Binary operation $xy$, has identity, but not associativity. Is the inverse unique?
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Let $S$ be a set with a binary operation $xy$ defined on it, with a neutral element, but not satisfying associativity. I want to prove that the inverse isn't necessarily unique.
My attempt to answer this is in the answer below.
abstract-algebra binary-operations
asked Jan 16 '15 at 13:00
Makoto K.
292111
add a comment |
up vote
1
down vote
favorite
1
Let $S$ be a set with a binary operation $xy$ defined on it, with a neutral element, but not satisfying associativity. I want to prove that the inverse isn't necessarily unique.
My attempt to answer this is in the answer below.
abstract-algebra binary-operations
asked Jan 16 '15 at 13:00
Makoto K.
292111
Presumably, you mean, when an inverse exists? Is the neutral element both a right and left neutral element?
– Thomas Andrews
Jan 16 '15 at 13:03
@ThomasAndrews Yes, that is the only way I know it so far
– Makoto K.
Jan 16 '15 at 13:05
add a comment |
up vote
1
down vote
favorite
1
up vote
1
down vote
favorite
1
1
Let $S$ be a set with a binary operation $xy$ defined on it, with a neutral element, but not satisfying associativity. I want to prove that the inverse isn't necessarily unique.
My attempt to answer this is in the answer below.
abstract-algebra binary-operations
asked Jan 16 '15 at 13:00
Makoto K.
292111
Let $S$ be a set with a binary operation $xy$ defined on it, with a neutral element, but not satisfying associativity. I want to prove that the inverse isn't necessarily unique.
My attempt to answer this is in the answer below.
abstract-algebra binary-operations
abstract-algebra binary-operations
asked Jan 16 '15 at 13:00
Makoto K.
292111
asked Jan 16 '15 at 13:00
Makoto K.
292111
asked Jan 16 '15 at 13:00
Makoto K.
292111
asked Jan 16 '15 at 13:00
Makoto K.
292111
asked Jan 16 '15 at 13:00
Makoto K.
292111
292111
Presumably, you mean, when an inverse exists? Is the neutral element both a right and left neutral element?
– Thomas Andrews
Jan 16 '15 at 13:03
@ThomasAndrews Yes, that is the only way I know it so far
– Makoto K.
Jan 16 '15 at 13:05
add a comment |
Presumably, you mean, when an inverse exists? Is the neutral element both a right and left neutral element?
– Thomas Andrews
Jan 16 '15 at 13:03
@ThomasAndrews Yes, that is the only way I know it so far
– Makoto K.
Jan 16 '15 at 13:05
Presumably, you mean, when an inverse exists? Is the neutral element both a right and left neutral element?
– Thomas Andrews
Jan 16 '15 at 13:03
Presumably, you mean, when an inverse exists? Is the neutral element both a right and left neutral element?
– Thomas Andrews
Jan 16 '15 at 13:03
@ThomasAndrews Yes, that is the only way I know it so far
– Makoto K.
Jan 16 '15 at 13:05
@ThomasAndrews Yes, that is the only way I know it so far
– Makoto K.
Jan 16 '15 at 13:05
add a comment |
1 Answer
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Let $S={1,2,3}$ and define $star$ so that $xstar 1 = 1star x=x$ and $xstar y=1$ if $x,yin{2,3}$. Then $(S,star)$ has a neutral element, $1$, but $2star 3=2star 2=1$, so inverse elements are not unique.
answered Jan 16 '15 at 13:07
Thomas Andrews
129k11145294
What is $f$? There is no $f$ here. It is a binary operation on a set with an identity.
– Thomas Andrews
Jan 16 '15 at 13:10
@MakotoK. No, the map is $Stimes Sto S$ (since it is a binary operation).
– Tobias Kildetoft
Jan 16 '15 at 13:12
1
The point is, if someone asks you whether, "Are all natural numbers even," you don't show that it is impossible to prove, you just say, "$1$ is a natural number and it is not even, so the statement is false."
– Thomas Andrews
Jan 16 '15 at 13:12
@ThomasAndrews Thank you
– Makoto K.
Jan 16 '15 at 13:12
2
I didn't deliberately make it minimal, I just realized I couldn't do it with two elements, and was looking for something simple.
– Thomas Andrews
Jan 16 '15 at 13:27
|
show 5 more comments
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1 Answer
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1 Answer
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active
oldest
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active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Let $S={1,2,3}$ and define $star$ so that $xstar 1 = 1star x=x$ and $xstar y=1$ if $x,yin{2,3}$. Then $(S,star)$ has a neutral element, $1$, but $2star 3=2star 2=1$, so inverse elements are not unique.
answered Jan 16 '15 at 13:07
Thomas Andrews
129k11145294
What is $f$? There is no $f$ here. It is a binary operation on a set with an identity.
– Thomas Andrews
Jan 16 '15 at 13:10
@MakotoK. No, the map is $Stimes Sto S$ (since it is a binary operation).
– Tobias Kildetoft
Jan 16 '15 at 13:12
1
The point is, if someone asks you whether, "Are all natural numbers even," you don't show that it is impossible to prove, you just say, "$1$ is a natural number and it is not even, so the statement is false."
– Thomas Andrews
Jan 16 '15 at 13:12
@ThomasAndrews Thank you
– Makoto K.
Jan 16 '15 at 13:12
2
I didn't deliberately make it minimal, I just realized I couldn't do it with two elements, and was looking for something simple.
– Thomas Andrews
Jan 16 '15 at 13:27
|
show 5 more comments
up vote
4
down vote
accepted
Let $S={1,2,3}$ and define $star$ so that $xstar 1 = 1star x=x$ and $xstar y=1$ if $x,yin{2,3}$. Then $(S,star)$ has a neutral element, $1$, but $2star 3=2star 2=1$, so inverse elements are not unique.
answered Jan 16 '15 at 13:07
Thomas Andrews
129k11145294
What is $f$? There is no $f$ here. It is a binary operation on a set with an identity.
– Thomas Andrews
Jan 16 '15 at 13:10
@MakotoK. No, the map is $Stimes Sto S$ (since it is a binary operation).
– Tobias Kildetoft
Jan 16 '15 at 13:12
1
The point is, if someone asks you whether, "Are all natural numbers even," you don't show that it is impossible to prove, you just say, "$1$ is a natural number and it is not even, so the statement is false."
– Thomas Andrews
Jan 16 '15 at 13:12
@ThomasAndrews Thank you
– Makoto K.
Jan 16 '15 at 13:12
2
I didn't deliberately make it minimal, I just realized I couldn't do it with two elements, and was looking for something simple.
– Thomas Andrews
Jan 16 '15 at 13:27
|
show 5 more comments
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Let $S={1,2,3}$ and define $star$ so that $xstar 1 = 1star x=x$ and $xstar y=1$ if $x,yin{2,3}$. Then $(S,star)$ has a neutral element, $1$, but $2star 3=2star 2=1$, so inverse elements are not unique.
answered Jan 16 '15 at 13:07
Thomas Andrews
129k11145294
Let $S={1,2,3}$ and define $star$ so that $xstar 1 = 1star x=x$ and $xstar y=1$ if $x,yin{2,3}$. Then $(S,star)$ has a neutral element, $1$, but $2star 3=2star 2=1$, so inverse elements are not unique.
answered Jan 16 '15 at 13:07
Thomas Andrews
129k11145294
edited Nov 24 at 1:58
answered Jan 16 '15 at 13:07
Thomas Andrews
129k11145294
answered Jan 16 '15 at 13:07
Thomas Andrews
129k11145294
answered Jan 16 '15 at 13:07
Thomas Andrews
129k11145294
129k11145294
What is $f$? There is no $f$ here. It is a binary operation on a set with an identity.
– Thomas Andrews
Jan 16 '15 at 13:10
@MakotoK. No, the map is $Stimes Sto S$ (since it is a binary operation).
– Tobias Kildetoft
Jan 16 '15 at 13:12
1
The point is, if someone asks you whether, "Are all natural numbers even," you don't show that it is impossible to prove, you just say, "$1$ is a natural number and it is not even, so the statement is false."
– Thomas Andrews
Jan 16 '15 at 13:12
@ThomasAndrews Thank you
– Makoto K.
Jan 16 '15 at 13:12
2
I didn't deliberately make it minimal, I just realized I couldn't do it with two elements, and was looking for something simple.
– Thomas Andrews
Jan 16 '15 at 13:27
|
show 5 more comments
What is $f$? There is no $f$ here. It is a binary operation on a set with an identity.
– Thomas Andrews
Jan 16 '15 at 13:10
@MakotoK. No, the map is $Stimes Sto S$ (since it is a binary operation).
– Tobias Kildetoft
Jan 16 '15 at 13:12
1
The point is, if someone asks you whether, "Are all natural numbers even," you don't show that it is impossible to prove, you just say, "$1$ is a natural number and it is not even, so the statement is false."
– Thomas Andrews
Jan 16 '15 at 13:12
@ThomasAndrews Thank you
– Makoto K.
Jan 16 '15 at 13:12
2
I didn't deliberately make it minimal, I just realized I couldn't do it with two elements, and was looking for something simple.
– Thomas Andrews
Jan 16 '15 at 13:27
What is $f$? There is no $f$ here. It is a binary operation on a set with an identity.
– Thomas Andrews
Jan 16 '15 at 13:10
What is $f$? There is no $f$ here. It is a binary operation on a set with an identity.
– Thomas Andrews
Jan 16 '15 at 13:10
@MakotoK. No, the map is $Stimes Sto S$ (since it is a binary operation).
– Tobias Kildetoft
Jan 16 '15 at 13:12
@MakotoK. No, the map is $Stimes Sto S$ (since it is a binary operation).
– Tobias Kildetoft
Jan 16 '15 at 13:12
1
1
The point is, if someone asks you whether, "Are all natural numbers even," you don't show that it is impossible to prove, you just say, "$1$ is a natural number and it is not even, so the statement is false."
– Thomas Andrews
Jan 16 '15 at 13:12
The point is, if someone asks you whether, "Are all natural numbers even," you don't show that it is impossible to prove, you just say, "$1$ is a natural number and it is not even, so the statement is false."
– Thomas Andrews
Jan 16 '15 at 13:12
@ThomasAndrews Thank you
– Makoto K.
Jan 16 '15 at 13:12
@ThomasAndrews Thank you
– Makoto K.
Jan 16 '15 at 13:12
2
2
I didn't deliberately make it minimal, I just realized I couldn't do it with two elements, and was looking for something simple.
– Thomas Andrews
Jan 16 '15 at 13:27
I didn't deliberately make it minimal, I just realized I couldn't do it with two elements, and was looking for something simple.
– Thomas Andrews
Jan 16 '15 at 13:27
|
show 5 more comments
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Presumably, you mean, when an inverse exists? Is the neutral element both a right and left neutral element?
– Thomas Andrews
Jan 16 '15 at 13:03
@ThomasAndrews Yes, that is the only way I know it so far
– Makoto K.
Jan 16 '15 at 13:05