Is there any unique function $c=f(a, b)$ such that if I know only $c$, I can deduce unique $a$ and $b$ from it? [duplicate]
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Injective function from $mathbbR^2$ to $mathbbR$?
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I want to know a relation $c = f(a, b)$ (where $a, b, c$ are real numbers) such that if I only know $c$ I can deduce $a, b$ inputs from it.
As an example consider $f(a, b) = a+b$, then $f(2,3) = 5$ but if I know $5$ as an output only, I cannot get $2, 3$ uniquely because $f(1,4), f(5,0)$ all of them give the same result $5$.
Does there exist such function which can make me guess input $a, b$ as unique?
also a, b belongs to every real number R. ie. I should be able to pick any a, b from R.
relations function-and-relation-composition
marked as duplicate by GoodDeeds, Jendrik Stelzner, José Carlos Santos, Ennar, user99914 Sep 9 at 19:35
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.
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up vote
2
down vote
favorite
This question already has an answer here:
Injective function from $mathbbR^2$ to $mathbbR$?
3 answers
I want to know a relation $c = f(a, b)$ (where $a, b, c$ are real numbers) such that if I only know $c$ I can deduce $a, b$ inputs from it.
As an example consider $f(a, b) = a+b$, then $f(2,3) = 5$ but if I know $5$ as an output only, I cannot get $2, 3$ uniquely because $f(1,4), f(5,0)$ all of them give the same result $5$.
Does there exist such function which can make me guess input $a, b$ as unique?
also a, b belongs to every real number R. ie. I should be able to pick any a, b from R.
relations function-and-relation-composition
marked as duplicate by GoodDeeds, Jendrik Stelzner, José Carlos Santos, Ennar, user99914 Sep 9 at 19:35
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 |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This question already has an answer here:
Injective function from $mathbbR^2$ to $mathbbR$?
3 answers
I want to know a relation $c = f(a, b)$ (where $a, b, c$ are real numbers) such that if I only know $c$ I can deduce $a, b$ inputs from it.
As an example consider $f(a, b) = a+b$, then $f(2,3) = 5$ but if I know $5$ as an output only, I cannot get $2, 3$ uniquely because $f(1,4), f(5,0)$ all of them give the same result $5$.
Does there exist such function which can make me guess input $a, b$ as unique?
also a, b belongs to every real number R. ie. I should be able to pick any a, b from R.
relations function-and-relation-composition
This question already has an answer here:
Injective function from $mathbbR^2$ to $mathbbR$?
3 answers
I want to know a relation $c = f(a, b)$ (where $a, b, c$ are real numbers) such that if I only know $c$ I can deduce $a, b$ inputs from it.
As an example consider $f(a, b) = a+b$, then $f(2,3) = 5$ but if I know $5$ as an output only, I cannot get $2, 3$ uniquely because $f(1,4), f(5,0)$ all of them give the same result $5$.
Does there exist such function which can make me guess input $a, b$ as unique?
also a, b belongs to every real number R. ie. I should be able to pick any a, b from R.
This question already has an answer here:
Injective function from $mathbbR^2$ to $mathbbR$?
3 answers
relations function-and-relation-composition
relations function-and-relation-composition
edited Sep 9 at 9:44
asked Sep 9 at 9:25
user8428217
112
112
marked as duplicate by GoodDeeds, Jendrik Stelzner, José Carlos Santos, Ennar, user99914 Sep 9 at 19:35
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 GoodDeeds, Jendrik Stelzner, José Carlos Santos, Ennar, user99914 Sep 9 at 19:35
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.
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Sure, for example $f$ might be the function that intertwines the decimal digits of its arguments so that for example
$$ beginalignf(sqrt2000000,pi)&=f(1414.213562373095ldots, 3.141592653589ldots)\&=10401043.211431556922367533059859ldotsendalign$$
where reconstructing the digits of the arguments is readily possible.
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What you are asking for is an injective (or even more, bijective) function from $mathbbR^2$ to $mathbbR.$
In general, there are many ones because it can be shown that they have the same cardinality (see Hagen von Eitzen's brilliant answer).
But it can also be shown that there are not continuous ones.
Indeed, consider the unit circle $S^1$ centered in the origin, and the restriction of $f$ to the circle: it is a continuous function on a compact set, so it has a point of maximum, say $M$, such that $f(M)geq f(x) forall x in S^1$ and a point of minimum, say $m$. Clearly they're not the same, because otherwise we obtain the whole circle with the same value, against the hypotheses on $f$.
So, because they're different you can joint them with two different arcs of the unit circle. On these arcs, by the intermediate value property, we can surely find two different points $c$ and $c'$ such that $f(c)-f(c')=fracf(M)+f(m)2.$
This goes against the hypothesis of the injective function $f$.
If you're interested in this kind of theorems, you can read the Borsuk-Ulam theorem: it's not really easy to prove but it's great.
ok i get it, but what if I concatenate a and b for example f(2,3) =23, then wont it be unique for every a and b?
â user8428217
Sep 9 at 10:42
1
Suppose that $a$ and $b$ are infinite decimals like $pi$ or $sqrt2$. How are you going to "concatenate" those? But the short answer is "no, that can't work, because Riccardo has just proven that it cannot."
â John Hughes
Sep 9 at 10:44
Notice that I used a lot the continuity, whereas concatenation is not continuous :)
â Riccardo Ceccon
Sep 9 at 10:46
1
ok. now it is clear. thanks
â user8428217
Sep 9 at 10:55
Please eave an upvote if you want :)
â Riccardo Ceccon
Sep 9 at 11:18
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Sure, for example $f$ might be the function that intertwines the decimal digits of its arguments so that for example
$$ beginalignf(sqrt2000000,pi)&=f(1414.213562373095ldots, 3.141592653589ldots)\&=10401043.211431556922367533059859ldotsendalign$$
where reconstructing the digits of the arguments is readily possible.
add a comment |Â
up vote
1
down vote
Sure, for example $f$ might be the function that intertwines the decimal digits of its arguments so that for example
$$ beginalignf(sqrt2000000,pi)&=f(1414.213562373095ldots, 3.141592653589ldots)\&=10401043.211431556922367533059859ldotsendalign$$
where reconstructing the digits of the arguments is readily possible.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Sure, for example $f$ might be the function that intertwines the decimal digits of its arguments so that for example
$$ beginalignf(sqrt2000000,pi)&=f(1414.213562373095ldots, 3.141592653589ldots)\&=10401043.211431556922367533059859ldotsendalign$$
where reconstructing the digits of the arguments is readily possible.
Sure, for example $f$ might be the function that intertwines the decimal digits of its arguments so that for example
$$ beginalignf(sqrt2000000,pi)&=f(1414.213562373095ldots, 3.141592653589ldots)\&=10401043.211431556922367533059859ldotsendalign$$
where reconstructing the digits of the arguments is readily possible.
answered Sep 9 at 9:38
Hagen von Eitzen
267k21260483
267k21260483
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up vote
0
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What you are asking for is an injective (or even more, bijective) function from $mathbbR^2$ to $mathbbR.$
In general, there are many ones because it can be shown that they have the same cardinality (see Hagen von Eitzen's brilliant answer).
But it can also be shown that there are not continuous ones.
Indeed, consider the unit circle $S^1$ centered in the origin, and the restriction of $f$ to the circle: it is a continuous function on a compact set, so it has a point of maximum, say $M$, such that $f(M)geq f(x) forall x in S^1$ and a point of minimum, say $m$. Clearly they're not the same, because otherwise we obtain the whole circle with the same value, against the hypotheses on $f$.
So, because they're different you can joint them with two different arcs of the unit circle. On these arcs, by the intermediate value property, we can surely find two different points $c$ and $c'$ such that $f(c)-f(c')=fracf(M)+f(m)2.$
This goes against the hypothesis of the injective function $f$.
If you're interested in this kind of theorems, you can read the Borsuk-Ulam theorem: it's not really easy to prove but it's great.
ok i get it, but what if I concatenate a and b for example f(2,3) =23, then wont it be unique for every a and b?
â user8428217
Sep 9 at 10:42
1
Suppose that $a$ and $b$ are infinite decimals like $pi$ or $sqrt2$. How are you going to "concatenate" those? But the short answer is "no, that can't work, because Riccardo has just proven that it cannot."
â John Hughes
Sep 9 at 10:44
Notice that I used a lot the continuity, whereas concatenation is not continuous :)
â Riccardo Ceccon
Sep 9 at 10:46
1
ok. now it is clear. thanks
â user8428217
Sep 9 at 10:55
Please eave an upvote if you want :)
â Riccardo Ceccon
Sep 9 at 11:18
add a comment |Â
up vote
0
down vote
What you are asking for is an injective (or even more, bijective) function from $mathbbR^2$ to $mathbbR.$
In general, there are many ones because it can be shown that they have the same cardinality (see Hagen von Eitzen's brilliant answer).
But it can also be shown that there are not continuous ones.
Indeed, consider the unit circle $S^1$ centered in the origin, and the restriction of $f$ to the circle: it is a continuous function on a compact set, so it has a point of maximum, say $M$, such that $f(M)geq f(x) forall x in S^1$ and a point of minimum, say $m$. Clearly they're not the same, because otherwise we obtain the whole circle with the same value, against the hypotheses on $f$.
So, because they're different you can joint them with two different arcs of the unit circle. On these arcs, by the intermediate value property, we can surely find two different points $c$ and $c'$ such that $f(c)-f(c')=fracf(M)+f(m)2.$
This goes against the hypothesis of the injective function $f$.
If you're interested in this kind of theorems, you can read the Borsuk-Ulam theorem: it's not really easy to prove but it's great.
ok i get it, but what if I concatenate a and b for example f(2,3) =23, then wont it be unique for every a and b?
â user8428217
Sep 9 at 10:42
1
Suppose that $a$ and $b$ are infinite decimals like $pi$ or $sqrt2$. How are you going to "concatenate" those? But the short answer is "no, that can't work, because Riccardo has just proven that it cannot."
â John Hughes
Sep 9 at 10:44
Notice that I used a lot the continuity, whereas concatenation is not continuous :)
â Riccardo Ceccon
Sep 9 at 10:46
1
ok. now it is clear. thanks
â user8428217
Sep 9 at 10:55
Please eave an upvote if you want :)
â Riccardo Ceccon
Sep 9 at 11:18
add a comment |Â
up vote
0
down vote
up vote
0
down vote
What you are asking for is an injective (or even more, bijective) function from $mathbbR^2$ to $mathbbR.$
In general, there are many ones because it can be shown that they have the same cardinality (see Hagen von Eitzen's brilliant answer).
But it can also be shown that there are not continuous ones.
Indeed, consider the unit circle $S^1$ centered in the origin, and the restriction of $f$ to the circle: it is a continuous function on a compact set, so it has a point of maximum, say $M$, such that $f(M)geq f(x) forall x in S^1$ and a point of minimum, say $m$. Clearly they're not the same, because otherwise we obtain the whole circle with the same value, against the hypotheses on $f$.
So, because they're different you can joint them with two different arcs of the unit circle. On these arcs, by the intermediate value property, we can surely find two different points $c$ and $c'$ such that $f(c)-f(c')=fracf(M)+f(m)2.$
This goes against the hypothesis of the injective function $f$.
If you're interested in this kind of theorems, you can read the Borsuk-Ulam theorem: it's not really easy to prove but it's great.
What you are asking for is an injective (or even more, bijective) function from $mathbbR^2$ to $mathbbR.$
In general, there are many ones because it can be shown that they have the same cardinality (see Hagen von Eitzen's brilliant answer).
But it can also be shown that there are not continuous ones.
Indeed, consider the unit circle $S^1$ centered in the origin, and the restriction of $f$ to the circle: it is a continuous function on a compact set, so it has a point of maximum, say $M$, such that $f(M)geq f(x) forall x in S^1$ and a point of minimum, say $m$. Clearly they're not the same, because otherwise we obtain the whole circle with the same value, against the hypotheses on $f$.
So, because they're different you can joint them with two different arcs of the unit circle. On these arcs, by the intermediate value property, we can surely find two different points $c$ and $c'$ such that $f(c)-f(c')=fracf(M)+f(m)2.$
This goes against the hypothesis of the injective function $f$.
If you're interested in this kind of theorems, you can read the Borsuk-Ulam theorem: it's not really easy to prove but it's great.
answered Sep 9 at 9:57
Riccardo Ceccon
875320
875320
ok i get it, but what if I concatenate a and b for example f(2,3) =23, then wont it be unique for every a and b?
â user8428217
Sep 9 at 10:42
1
Suppose that $a$ and $b$ are infinite decimals like $pi$ or $sqrt2$. How are you going to "concatenate" those? But the short answer is "no, that can't work, because Riccardo has just proven that it cannot."
â John Hughes
Sep 9 at 10:44
Notice that I used a lot the continuity, whereas concatenation is not continuous :)
â Riccardo Ceccon
Sep 9 at 10:46
1
ok. now it is clear. thanks
â user8428217
Sep 9 at 10:55
Please eave an upvote if you want :)
â Riccardo Ceccon
Sep 9 at 11:18
add a comment |Â
ok i get it, but what if I concatenate a and b for example f(2,3) =23, then wont it be unique for every a and b?
â user8428217
Sep 9 at 10:42
1
Suppose that $a$ and $b$ are infinite decimals like $pi$ or $sqrt2$. How are you going to "concatenate" those? But the short answer is "no, that can't work, because Riccardo has just proven that it cannot."
â John Hughes
Sep 9 at 10:44
Notice that I used a lot the continuity, whereas concatenation is not continuous :)
â Riccardo Ceccon
Sep 9 at 10:46
1
ok. now it is clear. thanks
â user8428217
Sep 9 at 10:55
Please eave an upvote if you want :)
â Riccardo Ceccon
Sep 9 at 11:18
ok i get it, but what if I concatenate a and b for example f(2,3) =23, then wont it be unique for every a and b?
â user8428217
Sep 9 at 10:42
ok i get it, but what if I concatenate a and b for example f(2,3) =23, then wont it be unique for every a and b?
â user8428217
Sep 9 at 10:42
1
1
Suppose that $a$ and $b$ are infinite decimals like $pi$ or $sqrt2$. How are you going to "concatenate" those? But the short answer is "no, that can't work, because Riccardo has just proven that it cannot."
â John Hughes
Sep 9 at 10:44
Suppose that $a$ and $b$ are infinite decimals like $pi$ or $sqrt2$. How are you going to "concatenate" those? But the short answer is "no, that can't work, because Riccardo has just proven that it cannot."
â John Hughes
Sep 9 at 10:44
Notice that I used a lot the continuity, whereas concatenation is not continuous :)
â Riccardo Ceccon
Sep 9 at 10:46
Notice that I used a lot the continuity, whereas concatenation is not continuous :)
â Riccardo Ceccon
Sep 9 at 10:46
1
1
ok. now it is clear. thanks
â user8428217
Sep 9 at 10:55
ok. now it is clear. thanks
â user8428217
Sep 9 at 10:55
Please eave an upvote if you want :)
â Riccardo Ceccon
Sep 9 at 11:18
Please eave an upvote if you want :)
â Riccardo Ceccon
Sep 9 at 11:18
add a comment |Â