Invent Binary Operator $*$ on Reals that Can Create $+$, $-$, $times$, $div$ [duplicate]
$begingroup$
This question already has an answer here:
Is there an numeric arithmetic with a single operator?
3 answers
Exact Question: Invent a single binary operator $*$ such that for every real numbers $a$ and $b$, the operations $a + b$, $a - b$, $a times b$, $a div b$ can be created by applying $*$ (multiple times), starting with only $a$'s and $b$'s
From my interpretation, you can apply $*$ recursively some number of times with carefully selected parameters to produce the desired outcome.
I thought the operator should be a combination of $a - b$ and $a cdot b^{-1}$ Since $-$ and $div$ can produce $+$ and $times$ respectively
Please do not tell me the full answer. Give me a hint to point me towards the right path
abstract-algebra functions contest-math recursion
asked Dec 21 '18 at 5:21
BrianHBrianH
658
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marked as duplicate by Community♦ Dec 21 '18 at 16:45
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Is there an numeric arithmetic with a single operator?
3 answers
Exact Question: Invent a single binary operator $*$ such that for every real numbers $a$ and $b$, the operations $a + b$, $a - b$, $a times b$, $a div b$ can be created by applying $*$ (multiple times), starting with only $a$'s and $b$'s
From my interpretation, you can apply $*$ recursively some number of times with carefully selected parameters to produce the desired outcome.
I thought the operator should be a combination of $a - b$ and $a cdot b^{-1}$ Since $-$ and $div$ can produce $+$ and $times$ respectively
Please do not tell me the full answer. Give me a hint to point me towards the right path
abstract-algebra functions contest-math recursion
asked Dec 21 '18 at 5:21
BrianHBrianH
658
$endgroup$
marked as duplicate by Community♦ Dec 21 '18 at 16:45
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Hint: First derive all operations from subtraction and reciprocal and find a single binary operation that gets you these.
$endgroup$
– John
Dec 21 '18 at 5:40
$begingroup$
@John I realize that, as I've stated in my post as well. But could you give a hint on how to combine them into expression/operator?
$endgroup$
– BrianH
Dec 21 '18 at 5:43
$begingroup$
That would involve giving away what $star$ is. But you said you only wanted a hint.
$endgroup$
– John
Dec 21 '18 at 5:45
add a comment |
$begingroup$
This question already has an answer here:
Is there an numeric arithmetic with a single operator?
3 answers
Exact Question: Invent a single binary operator $*$ such that for every real numbers $a$ and $b$, the operations $a + b$, $a - b$, $a times b$, $a div b$ can be created by applying $*$ (multiple times), starting with only $a$'s and $b$'s
From my interpretation, you can apply $*$ recursively some number of times with carefully selected parameters to produce the desired outcome.
I thought the operator should be a combination of $a - b$ and $a cdot b^{-1}$ Since $-$ and $div$ can produce $+$ and $times$ respectively
Please do not tell me the full answer. Give me a hint to point me towards the right path
abstract-algebra functions contest-math recursion
asked Dec 21 '18 at 5:21
BrianHBrianH
658
$endgroup$
This question already has an answer here:
Is there an numeric arithmetic with a single operator?
3 answers
Exact Question: Invent a single binary operator $*$ such that for every real numbers $a$ and $b$, the operations $a + b$, $a - b$, $a times b$, $a div b$ can be created by applying $*$ (multiple times), starting with only $a$'s and $b$'s
From my interpretation, you can apply $*$ recursively some number of times with carefully selected parameters to produce the desired outcome.
I thought the operator should be a combination of $a - b$ and $a cdot b^{-1}$ Since $-$ and $div$ can produce $+$ and $times$ respectively
Please do not tell me the full answer. Give me a hint to point me towards the right path
This question already has an answer here:
Is there an numeric arithmetic with a single operator?
3 answers
abstract-algebra functions contest-math recursion
abstract-algebra functions contest-math recursion
asked Dec 21 '18 at 5:21
BrianHBrianH
658
asked Dec 21 '18 at 5:21
BrianHBrianH
658
asked Dec 21 '18 at 5:21
BrianHBrianH
658
asked Dec 21 '18 at 5:21
BrianHBrianH
658
asked Dec 21 '18 at 5:21
BrianHBrianH
658
658
marked as duplicate by Community♦ Dec 21 '18 at 16:45
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Community♦ Dec 21 '18 at 16:45
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Hint: First derive all operations from subtraction and reciprocal and find a single binary operation that gets you these.
$endgroup$
– John
Dec 21 '18 at 5:40
$begingroup$
@John I realize that, as I've stated in my post as well. But could you give a hint on how to combine them into expression/operator?
$endgroup$
– BrianH
Dec 21 '18 at 5:43
$begingroup$
That would involve giving away what $star$ is. But you said you only wanted a hint.
$endgroup$
– John
Dec 21 '18 at 5:45
add a comment |
$begingroup$
Hint: First derive all operations from subtraction and reciprocal and find a single binary operation that gets you these.
$endgroup$
– John
Dec 21 '18 at 5:40
$begingroup$
@John I realize that, as I've stated in my post as well. But could you give a hint on how to combine them into expression/operator?
$endgroup$
– BrianH
Dec 21 '18 at 5:43
$begingroup$
That would involve giving away what $star$ is. But you said you only wanted a hint.
$endgroup$
– John
Dec 21 '18 at 5:45
$begingroup$
Hint: First derive all operations from subtraction and reciprocal and find a single binary operation that gets you these.
$endgroup$
– John
Dec 21 '18 at 5:40
$begingroup$
Hint: First derive all operations from subtraction and reciprocal and find a single binary operation that gets you these.
$endgroup$
– John
Dec 21 '18 at 5:40
$begingroup$
@John I realize that, as I've stated in my post as well. But could you give a hint on how to combine them into expression/operator?
$endgroup$
– BrianH
Dec 21 '18 at 5:43
$begingroup$
@John I realize that, as I've stated in my post as well. But could you give a hint on how to combine them into expression/operator?
$endgroup$
– BrianH
Dec 21 '18 at 5:43
$begingroup$
That would involve giving away what $star$ is. But you said you only wanted a hint.
$endgroup$
– John
Dec 21 '18 at 5:45
$begingroup$
That would involve giving away what $star$ is. But you said you only wanted a hint.
$endgroup$
– John
Dec 21 '18 at 5:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'm not sure that this is possible. You wish to define a binary operator $*:mathbb{R}times mathbb{R}rightarrowmathbb{R}$ that has some properties, but in particular you must define what $0*0$ is equal to. This means that however $*$ is defined in terms of the four fundamental operations, it cannot involve division, since $0/0$ is an undefined quantity. But if we cannot involve division in the definition, then there is no way to get back $1/2$ from any number of applications (and groupings) of $*$ to $1$ and $2$, since $mathbb{Z}$ is a ring under $+$ and $times$.
Where does that leave us? The definition of $*$ we seek must use an operation beyond the four fundamental operations, and still must be able to recreate the effects of all four under some number of applications (and groupings)... which just strikes me as impossible.
answered Dec 21 '18 at 5:34
DanielJackDanielJack
661
$endgroup$
$begingroup$
What if we assume for the sake of simplicity and define * on $mathbb{R} setminus {0}$. I mean 0 + 0 = 0, 0 - 0 = 0, 0 $cdot$ 0 = 0, 0/0 is irrelavent. So maybe glipmse over the fact?
$endgroup$
– BrianH
Dec 21 '18 at 5:41
$begingroup$
@BrianH That is a possibility; $0$ does present the obstructions I raise here, but I will have to think longer to come up with an answer in this restricted version of the problem.
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– DanielJack
Dec 21 '18 at 5:43
$begingroup$
@BrianH After thinking a bit more, I remain unconvinced that this is solvable. If division occurs anywhere in the definition of $*$ then whatever function of $a$ and $b$ we are dividing by must never be equal to $0$. Yes, such functions are easy to build, but to then get back values like $a+b$ at the end of the day? I just don't buy it. Where did you find this problem? Is it from a class? Textbook? You used the label contest math, so is there a particular year and contest this is from?
$endgroup$
– DanielJack
Dec 21 '18 at 5:52
$begingroup$
This is from a class called "Putnam Seminar" taught by Po Shen Loh, renown math olympiad coach (en.wikipedia.org/wiki/Po-Shen_Loh). I am not sure where he got the problem, but I believe there is an answer. He may have been a little less formal with the delivery of the problem...
$endgroup$
– BrianH
Dec 21 '18 at 5:58
$begingroup$
It is in A problem seminar by Donald J. Newman.
$endgroup$
– John
Dec 21 '18 at 5:59
|
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I'm not sure that this is possible. You wish to define a binary operator $*:mathbb{R}times mathbb{R}rightarrowmathbb{R}$ that has some properties, but in particular you must define what $0*0$ is equal to. This means that however $*$ is defined in terms of the four fundamental operations, it cannot involve division, since $0/0$ is an undefined quantity. But if we cannot involve division in the definition, then there is no way to get back $1/2$ from any number of applications (and groupings) of $*$ to $1$ and $2$, since $mathbb{Z}$ is a ring under $+$ and $times$.
Where does that leave us? The definition of $*$ we seek must use an operation beyond the four fundamental operations, and still must be able to recreate the effects of all four under some number of applications (and groupings)... which just strikes me as impossible.
answered Dec 21 '18 at 5:34
DanielJackDanielJack
661
$endgroup$
$begingroup$
What if we assume for the sake of simplicity and define * on $mathbb{R} setminus {0}$. I mean 0 + 0 = 0, 0 - 0 = 0, 0 $cdot$ 0 = 0, 0/0 is irrelavent. So maybe glipmse over the fact?
$endgroup$
– BrianH
Dec 21 '18 at 5:41
$begingroup$
@BrianH That is a possibility; $0$ does present the obstructions I raise here, but I will have to think longer to come up with an answer in this restricted version of the problem.
$endgroup$
– DanielJack
Dec 21 '18 at 5:43
$begingroup$
@BrianH After thinking a bit more, I remain unconvinced that this is solvable. If division occurs anywhere in the definition of $*$ then whatever function of $a$ and $b$ we are dividing by must never be equal to $0$. Yes, such functions are easy to build, but to then get back values like $a+b$ at the end of the day? I just don't buy it. Where did you find this problem? Is it from a class? Textbook? You used the label contest math, so is there a particular year and contest this is from?
$endgroup$
– DanielJack
Dec 21 '18 at 5:52
$begingroup$
This is from a class called "Putnam Seminar" taught by Po Shen Loh, renown math olympiad coach (en.wikipedia.org/wiki/Po-Shen_Loh). I am not sure where he got the problem, but I believe there is an answer. He may have been a little less formal with the delivery of the problem...
$endgroup$
– BrianH
Dec 21 '18 at 5:58
$begingroup$
It is in A problem seminar by Donald J. Newman.
$endgroup$
– John
Dec 21 '18 at 5:59
|
show 4 more comments
$begingroup$
I'm not sure that this is possible. You wish to define a binary operator $*:mathbb{R}times mathbb{R}rightarrowmathbb{R}$ that has some properties, but in particular you must define what $0*0$ is equal to. This means that however $*$ is defined in terms of the four fundamental operations, it cannot involve division, since $0/0$ is an undefined quantity. But if we cannot involve division in the definition, then there is no way to get back $1/2$ from any number of applications (and groupings) of $*$ to $1$ and $2$, since $mathbb{Z}$ is a ring under $+$ and $times$.
Where does that leave us? The definition of $*$ we seek must use an operation beyond the four fundamental operations, and still must be able to recreate the effects of all four under some number of applications (and groupings)... which just strikes me as impossible.
answered Dec 21 '18 at 5:34
DanielJackDanielJack
661
$endgroup$
$begingroup$
What if we assume for the sake of simplicity and define * on $mathbb{R} setminus {0}$. I mean 0 + 0 = 0, 0 - 0 = 0, 0 $cdot$ 0 = 0, 0/0 is irrelavent. So maybe glipmse over the fact?
$endgroup$
– BrianH
Dec 21 '18 at 5:41
$begingroup$
@BrianH That is a possibility; $0$ does present the obstructions I raise here, but I will have to think longer to come up with an answer in this restricted version of the problem.
$endgroup$
– DanielJack
Dec 21 '18 at 5:43
$begingroup$
@BrianH After thinking a bit more, I remain unconvinced that this is solvable. If division occurs anywhere in the definition of $*$ then whatever function of $a$ and $b$ we are dividing by must never be equal to $0$. Yes, such functions are easy to build, but to then get back values like $a+b$ at the end of the day? I just don't buy it. Where did you find this problem? Is it from a class? Textbook? You used the label contest math, so is there a particular year and contest this is from?
$endgroup$
– DanielJack
Dec 21 '18 at 5:52
$begingroup$
This is from a class called "Putnam Seminar" taught by Po Shen Loh, renown math olympiad coach (en.wikipedia.org/wiki/Po-Shen_Loh). I am not sure where he got the problem, but I believe there is an answer. He may have been a little less formal with the delivery of the problem...
$endgroup$
– BrianH
Dec 21 '18 at 5:58
$begingroup$
It is in A problem seminar by Donald J. Newman.
$endgroup$
– John
Dec 21 '18 at 5:59
|
show 4 more comments
$begingroup$
I'm not sure that this is possible. You wish to define a binary operator $*:mathbb{R}times mathbb{R}rightarrowmathbb{R}$ that has some properties, but in particular you must define what $0*0$ is equal to. This means that however $*$ is defined in terms of the four fundamental operations, it cannot involve division, since $0/0$ is an undefined quantity. But if we cannot involve division in the definition, then there is no way to get back $1/2$ from any number of applications (and groupings) of $*$ to $1$ and $2$, since $mathbb{Z}$ is a ring under $+$ and $times$.
Where does that leave us? The definition of $*$ we seek must use an operation beyond the four fundamental operations, and still must be able to recreate the effects of all four under some number of applications (and groupings)... which just strikes me as impossible.
answered Dec 21 '18 at 5:34
DanielJackDanielJack
661
$endgroup$
I'm not sure that this is possible. You wish to define a binary operator $*:mathbb{R}times mathbb{R}rightarrowmathbb{R}$ that has some properties, but in particular you must define what $0*0$ is equal to. This means that however $*$ is defined in terms of the four fundamental operations, it cannot involve division, since $0/0$ is an undefined quantity. But if we cannot involve division in the definition, then there is no way to get back $1/2$ from any number of applications (and groupings) of $*$ to $1$ and $2$, since $mathbb{Z}$ is a ring under $+$ and $times$.
Where does that leave us? The definition of $*$ we seek must use an operation beyond the four fundamental operations, and still must be able to recreate the effects of all four under some number of applications (and groupings)... which just strikes me as impossible.
answered Dec 21 '18 at 5:34
DanielJackDanielJack
661
answered Dec 21 '18 at 5:34
DanielJackDanielJack
661
answered Dec 21 '18 at 5:34
DanielJackDanielJack
661
answered Dec 21 '18 at 5:34
DanielJackDanielJack
661
661
$begingroup$
What if we assume for the sake of simplicity and define * on $mathbb{R} setminus {0}$. I mean 0 + 0 = 0, 0 - 0 = 0, 0 $cdot$ 0 = 0, 0/0 is irrelavent. So maybe glipmse over the fact?
$endgroup$
– BrianH
Dec 21 '18 at 5:41
$begingroup$
@BrianH That is a possibility; $0$ does present the obstructions I raise here, but I will have to think longer to come up with an answer in this restricted version of the problem.
$endgroup$
– DanielJack
Dec 21 '18 at 5:43
$begingroup$
@BrianH After thinking a bit more, I remain unconvinced that this is solvable. If division occurs anywhere in the definition of $*$ then whatever function of $a$ and $b$ we are dividing by must never be equal to $0$. Yes, such functions are easy to build, but to then get back values like $a+b$ at the end of the day? I just don't buy it. Where did you find this problem? Is it from a class? Textbook? You used the label contest math, so is there a particular year and contest this is from?
$endgroup$
– DanielJack
Dec 21 '18 at 5:52
$begingroup$
This is from a class called "Putnam Seminar" taught by Po Shen Loh, renown math olympiad coach (en.wikipedia.org/wiki/Po-Shen_Loh). I am not sure where he got the problem, but I believe there is an answer. He may have been a little less formal with the delivery of the problem...
$endgroup$
– BrianH
Dec 21 '18 at 5:58
$begingroup$
It is in A problem seminar by Donald J. Newman.
$endgroup$
– John
Dec 21 '18 at 5:59
|
show 4 more comments
$begingroup$
What if we assume for the sake of simplicity and define * on $mathbb{R} setminus {0}$. I mean 0 + 0 = 0, 0 - 0 = 0, 0 $cdot$ 0 = 0, 0/0 is irrelavent. So maybe glipmse over the fact?
$endgroup$
– BrianH
Dec 21 '18 at 5:41
$begingroup$
@BrianH That is a possibility; $0$ does present the obstructions I raise here, but I will have to think longer to come up with an answer in this restricted version of the problem.
$endgroup$
– DanielJack
Dec 21 '18 at 5:43
$begingroup$
@BrianH After thinking a bit more, I remain unconvinced that this is solvable. If division occurs anywhere in the definition of $*$ then whatever function of $a$ and $b$ we are dividing by must never be equal to $0$. Yes, such functions are easy to build, but to then get back values like $a+b$ at the end of the day? I just don't buy it. Where did you find this problem? Is it from a class? Textbook? You used the label contest math, so is there a particular year and contest this is from?
$endgroup$
– DanielJack
Dec 21 '18 at 5:52
$begingroup$
This is from a class called "Putnam Seminar" taught by Po Shen Loh, renown math olympiad coach (en.wikipedia.org/wiki/Po-Shen_Loh). I am not sure where he got the problem, but I believe there is an answer. He may have been a little less formal with the delivery of the problem...
$endgroup$
– BrianH
Dec 21 '18 at 5:58
$begingroup$
It is in A problem seminar by Donald J. Newman.
$endgroup$
– John
Dec 21 '18 at 5:59
$begingroup$
What if we assume for the sake of simplicity and define * on $mathbb{R} setminus {0}$. I mean 0 + 0 = 0, 0 - 0 = 0, 0 $cdot$ 0 = 0, 0/0 is irrelavent. So maybe glipmse over the fact?
$endgroup$
– BrianH
Dec 21 '18 at 5:41
$begingroup$
What if we assume for the sake of simplicity and define * on $mathbb{R} setminus {0}$. I mean 0 + 0 = 0, 0 - 0 = 0, 0 $cdot$ 0 = 0, 0/0 is irrelavent. So maybe glipmse over the fact?
$endgroup$
– BrianH
Dec 21 '18 at 5:41
$begingroup$
@BrianH That is a possibility; $0$ does present the obstructions I raise here, but I will have to think longer to come up with an answer in this restricted version of the problem.
$endgroup$
– DanielJack
Dec 21 '18 at 5:43
$begingroup$
@BrianH That is a possibility; $0$ does present the obstructions I raise here, but I will have to think longer to come up with an answer in this restricted version of the problem.
$endgroup$
– DanielJack
Dec 21 '18 at 5:43
$begingroup$
@BrianH After thinking a bit more, I remain unconvinced that this is solvable. If division occurs anywhere in the definition of $*$ then whatever function of $a$ and $b$ we are dividing by must never be equal to $0$. Yes, such functions are easy to build, but to then get back values like $a+b$ at the end of the day? I just don't buy it. Where did you find this problem? Is it from a class? Textbook? You used the label contest math, so is there a particular year and contest this is from?
$endgroup$
– DanielJack
Dec 21 '18 at 5:52
$begingroup$
@BrianH After thinking a bit more, I remain unconvinced that this is solvable. If division occurs anywhere in the definition of $*$ then whatever function of $a$ and $b$ we are dividing by must never be equal to $0$. Yes, such functions are easy to build, but to then get back values like $a+b$ at the end of the day? I just don't buy it. Where did you find this problem? Is it from a class? Textbook? You used the label contest math, so is there a particular year and contest this is from?
$endgroup$
– DanielJack
Dec 21 '18 at 5:52
$begingroup$
This is from a class called "Putnam Seminar" taught by Po Shen Loh, renown math olympiad coach (en.wikipedia.org/wiki/Po-Shen_Loh). I am not sure where he got the problem, but I believe there is an answer. He may have been a little less formal with the delivery of the problem...
$endgroup$
– BrianH
Dec 21 '18 at 5:58
$begingroup$
This is from a class called "Putnam Seminar" taught by Po Shen Loh, renown math olympiad coach (en.wikipedia.org/wiki/Po-Shen_Loh). I am not sure where he got the problem, but I believe there is an answer. He may have been a little less formal with the delivery of the problem...
$endgroup$
– BrianH
Dec 21 '18 at 5:58
$begingroup$
It is in A problem seminar by Donald J. Newman.
$endgroup$
– John
Dec 21 '18 at 5:59
$begingroup$
It is in A problem seminar by Donald J. Newman.
$endgroup$
– John
Dec 21 '18 at 5:59
|
show 4 more comments
$begingroup$
Hint: First derive all operations from subtraction and reciprocal and find a single binary operation that gets you these.
$endgroup$
– John
Dec 21 '18 at 5:40
$begingroup$
@John I realize that, as I've stated in my post as well. But could you give a hint on how to combine them into expression/operator?
$endgroup$
– BrianH
Dec 21 '18 at 5:43
$begingroup$
That would involve giving away what $star$ is. But you said you only wanted a hint.
$endgroup$
– John
Dec 21 '18 at 5:45