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Define $*$ on $mathbbZ$ by $a*b = a+b$. Show $*$ is a binary operation on $mathbbZ^+$.
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Let $S = mathbbZ^+$. Define $*$ on $mathbbZ$ by $a*b = a+b$. Show $*$ is a binary operation on $mathbbZ^+$.
In our course on Abstract Algebra our book says the following in it.
$*$ is a binary operation on $mathbbZ^+$. Because
- $*$ is well defined: if $(a, b) = (c, d)$, then $a = c$ and $b=d$. Hence we have $$*(a, b)= a+b = c+d = *(c, d)$$
- $mathbbZ^+$ is closed under $*$. If $a, b in mathbbZ^+$, then $a*b = a+b in mathbbZ^+$
The above seems wrong to me. I see two errors. Firstly if $*$ is a well-defined function then there is no need to show that $mathbbZ^+$ is closed under $*$ and secondly they haven't actually shown well definedness of $*$, since all they've done is just take two equal elements in $mathbbZ^+ times mathbbZ^+$ and shown their outputs are equal.
Now the way I see it, all I have to show is that $* : mathbbZ^+ times mathbbZ^+ to mathbbZ^+$ defined by $$*(a, b) = a*b = a+b$$ is actually a well-defined function. To do this (in more familiar notation compared to the answer given here: https://math.stackexchange.com/a/313182/266135) I need to show the following
- $*(a, b) = a+b in mathbbZ^+$ for all $(a, b) in mathbbZ^+ times mathbbZ^+$
- $forall (x, y) in mathbbZ^+ times mathbbZ^+$, there exists a $z in mathbbZ^+$ such that $*(x, y) = x+y =z$
- If $*(a, b) =c$ and $*(a, b)=d$ then $c=d$.
Now 1. above is satisfied easily since we know that by properties of the integers that the addition of any two positive integers is again a positive integer. Since 1. is satisfied 2. is redudant and there's nothing to prove (please correct me if I am wrong on this). So all that's left to prove is 3.
To prove 3. all I'd need to do is suppose that $*(a, b) = c$ and $*(a, b) = d$, then I'd have $a+b=c$ and $a+b=d$, but then trivially I'd have $c=d$ (because if you add the same two positive integers $a, b$ together then you get the same unique result $a+b$ always).
Am I correct in everything I've said?
functions elementary-set-theory binary-operations
asked Jul 24 at 20:27
Perturbative
3,49911039
add a comment |Â
up vote
1
down vote
favorite
Let $S = mathbbZ^+$. Define $*$ on $mathbbZ$ by $a*b = a+b$. Show $*$ is a binary operation on $mathbbZ^+$.
In our course on Abstract Algebra our book says the following in it.
$*$ is a binary operation on $mathbbZ^+$. Because
- $*$ is well defined: if $(a, b) = (c, d)$, then $a = c$ and $b=d$. Hence we have $$*(a, b)= a+b = c+d = *(c, d)$$
- $mathbbZ^+$ is closed under $*$. If $a, b in mathbbZ^+$, then $a*b = a+b in mathbbZ^+$
The above seems wrong to me. I see two errors. Firstly if $*$ is a well-defined function then there is no need to show that $mathbbZ^+$ is closed under $*$ and secondly they haven't actually shown well definedness of $*$, since all they've done is just take two equal elements in $mathbbZ^+ times mathbbZ^+$ and shown their outputs are equal.
Now the way I see it, all I have to show is that $* : mathbbZ^+ times mathbbZ^+ to mathbbZ^+$ defined by $$*(a, b) = a*b = a+b$$ is actually a well-defined function. To do this (in more familiar notation compared to the answer given here: https://math.stackexchange.com/a/313182/266135) I need to show the following
- $*(a, b) = a+b in mathbbZ^+$ for all $(a, b) in mathbbZ^+ times mathbbZ^+$
- $forall (x, y) in mathbbZ^+ times mathbbZ^+$, there exists a $z in mathbbZ^+$ such that $*(x, y) = x+y =z$
- If $*(a, b) =c$ and $*(a, b)=d$ then $c=d$.
Now 1. above is satisfied easily since we know that by properties of the integers that the addition of any two positive integers is again a positive integer. Since 1. is satisfied 2. is redudant and there's nothing to prove (please correct me if I am wrong on this). So all that's left to prove is 3.
To prove 3. all I'd need to do is suppose that $*(a, b) = c$ and $*(a, b) = d$, then I'd have $a+b=c$ and $a+b=d$, but then trivially I'd have $c=d$ (because if you add the same two positive integers $a, b$ together then you get the same unique result $a+b$ always).
Am I correct in everything I've said?
functions elementary-set-theory binary-operations
asked Jul 24 at 20:27
Perturbative
3,49911039
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $S = mathbbZ^+$. Define $*$ on $mathbbZ$ by $a*b = a+b$. Show $*$ is a binary operation on $mathbbZ^+$.
In our course on Abstract Algebra our book says the following in it.
$*$ is a binary operation on $mathbbZ^+$. Because
- $*$ is well defined: if $(a, b) = (c, d)$, then $a = c$ and $b=d$. Hence we have $$*(a, b)= a+b = c+d = *(c, d)$$
- $mathbbZ^+$ is closed under $*$. If $a, b in mathbbZ^+$, then $a*b = a+b in mathbbZ^+$
The above seems wrong to me. I see two errors. Firstly if $*$ is a well-defined function then there is no need to show that $mathbbZ^+$ is closed under $*$ and secondly they haven't actually shown well definedness of $*$, since all they've done is just take two equal elements in $mathbbZ^+ times mathbbZ^+$ and shown their outputs are equal.
Now the way I see it, all I have to show is that $* : mathbbZ^+ times mathbbZ^+ to mathbbZ^+$ defined by $$*(a, b) = a*b = a+b$$ is actually a well-defined function. To do this (in more familiar notation compared to the answer given here: https://math.stackexchange.com/a/313182/266135) I need to show the following
- $*(a, b) = a+b in mathbbZ^+$ for all $(a, b) in mathbbZ^+ times mathbbZ^+$
- $forall (x, y) in mathbbZ^+ times mathbbZ^+$, there exists a $z in mathbbZ^+$ such that $*(x, y) = x+y =z$
- If $*(a, b) =c$ and $*(a, b)=d$ then $c=d$.
Now 1. above is satisfied easily since we know that by properties of the integers that the addition of any two positive integers is again a positive integer. Since 1. is satisfied 2. is redudant and there's nothing to prove (please correct me if I am wrong on this). So all that's left to prove is 3.
To prove 3. all I'd need to do is suppose that $*(a, b) = c$ and $*(a, b) = d$, then I'd have $a+b=c$ and $a+b=d$, but then trivially I'd have $c=d$ (because if you add the same two positive integers $a, b$ together then you get the same unique result $a+b$ always).
Am I correct in everything I've said?
functions elementary-set-theory binary-operations
asked Jul 24 at 20:27
Perturbative
3,49911039
Let $S = mathbbZ^+$. Define $*$ on $mathbbZ$ by $a*b = a+b$. Show $*$ is a binary operation on $mathbbZ^+$.
In our course on Abstract Algebra our book says the following in it.
$*$ is a binary operation on $mathbbZ^+$. Because
- $*$ is well defined: if $(a, b) = (c, d)$, then $a = c$ and $b=d$. Hence we have $$*(a, b)= a+b = c+d = *(c, d)$$
- $mathbbZ^+$ is closed under $*$. If $a, b in mathbbZ^+$, then $a*b = a+b in mathbbZ^+$
The above seems wrong to me. I see two errors. Firstly if $*$ is a well-defined function then there is no need to show that $mathbbZ^+$ is closed under $*$ and secondly they haven't actually shown well definedness of $*$, since all they've done is just take two equal elements in $mathbbZ^+ times mathbbZ^+$ and shown their outputs are equal.
Now the way I see it, all I have to show is that $* : mathbbZ^+ times mathbbZ^+ to mathbbZ^+$ defined by $$*(a, b) = a*b = a+b$$ is actually a well-defined function. To do this (in more familiar notation compared to the answer given here: https://math.stackexchange.com/a/313182/266135) I need to show the following
- $*(a, b) = a+b in mathbbZ^+$ for all $(a, b) in mathbbZ^+ times mathbbZ^+$
- $forall (x, y) in mathbbZ^+ times mathbbZ^+$, there exists a $z in mathbbZ^+$ such that $*(x, y) = x+y =z$
- If $*(a, b) =c$ and $*(a, b)=d$ then $c=d$.
Now 1. above is satisfied easily since we know that by properties of the integers that the addition of any two positive integers is again a positive integer. Since 1. is satisfied 2. is redudant and there's nothing to prove (please correct me if I am wrong on this). So all that's left to prove is 3.
To prove 3. all I'd need to do is suppose that $*(a, b) = c$ and $*(a, b) = d$, then I'd have $a+b=c$ and $a+b=d$, but then trivially I'd have $c=d$ (because if you add the same two positive integers $a, b$ together then you get the same unique result $a+b$ always).
Am I correct in everything I've said?
functions elementary-set-theory binary-operations
asked Jul 24 at 20:27
Perturbative
3,49911039
asked Jul 24 at 20:27
Perturbative
3,49911039
asked Jul 24 at 20:27
Perturbative
3,49911039
asked Jul 24 at 20:27
Perturbative
3,49911039
3,49911039
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
You're right that what you quote from the book doesn't seem very enlightening. It even looks likely that the author is somehow confusing the situation for the case where showing well-definedness is a meaningful task (such as when defining arithmetic on congruence classes for modular arithmetic).
So you're basically correct in your criticism, but your own proposal is still more complex that it needs to be.
It seems that your properties 1 and 2 are just slightly different ways of saying the same thing; you certainly don't need to prove them separately.
Your property 3 just says that $*$ is actually a function. It would be relevant to prove something like this if you have attempted to define a function by a description where you (apparently) need to make some kind of arbitrary choice to apply the description to find a function value. Then one needs to show that the value you eventually get is the same no matter which choices you made along the way. But there's nothing of that sort going on, so demanding (or providing) a proof that the same inputs give the same outputs are pretty pointless here.
The exception to this is if you have actually defined $mathbb Z$ as equivalence classes of pairs of naturals (as first done by Hamilton). If so, you may need to show that whatever your definition of adding such equivalence classes is, is independent of which representatives you choose for each of the operands. But that is probably not the point here, where it seems that you already know what $+$ means and are just asked to show that you can choose to call it $*$ instead.
answered Jul 24 at 20:35
Henning Makholm
225k16290516
add a comment |Â
up vote
1
down vote
The proof in the book is correct, and so is yours. An $n$-ary operation on a set $A$ is a function $f: A^nrightarrow A$. This can be rephrased by saying that it is "well-defined" (at a given point $(a_1, ldots, a_n)$ it has a unique value), and that $A$ is closed under the function (the image is contained in $A$).
And this is exactly what is checked, and it is completely correct.
Your first point means that the image is contained in $A$.
The second and third together means that $star$ is a funcion.
answered Jul 24 at 20:32
A. Pongrácz
1,804116
Your last paragraph is incorrect. If I have a function $f : X to Y$, $f$ being surjective means that $forall y in Y$ there exists an $x in X$ such that $f(x)=y$, the quanitfiers in point 2. in my question are swapped around and not the same as surjectivity
– Perturbative
Jul 25 at 5:34
You are right, my mistake. In that case, your approach is equally correct to the one in the book. I edit my answer.
– A. Pongrácz
Jul 25 at 6:15
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You're right that what you quote from the book doesn't seem very enlightening. It even looks likely that the author is somehow confusing the situation for the case where showing well-definedness is a meaningful task (such as when defining arithmetic on congruence classes for modular arithmetic).
So you're basically correct in your criticism, but your own proposal is still more complex that it needs to be.
It seems that your properties 1 and 2 are just slightly different ways of saying the same thing; you certainly don't need to prove them separately.
Your property 3 just says that $*$ is actually a function. It would be relevant to prove something like this if you have attempted to define a function by a description where you (apparently) need to make some kind of arbitrary choice to apply the description to find a function value. Then one needs to show that the value you eventually get is the same no matter which choices you made along the way. But there's nothing of that sort going on, so demanding (or providing) a proof that the same inputs give the same outputs are pretty pointless here.
The exception to this is if you have actually defined $mathbb Z$ as equivalence classes of pairs of naturals (as first done by Hamilton). If so, you may need to show that whatever your definition of adding such equivalence classes is, is independent of which representatives you choose for each of the operands. But that is probably not the point here, where it seems that you already know what $+$ means and are just asked to show that you can choose to call it $*$ instead.
answered Jul 24 at 20:35
Henning Makholm
225k16290516
add a comment |Â
up vote
2
down vote
accepted
You're right that what you quote from the book doesn't seem very enlightening. It even looks likely that the author is somehow confusing the situation for the case where showing well-definedness is a meaningful task (such as when defining arithmetic on congruence classes for modular arithmetic).
So you're basically correct in your criticism, but your own proposal is still more complex that it needs to be.
It seems that your properties 1 and 2 are just slightly different ways of saying the same thing; you certainly don't need to prove them separately.
Your property 3 just says that $*$ is actually a function. It would be relevant to prove something like this if you have attempted to define a function by a description where you (apparently) need to make some kind of arbitrary choice to apply the description to find a function value. Then one needs to show that the value you eventually get is the same no matter which choices you made along the way. But there's nothing of that sort going on, so demanding (or providing) a proof that the same inputs give the same outputs are pretty pointless here.
The exception to this is if you have actually defined $mathbb Z$ as equivalence classes of pairs of naturals (as first done by Hamilton). If so, you may need to show that whatever your definition of adding such equivalence classes is, is independent of which representatives you choose for each of the operands. But that is probably not the point here, where it seems that you already know what $+$ means and are just asked to show that you can choose to call it $*$ instead.
answered Jul 24 at 20:35
Henning Makholm
225k16290516
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You're right that what you quote from the book doesn't seem very enlightening. It even looks likely that the author is somehow confusing the situation for the case where showing well-definedness is a meaningful task (such as when defining arithmetic on congruence classes for modular arithmetic).
So you're basically correct in your criticism, but your own proposal is still more complex that it needs to be.
It seems that your properties 1 and 2 are just slightly different ways of saying the same thing; you certainly don't need to prove them separately.
Your property 3 just says that $*$ is actually a function. It would be relevant to prove something like this if you have attempted to define a function by a description where you (apparently) need to make some kind of arbitrary choice to apply the description to find a function value. Then one needs to show that the value you eventually get is the same no matter which choices you made along the way. But there's nothing of that sort going on, so demanding (or providing) a proof that the same inputs give the same outputs are pretty pointless here.
The exception to this is if you have actually defined $mathbb Z$ as equivalence classes of pairs of naturals (as first done by Hamilton). If so, you may need to show that whatever your definition of adding such equivalence classes is, is independent of which representatives you choose for each of the operands. But that is probably not the point here, where it seems that you already know what $+$ means and are just asked to show that you can choose to call it $*$ instead.
answered Jul 24 at 20:35
Henning Makholm
225k16290516
You're right that what you quote from the book doesn't seem very enlightening. It even looks likely that the author is somehow confusing the situation for the case where showing well-definedness is a meaningful task (such as when defining arithmetic on congruence classes for modular arithmetic).
So you're basically correct in your criticism, but your own proposal is still more complex that it needs to be.
It seems that your properties 1 and 2 are just slightly different ways of saying the same thing; you certainly don't need to prove them separately.
Your property 3 just says that $*$ is actually a function. It would be relevant to prove something like this if you have attempted to define a function by a description where you (apparently) need to make some kind of arbitrary choice to apply the description to find a function value. Then one needs to show that the value you eventually get is the same no matter which choices you made along the way. But there's nothing of that sort going on, so demanding (or providing) a proof that the same inputs give the same outputs are pretty pointless here.
The exception to this is if you have actually defined $mathbb Z$ as equivalence classes of pairs of naturals (as first done by Hamilton). If so, you may need to show that whatever your definition of adding such equivalence classes is, is independent of which representatives you choose for each of the operands. But that is probably not the point here, where it seems that you already know what $+$ means and are just asked to show that you can choose to call it $*$ instead.
answered Jul 24 at 20:35
Henning Makholm
225k16290516
edited Jul 24 at 20:48
answered Jul 24 at 20:35
Henning Makholm
225k16290516
answered Jul 24 at 20:35
Henning Makholm
225k16290516
answered Jul 24 at 20:35
Henning Makholm
225k16290516
225k16290516
add a comment |Â
add a comment |Â
up vote
1
down vote
The proof in the book is correct, and so is yours. An $n$-ary operation on a set $A$ is a function $f: A^nrightarrow A$. This can be rephrased by saying that it is "well-defined" (at a given point $(a_1, ldots, a_n)$ it has a unique value), and that $A$ is closed under the function (the image is contained in $A$).
And this is exactly what is checked, and it is completely correct.
Your first point means that the image is contained in $A$.
The second and third together means that $star$ is a funcion.
answered Jul 24 at 20:32
A. Pongrácz
1,804116
Your last paragraph is incorrect. If I have a function $f : X to Y$, $f$ being surjective means that $forall y in Y$ there exists an $x in X$ such that $f(x)=y$, the quanitfiers in point 2. in my question are swapped around and not the same as surjectivity
– Perturbative
Jul 25 at 5:34
You are right, my mistake. In that case, your approach is equally correct to the one in the book. I edit my answer.
– A. Pongrácz
Jul 25 at 6:15
add a comment |Â
up vote
1
down vote
The proof in the book is correct, and so is yours. An $n$-ary operation on a set $A$ is a function $f: A^nrightarrow A$. This can be rephrased by saying that it is "well-defined" (at a given point $(a_1, ldots, a_n)$ it has a unique value), and that $A$ is closed under the function (the image is contained in $A$).
And this is exactly what is checked, and it is completely correct.
Your first point means that the image is contained in $A$.
The second and third together means that $star$ is a funcion.
answered Jul 24 at 20:32
A. Pongrácz
1,804116
Your last paragraph is incorrect. If I have a function $f : X to Y$, $f$ being surjective means that $forall y in Y$ there exists an $x in X$ such that $f(x)=y$, the quanitfiers in point 2. in my question are swapped around and not the same as surjectivity
– Perturbative
Jul 25 at 5:34
You are right, my mistake. In that case, your approach is equally correct to the one in the book. I edit my answer.
– A. Pongrácz
Jul 25 at 6:15
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The proof in the book is correct, and so is yours. An $n$-ary operation on a set $A$ is a function $f: A^nrightarrow A$. This can be rephrased by saying that it is "well-defined" (at a given point $(a_1, ldots, a_n)$ it has a unique value), and that $A$ is closed under the function (the image is contained in $A$).
And this is exactly what is checked, and it is completely correct.
Your first point means that the image is contained in $A$.
The second and third together means that $star$ is a funcion.
answered Jul 24 at 20:32
A. Pongrácz
1,804116
The proof in the book is correct, and so is yours. An $n$-ary operation on a set $A$ is a function $f: A^nrightarrow A$. This can be rephrased by saying that it is "well-defined" (at a given point $(a_1, ldots, a_n)$ it has a unique value), and that $A$ is closed under the function (the image is contained in $A$).
And this is exactly what is checked, and it is completely correct.
Your first point means that the image is contained in $A$.
The second and third together means that $star$ is a funcion.
answered Jul 24 at 20:32
A. Pongrácz
1,804116
edited Jul 25 at 6:18
answered Jul 24 at 20:32
A. Pongrácz
1,804116
answered Jul 24 at 20:32
A. Pongrácz
1,804116
answered Jul 24 at 20:32
A. Pongrácz
1,804116
1,804116
Your last paragraph is incorrect. If I have a function $f : X to Y$, $f$ being surjective means that $forall y in Y$ there exists an $x in X$ such that $f(x)=y$, the quanitfiers in point 2. in my question are swapped around and not the same as surjectivity
– Perturbative
Jul 25 at 5:34
You are right, my mistake. In that case, your approach is equally correct to the one in the book. I edit my answer.
– A. Pongrácz
Jul 25 at 6:15
add a comment |Â
Your last paragraph is incorrect. If I have a function $f : X to Y$, $f$ being surjective means that $forall y in Y$ there exists an $x in X$ such that $f(x)=y$, the quanitfiers in point 2. in my question are swapped around and not the same as surjectivity
– Perturbative
Jul 25 at 5:34
You are right, my mistake. In that case, your approach is equally correct to the one in the book. I edit my answer.
– A. Pongrácz
Jul 25 at 6:15
Your last paragraph is incorrect. If I have a function $f : X to Y$, $f$ being surjective means that $forall y in Y$ there exists an $x in X$ such that $f(x)=y$, the quanitfiers in point 2. in my question are swapped around and not the same as surjectivity
– Perturbative
Jul 25 at 5:34
Your last paragraph is incorrect. If I have a function $f : X to Y$, $f$ being surjective means that $forall y in Y$ there exists an $x in X$ such that $f(x)=y$, the quanitfiers in point 2. in my question are swapped around and not the same as surjectivity
– Perturbative
Jul 25 at 5:34
You are right, my mistake. In that case, your approach is equally correct to the one in the book. I edit my answer.
– A. Pongrácz
Jul 25 at 6:15
You are right, my mistake. In that case, your approach is equally correct to the one in the book. I edit my answer.
– A. Pongrácz
Jul 25 at 6:15
add a comment |Â
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