
Aug 28 at 19:55
2
To define a vector space, you need a set, a field, an internal law on the set (the addition) and a scalar multiplication. You haven’t told us what are the set, the field and the scalar multiplication. It would be good to provide us with those information!
– mathcounterexamples.net
Aug 28 at 20:02
1
After verifying you do indeed get a real vector space, note that $mathbb R^2to V$ given by $(x,y) mapsto (exp x, exp y)$ is an isomorphism of real vector spaces, where $mathbb R^2$ here carries the usual structure.
– Christoph
Aug 28 at 20:42