$(mathbb{R}, oplus, odot)$ is a field? Given $a oplus b=a+b+1$ and $a odot b=a+b+ab$.
We are given that $a oplus b = a+b+1$ and $a odot b= a+b+ab$ and I need to prove that it is a field. Do I just need to know that there is an inverse for $a odot b$? So if I could let $a+b+ab=c$ then I want to find a $-c$ such that $c odot -c=1$? Then the inverse, $-c=frac{1}{a+b+ab}$ then? This would be that for every c I have a multiplicative inverse and since the set is a infinite I will not have any zero divisors and I have an integral domain. And since addition and multiplication is commutative here I also have a field.
Am I on the right track?
abstract-algebra proof-verification ring-theory
edited Nov 29 at 20:31
Ravi
7,2301638
asked Nov 29 at 19:59
dls
947
add a comment |
We are given that $a oplus b = a+b+1$ and $a odot b= a+b+ab$ and I need to prove that it is a field. Do I just need to know that there is an inverse for $a odot b$? So if I could let $a+b+ab=c$ then I want to find a $-c$ such that $c odot -c=1$? Then the inverse, $-c=frac{1}{a+b+ab}$ then? This would be that for every c I have a multiplicative inverse and since the set is a infinite I will not have any zero divisors and I have an integral domain. And since addition and multiplication is commutative here I also have a field.
Am I on the right track?
abstract-algebra proof-verification ring-theory
edited Nov 29 at 20:31
Ravi
7,2301638
asked Nov 29 at 19:59
dls
947
associativity is not obvious for either operation
– Will Jagy
Nov 29 at 20:01
Is 1 the correct identity? $a odot 1 = a+ 1 + a ne a$
– Doug M
Nov 29 at 20:04
You must also show that multiplication distributes over addition.
– lulu
Nov 29 at 20:12
add a comment |
We are given that $a oplus b = a+b+1$ and $a odot b= a+b+ab$ and I need to prove that it is a field. Do I just need to know that there is an inverse for $a odot b$? So if I could let $a+b+ab=c$ then I want to find a $-c$ such that $c odot -c=1$? Then the inverse, $-c=frac{1}{a+b+ab}$ then? This would be that for every c I have a multiplicative inverse and since the set is a infinite I will not have any zero divisors and I have an integral domain. And since addition and multiplication is commutative here I also have a field.
Am I on the right track?
abstract-algebra proof-verification ring-theory
edited Nov 29 at 20:31
Ravi
7,2301638
asked Nov 29 at 19:59
dls
947
We are given that $a oplus b = a+b+1$ and $a odot b= a+b+ab$ and I need to prove that it is a field. Do I just need to know that there is an inverse for $a odot b$? So if I could let $a+b+ab=c$ then I want to find a $-c$ such that $c odot -c=1$? Then the inverse, $-c=frac{1}{a+b+ab}$ then? This would be that for every c I have a multiplicative inverse and since the set is a infinite I will not have any zero divisors and I have an integral domain. And since addition and multiplication is commutative here I also have a field.
Am I on the right track?
abstract-algebra proof-verification ring-theory
abstract-algebra proof-verification ring-theory
edited Nov 29 at 20:31
Ravi
7,2301638
asked Nov 29 at 19:59
dls
947
edited Nov 29 at 20:31
Ravi
7,2301638
asked Nov 29 at 19:59
dls
947
edited Nov 29 at 20:31
Ravi
7,2301638
edited Nov 29 at 20:31
Ravi
7,2301638
edited Nov 29 at 20:31
Ravi
7,2301638
7,2301638
asked Nov 29 at 19:59
dls
947
asked Nov 29 at 19:59
dls
947
asked Nov 29 at 19:59
dls
947
947
associativity is not obvious for either operation
– Will Jagy
Nov 29 at 20:01
Is 1 the correct identity? $a odot 1 = a+ 1 + a ne a$
– Doug M
Nov 29 at 20:04
You must also show that multiplication distributes over addition.
– lulu
Nov 29 at 20:12
add a comment |
associativity is not obvious for either operation
– Will Jagy
Nov 29 at 20:01
Is 1 the correct identity? $a odot 1 = a+ 1 + a ne a$
– Doug M
Nov 29 at 20:04
You must also show that multiplication distributes over addition.
– lulu
Nov 29 at 20:12
associativity is not obvious for either operation
– Will Jagy
Nov 29 at 20:01
associativity is not obvious for either operation
– Will Jagy
Nov 29 at 20:01
Is 1 the correct identity? $a odot 1 = a+ 1 + a ne a$
– Doug M
Nov 29 at 20:04
Is 1 the correct identity? $a odot 1 = a+ 1 + a ne a$
– Doug M
Nov 29 at 20:04
You must also show that multiplication distributes over addition.
– lulu
Nov 29 at 20:12
You must also show that multiplication distributes over addition.
– lulu
Nov 29 at 20:12
add a comment |
2 Answers
2
active
oldest
votes
Hint: Define $f:Bbb RtoBbb R$ by $f(t)=t+1$. Then $$f(xoplus y)=f(x)+f(y)$$and $$f(xodot y)=f(x)f(y).$$(So $f$ is an isomorphism, hence yes $(Bbb R,oplus,odot)$ is a field.)
Detail: For the benefit of readers new to the notion of isomorphism: One might say one knows about field isomorphisms, but question how that can be relevant before we know that $(Bbb R,oplus,odot)$ is a field.
That's a reasonable complaint. But the notion of "isomorphism" is more general than that - in general ann isomorphism between two structures is a bijection that preserves whatever operations and relations are relevant in the given context.
To be more explicit than people usually are, let's say an MAS ("Multiplication-Addition-Structure") is a set together with two binary operations of addition and multiplication. That's it, no further axioms. An MAS may or may not be a field. A bijection satisfying the two equations above is an MAS isomorphism, and it's obvious that if two MASs are isomorphic and one is a field then so is the other. The $f$ above is an MAS isomorphismm, qed.
To be even more explicit, the fact that $oplus$ is commutative follows from the fact that $+$ is commutative and $f$ is an MAS isomorphism: $$aoplus b=f^{-1}(f(a)+f(b))
=f^{-1}(f(b)+f(a))=boplus a.$$Similarly for all the other field axioms.
answered Nov 29 at 20:27
David C. Ullrich
58.1k43891
I don't really know what you mean by defining $f(t)=t+1$. Intuitively, your hint seems to define a+b=t? but this only seems to work the addition operation? I have difficulty seeing how it works for the multiplication property. Maybe I'm picturing it wrong. Could I have a little more clarification regarding $f(t)=t+1$ is helpful in this case?
– dls
Dec 4 at 20:01
1
Have you learned what an isomorphism is?
– David C. Ullrich
Dec 5 at 13:11
if $f$ is an isomorphism it is then a field? I haven't heard of that property yet. We haven't talked about Ring Homomorphisms yet however we have talked about Group Homomorphisms. I'm assuming there are many similar properties?
– dls
Dec 6 at 16:54
1
@dls See the Detail I added.
– David C. Ullrich
Dec 6 at 18:43
add a comment |
You need to verify all the field axioms. The two operations are clearly commutative, but it is not obvious that $odot$ is associative. Then find the identities. It is not true that $1$ is the identity for $odot$, so your inverse for $c$ is not correct. Finally you need to verify distributivity.
answered Nov 29 at 20:24
Ross Millikan
291k23196370
add a comment |
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2 Answers
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2 Answers
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Hint: Define $f:Bbb RtoBbb R$ by $f(t)=t+1$. Then $$f(xoplus y)=f(x)+f(y)$$and $$f(xodot y)=f(x)f(y).$$(So $f$ is an isomorphism, hence yes $(Bbb R,oplus,odot)$ is a field.)
Detail: For the benefit of readers new to the notion of isomorphism: One might say one knows about field isomorphisms, but question how that can be relevant before we know that $(Bbb R,oplus,odot)$ is a field.
That's a reasonable complaint. But the notion of "isomorphism" is more general than that - in general ann isomorphism between two structures is a bijection that preserves whatever operations and relations are relevant in the given context.
To be more explicit than people usually are, let's say an MAS ("Multiplication-Addition-Structure") is a set together with two binary operations of addition and multiplication. That's it, no further axioms. An MAS may or may not be a field. A bijection satisfying the two equations above is an MAS isomorphism, and it's obvious that if two MASs are isomorphic and one is a field then so is the other. The $f$ above is an MAS isomorphismm, qed.
To be even more explicit, the fact that $oplus$ is commutative follows from the fact that $+$ is commutative and $f$ is an MAS isomorphism: $$aoplus b=f^{-1}(f(a)+f(b))
=f^{-1}(f(b)+f(a))=boplus a.$$Similarly for all the other field axioms.
answered Nov 29 at 20:27
David C. Ullrich
58.1k43891
I don't really know what you mean by defining $f(t)=t+1$. Intuitively, your hint seems to define a+b=t? but this only seems to work the addition operation? I have difficulty seeing how it works for the multiplication property. Maybe I'm picturing it wrong. Could I have a little more clarification regarding $f(t)=t+1$ is helpful in this case?
– dls
Dec 4 at 20:01
1
Have you learned what an isomorphism is?
– David C. Ullrich
Dec 5 at 13:11
if $f$ is an isomorphism it is then a field? I haven't heard of that property yet. We haven't talked about Ring Homomorphisms yet however we have talked about Group Homomorphisms. I'm assuming there are many similar properties?
– dls
Dec 6 at 16:54
1
@dls See the Detail I added.
– David C. Ullrich
Dec 6 at 18:43
add a comment |
Hint: Define $f:Bbb RtoBbb R$ by $f(t)=t+1$. Then $$f(xoplus y)=f(x)+f(y)$$and $$f(xodot y)=f(x)f(y).$$(So $f$ is an isomorphism, hence yes $(Bbb R,oplus,odot)$ is a field.)
Detail: For the benefit of readers new to the notion of isomorphism: One might say one knows about field isomorphisms, but question how that can be relevant before we know that $(Bbb R,oplus,odot)$ is a field.
That's a reasonable complaint. But the notion of "isomorphism" is more general than that - in general ann isomorphism between two structures is a bijection that preserves whatever operations and relations are relevant in the given context.
To be more explicit than people usually are, let's say an MAS ("Multiplication-Addition-Structure") is a set together with two binary operations of addition and multiplication. That's it, no further axioms. An MAS may or may not be a field. A bijection satisfying the two equations above is an MAS isomorphism, and it's obvious that if two MASs are isomorphic and one is a field then so is the other. The $f$ above is an MAS isomorphismm, qed.
To be even more explicit, the fact that $oplus$ is commutative follows from the fact that $+$ is commutative and $f$ is an MAS isomorphism: $$aoplus b=f^{-1}(f(a)+f(b))
=f^{-1}(f(b)+f(a))=boplus a.$$Similarly for all the other field axioms.
answered Nov 29 at 20:27
David C. Ullrich
58.1k43891
I don't really know what you mean by defining $f(t)=t+1$. Intuitively, your hint seems to define a+b=t? but this only seems to work the addition operation? I have difficulty seeing how it works for the multiplication property. Maybe I'm picturing it wrong. Could I have a little more clarification regarding $f(t)=t+1$ is helpful in this case?
– dls
Dec 4 at 20:01
1
Have you learned what an isomorphism is?
– David C. Ullrich
Dec 5 at 13:11
if $f$ is an isomorphism it is then a field? I haven't heard of that property yet. We haven't talked about Ring Homomorphisms yet however we have talked about Group Homomorphisms. I'm assuming there are many similar properties?
– dls
Dec 6 at 16:54
1
@dls See the Detail I added.
– David C. Ullrich
Dec 6 at 18:43
add a comment |
Hint: Define $f:Bbb RtoBbb R$ by $f(t)=t+1$. Then $$f(xoplus y)=f(x)+f(y)$$and $$f(xodot y)=f(x)f(y).$$(So $f$ is an isomorphism, hence yes $(Bbb R,oplus,odot)$ is a field.)
Detail: For the benefit of readers new to the notion of isomorphism: One might say one knows about field isomorphisms, but question how that can be relevant before we know that $(Bbb R,oplus,odot)$ is a field.
That's a reasonable complaint. But the notion of "isomorphism" is more general than that - in general ann isomorphism between two structures is a bijection that preserves whatever operations and relations are relevant in the given context.
To be more explicit than people usually are, let's say an MAS ("Multiplication-Addition-Structure") is a set together with two binary operations of addition and multiplication. That's it, no further axioms. An MAS may or may not be a field. A bijection satisfying the two equations above is an MAS isomorphism, and it's obvious that if two MASs are isomorphic and one is a field then so is the other. The $f$ above is an MAS isomorphismm, qed.
To be even more explicit, the fact that $oplus$ is commutative follows from the fact that $+$ is commutative and $f$ is an MAS isomorphism: $$aoplus b=f^{-1}(f(a)+f(b))
=f^{-1}(f(b)+f(a))=boplus a.$$Similarly for all the other field axioms.
answered Nov 29 at 20:27
David C. Ullrich
58.1k43891
Hint: Define $f:Bbb RtoBbb R$ by $f(t)=t+1$. Then $$f(xoplus y)=f(x)+f(y)$$and $$f(xodot y)=f(x)f(y).$$(So $f$ is an isomorphism, hence yes $(Bbb R,oplus,odot)$ is a field.)
Detail: For the benefit of readers new to the notion of isomorphism: One might say one knows about field isomorphisms, but question how that can be relevant before we know that $(Bbb R,oplus,odot)$ is a field.
That's a reasonable complaint. But the notion of "isomorphism" is more general than that - in general ann isomorphism between two structures is a bijection that preserves whatever operations and relations are relevant in the given context.
To be more explicit than people usually are, let's say an MAS ("Multiplication-Addition-Structure") is a set together with two binary operations of addition and multiplication. That's it, no further axioms. An MAS may or may not be a field. A bijection satisfying the two equations above is an MAS isomorphism, and it's obvious that if two MASs are isomorphic and one is a field then so is the other. The $f$ above is an MAS isomorphismm, qed.
To be even more explicit, the fact that $oplus$ is commutative follows from the fact that $+$ is commutative and $f$ is an MAS isomorphism: $$aoplus b=f^{-1}(f(a)+f(b))
=f^{-1}(f(b)+f(a))=boplus a.$$Similarly for all the other field axioms.
answered Nov 29 at 20:27
David C. Ullrich
58.1k43891
edited Dec 6 at 18:42
answered Nov 29 at 20:27
David C. Ullrich
58.1k43891
answered Nov 29 at 20:27
David C. Ullrich
58.1k43891
answered Nov 29 at 20:27
David C. Ullrich
58.1k43891
58.1k43891
I don't really know what you mean by defining $f(t)=t+1$. Intuitively, your hint seems to define a+b=t? but this only seems to work the addition operation? I have difficulty seeing how it works for the multiplication property. Maybe I'm picturing it wrong. Could I have a little more clarification regarding $f(t)=t+1$ is helpful in this case?
– dls
Dec 4 at 20:01
1
Have you learned what an isomorphism is?
– David C. Ullrich
Dec 5 at 13:11
if $f$ is an isomorphism it is then a field? I haven't heard of that property yet. We haven't talked about Ring Homomorphisms yet however we have talked about Group Homomorphisms. I'm assuming there are many similar properties?
– dls
Dec 6 at 16:54
1
@dls See the Detail I added.
– David C. Ullrich
Dec 6 at 18:43
add a comment |
I don't really know what you mean by defining $f(t)=t+1$. Intuitively, your hint seems to define a+b=t? but this only seems to work the addition operation? I have difficulty seeing how it works for the multiplication property. Maybe I'm picturing it wrong. Could I have a little more clarification regarding $f(t)=t+1$ is helpful in this case?
– dls
Dec 4 at 20:01
1
Have you learned what an isomorphism is?
– David C. Ullrich
Dec 5 at 13:11
if $f$ is an isomorphism it is then a field? I haven't heard of that property yet. We haven't talked about Ring Homomorphisms yet however we have talked about Group Homomorphisms. I'm assuming there are many similar properties?
– dls
Dec 6 at 16:54
1
@dls See the Detail I added.
– David C. Ullrich
Dec 6 at 18:43
I don't really know what you mean by defining $f(t)=t+1$. Intuitively, your hint seems to define a+b=t? but this only seems to work the addition operation? I have difficulty seeing how it works for the multiplication property. Maybe I'm picturing it wrong. Could I have a little more clarification regarding $f(t)=t+1$ is helpful in this case?
– dls
Dec 4 at 20:01
I don't really know what you mean by defining $f(t)=t+1$. Intuitively, your hint seems to define a+b=t? but this only seems to work the addition operation? I have difficulty seeing how it works for the multiplication property. Maybe I'm picturing it wrong. Could I have a little more clarification regarding $f(t)=t+1$ is helpful in this case?
– dls
Dec 4 at 20:01
1
1
Have you learned what an isomorphism is?
– David C. Ullrich
Dec 5 at 13:11
Have you learned what an isomorphism is?
– David C. Ullrich
Dec 5 at 13:11
if $f$ is an isomorphism it is then a field? I haven't heard of that property yet. We haven't talked about Ring Homomorphisms yet however we have talked about Group Homomorphisms. I'm assuming there are many similar properties?
– dls
Dec 6 at 16:54
if $f$ is an isomorphism it is then a field? I haven't heard of that property yet. We haven't talked about Ring Homomorphisms yet however we have talked about Group Homomorphisms. I'm assuming there are many similar properties?
– dls
Dec 6 at 16:54
1
1
@dls See the Detail I added.
– David C. Ullrich
Dec 6 at 18:43
@dls See the Detail I added.
– David C. Ullrich
Dec 6 at 18:43
add a comment |
You need to verify all the field axioms. The two operations are clearly commutative, but it is not obvious that $odot$ is associative. Then find the identities. It is not true that $1$ is the identity for $odot$, so your inverse for $c$ is not correct. Finally you need to verify distributivity.
answered Nov 29 at 20:24
Ross Millikan
291k23196370
add a comment |
You need to verify all the field axioms. The two operations are clearly commutative, but it is not obvious that $odot$ is associative. Then find the identities. It is not true that $1$ is the identity for $odot$, so your inverse for $c$ is not correct. Finally you need to verify distributivity.
answered Nov 29 at 20:24
Ross Millikan
291k23196370
add a comment |
You need to verify all the field axioms. The two operations are clearly commutative, but it is not obvious that $odot$ is associative. Then find the identities. It is not true that $1$ is the identity for $odot$, so your inverse for $c$ is not correct. Finally you need to verify distributivity.
answered Nov 29 at 20:24
Ross Millikan
291k23196370
You need to verify all the field axioms. The two operations are clearly commutative, but it is not obvious that $odot$ is associative. Then find the identities. It is not true that $1$ is the identity for $odot$, so your inverse for $c$ is not correct. Finally you need to verify distributivity.
answered Nov 29 at 20:24
Ross Millikan
291k23196370
answered Nov 29 at 20:24
Ross Millikan
291k23196370
answered Nov 29 at 20:24
Ross Millikan
291k23196370
answered Nov 29 at 20:24
Ross Millikan
291k23196370
291k23196370
add a comment |
add a comment |
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associativity is not obvious for either operation
– Will Jagy
Nov 29 at 20:01
Is 1 the correct identity? $a odot 1 = a+ 1 + a ne a$
– Doug M
Nov 29 at 20:04
You must also show that multiplication distributes over addition.
– lulu
Nov 29 at 20:12