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Conjugate of real number
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I'm slightly confused on the subject of conjugates and how to define them.
I know that for a complex number $ a - bi $ the conjugate is $ a + bi $ and similarly for $ 1 + sqrt 2 $ the conjugate is $ 1 - sqrt2 $ because when multiplied it gives a rational answer.
But how about for just a simple real number like 1 or 2, what would be the conjugate for this? Does a conjugate exist for a real number?
I'm new to this topic and have tried searching Maths SE and Google in vain; any help would be appreciated.
radicals real-numbers rational-numbers
asked Jul 30 at 12:05
Shawn Li
1537
add a comment |Â
up vote
10
down vote
favorite
I'm slightly confused on the subject of conjugates and how to define them.
I know that for a complex number $ a - bi $ the conjugate is $ a + bi $ and similarly for $ 1 + sqrt 2 $ the conjugate is $ 1 - sqrt2 $ because when multiplied it gives a rational answer.
But how about for just a simple real number like 1 or 2, what would be the conjugate for this? Does a conjugate exist for a real number?
I'm new to this topic and have tried searching Maths SE and Google in vain; any help would be appreciated.
radicals real-numbers rational-numbers
asked Jul 30 at 12:05
Shawn Li
1537
Since $mathbbR subset mathbbC$, any $a in mathbbR$ can be written in the following way: $a=a+icdot 0$.
– Iuli
Jul 30 at 12:10
14
There is complex conjugate for complex numbers, and there is also quadratic conjugate for algebraic numbers of the form $p+qsqrtd$ with $p,qinmathbbQ$ and $dinmathbbZ$ is not a square.
– Batominovski
Jul 30 at 12:11
add a comment |Â
up vote
10
down vote
favorite
up vote
10
down vote
favorite
I'm slightly confused on the subject of conjugates and how to define them.
I know that for a complex number $ a - bi $ the conjugate is $ a + bi $ and similarly for $ 1 + sqrt 2 $ the conjugate is $ 1 - sqrt2 $ because when multiplied it gives a rational answer.
But how about for just a simple real number like 1 or 2, what would be the conjugate for this? Does a conjugate exist for a real number?
I'm new to this topic and have tried searching Maths SE and Google in vain; any help would be appreciated.
radicals real-numbers rational-numbers
asked Jul 30 at 12:05
Shawn Li
1537
I'm slightly confused on the subject of conjugates and how to define them.
I know that for a complex number $ a - bi $ the conjugate is $ a + bi $ and similarly for $ 1 + sqrt 2 $ the conjugate is $ 1 - sqrt2 $ because when multiplied it gives a rational answer.
But how about for just a simple real number like 1 or 2, what would be the conjugate for this? Does a conjugate exist for a real number?
I'm new to this topic and have tried searching Maths SE and Google in vain; any help would be appreciated.
radicals real-numbers rational-numbers
asked Jul 30 at 12:05
Shawn Li
1537
asked Jul 30 at 12:05
Shawn Li
1537
asked Jul 30 at 12:05
Shawn Li
1537
asked Jul 30 at 12:05
Shawn Li
1537
1537
Since $mathbbR subset mathbbC$, any $a in mathbbR$ can be written in the following way: $a=a+icdot 0$.
– Iuli
Jul 30 at 12:10
14
There is complex conjugate for complex numbers, and there is also quadratic conjugate for algebraic numbers of the form $p+qsqrtd$ with $p,qinmathbbQ$ and $dinmathbbZ$ is not a square.
– Batominovski
Jul 30 at 12:11
add a comment |Â
Since $mathbbR subset mathbbC$, any $a in mathbbR$ can be written in the following way: $a=a+icdot 0$.
– Iuli
Jul 30 at 12:10
14
There is complex conjugate for complex numbers, and there is also quadratic conjugate for algebraic numbers of the form $p+qsqrtd$ with $p,qinmathbbQ$ and $dinmathbbZ$ is not a square.
– Batominovski
Jul 30 at 12:11
Since $mathbbR subset mathbbC$, any $a in mathbbR$ can be written in the following way: $a=a+icdot 0$.
– Iuli
Jul 30 at 12:10
Since $mathbbR subset mathbbC$, any $a in mathbbR$ can be written in the following way: $a=a+icdot 0$.
– Iuli
Jul 30 at 12:10
14
14
There is complex conjugate for complex numbers, and there is also quadratic conjugate for algebraic numbers of the form $p+qsqrtd$ with $p,qinmathbbQ$ and $dinmathbbZ$ is not a square.
– Batominovski
Jul 30 at 12:11
There is complex conjugate for complex numbers, and there is also quadratic conjugate for algebraic numbers of the form $p+qsqrtd$ with $p,qinmathbbQ$ and $dinmathbbZ$ is not a square.
– Batominovski
Jul 30 at 12:11
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
42
down vote
accepted
Careful! These are two different notions of conjugate.
First we have the complex conjugate, given by $overlinea+bi = a-bi$. Then, since we can write a real number $x$ as $x+0i$, the complex conjugate of a real number is itself.
There is also a second idea of a rational conjugate, where as in your example, if $a,b$ are rational and $d$ is squarefree, the conjugate of $a+bsqrtd$ is $a-bsqrtd$.
There is a connection between these two ideas. In general, given a field extension $E/F$, take an algebraic element $alpha$ of $E$, and let $m(x)$ be it's minimal polynomial over $F$. Then we call the other roots of $m$ in $E$ the conjugates of $alpha$.
In the case of the extensions $mathbbC/mathbbR$ and $mathbbQ(sqrtd)/ mathbbQ$ this agrees with the above.
edited Jul 30 at 19:59
Community♦
1
answered Jul 30 at 12:22
Daniel Mroz
851314
I can see that those are two different kinds of conjugates, but by treating $i$ and $sqrtd$ as (mere) indeterminate variables, don't we get a unified notion of conjugation, as in mapping from $alpha+beta y$ to $alpha-beta y$ when this mapping makes sense (as in well defined and preserves algebraic structure)? In both cases, $alpha$ gets mapped to $alpha$ and we declare that the conjugate of $alpha$ is itself.
– Frenzy Li
Jul 30 at 15:19
6
@FrenzyLi In fact, that's what Daniel Mroz is hinting at with his part on field extensions. Conjugation is an automorphism of the bigger field, leaving the smaller field fixed; an element of the Galois group.
– Babelfish
Jul 30 at 16:11
add a comment |Â
up vote
6
down vote
The notion of a conjugate comes up when you have a naturally defined function $c$ from a set to itself such that for all $x$ you have $c(c(x)) = x$. Then $c(x)$ is the conjugate of $x$.
That happens naturally when the numbers you are thinking about are of the form $a+bsqrtd$ for some $d$. Then you define the conjugating function by switching the sign of $b$.
In this case $2$ will be its own conjugate.
The conjugate of $1 + sqrt2$ is tricky. It's sometimes $1 - sqrt2$ but it's just $1 + sqrt2$ (itself) in the complex numbers, since it's a real number.
answered Jul 30 at 12:14
Ethan Bolker
35.7k54199
Perhaps outside the scope of the question, but I would say a conjugate must be with respect to some subset, i.e. $c(x)=x$ for every $x$ in the sub-set as well as $c(c(x))=x$ for every $x$ in the whole set. Assuming only members of the sub-set satisfy the first condition, we have $c(x cdot c(x))=c(x)cdot c(c(x))=xcdot c(x)$ which means $xcdot c(x)$ is always in our "special subset" (in other words - a conjugate of $L$ over $K$ is a non-trivial involution in $mathrmAut(L/K)$)
– Itai
Jul 30 at 13:27
I think you should clarify "numbers of the form a + b sqrt (d)", it's evidently not well defined when considering real numbers, or when d is a perfect square. Also the "It's sometimes..." sentence could be clarified, it's not like it is non-deterministic, depending on the context and object considered it will be clear what notion of conjugation one is referring to.
– Maxim
Jul 31 at 9:30
add a comment |Â
up vote
2
down vote
Usually our definition of "conjugate" refers to complex numbers: the conjugate of $a+bi$ is $a-bi$. You could say "complex conjugate" be be extra specific.
Note that $1+sqrt2$ is a real number, so its conjugate is $1+sqrt2$.
A nice way of thinking about conjugates is how they are related in the complex plane (on an Argand diagram). Given a complex number, reflect it across the horizontal (real) axis to get its conjugate. Since $1$, $2$ and $1+sqrt2$ all lie on the real line, they are their own conjugate.
Getting a little bit more advanced, you can define conjugates in different fields. The complex numbers are an extension of the real numbers with the number $i$. We can write this "field extension" as $mathbbC=mathbbR(i)$.
Working instead in $mathbbQ(sqrt2)$ (the rationals extended with $sqrt2$), you can define the conjugate of $a+bsqrt2$ as $a-bsqrt2$. But don't worry so much about this if you are new to the topic!
answered Jul 30 at 12:16
Malkin
914421
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
42
down vote
accepted
Careful! These are two different notions of conjugate.
First we have the complex conjugate, given by $overlinea+bi = a-bi$. Then, since we can write a real number $x$ as $x+0i$, the complex conjugate of a real number is itself.
There is also a second idea of a rational conjugate, where as in your example, if $a,b$ are rational and $d$ is squarefree, the conjugate of $a+bsqrtd$ is $a-bsqrtd$.
There is a connection between these two ideas. In general, given a field extension $E/F$, take an algebraic element $alpha$ of $E$, and let $m(x)$ be it's minimal polynomial over $F$. Then we call the other roots of $m$ in $E$ the conjugates of $alpha$.
In the case of the extensions $mathbbC/mathbbR$ and $mathbbQ(sqrtd)/ mathbbQ$ this agrees with the above.
edited Jul 30 at 19:59
Community♦
1
answered Jul 30 at 12:22
Daniel Mroz
851314
I can see that those are two different kinds of conjugates, but by treating $i$ and $sqrtd$ as (mere) indeterminate variables, don't we get a unified notion of conjugation, as in mapping from $alpha+beta y$ to $alpha-beta y$ when this mapping makes sense (as in well defined and preserves algebraic structure)? In both cases, $alpha$ gets mapped to $alpha$ and we declare that the conjugate of $alpha$ is itself.
– Frenzy Li
Jul 30 at 15:19
6
@FrenzyLi In fact, that's what Daniel Mroz is hinting at with his part on field extensions. Conjugation is an automorphism of the bigger field, leaving the smaller field fixed; an element of the Galois group.
– Babelfish
Jul 30 at 16:11
add a comment |Â
up vote
42
down vote
accepted
Careful! These are two different notions of conjugate.
First we have the complex conjugate, given by $overlinea+bi = a-bi$. Then, since we can write a real number $x$ as $x+0i$, the complex conjugate of a real number is itself.
There is also a second idea of a rational conjugate, where as in your example, if $a,b$ are rational and $d$ is squarefree, the conjugate of $a+bsqrtd$ is $a-bsqrtd$.
There is a connection between these two ideas. In general, given a field extension $E/F$, take an algebraic element $alpha$ of $E$, and let $m(x)$ be it's minimal polynomial over $F$. Then we call the other roots of $m$ in $E$ the conjugates of $alpha$.
In the case of the extensions $mathbbC/mathbbR$ and $mathbbQ(sqrtd)/ mathbbQ$ this agrees with the above.
edited Jul 30 at 19:59
Community♦
1
answered Jul 30 at 12:22
Daniel Mroz
851314
I can see that those are two different kinds of conjugates, but by treating $i$ and $sqrtd$ as (mere) indeterminate variables, don't we get a unified notion of conjugation, as in mapping from $alpha+beta y$ to $alpha-beta y$ when this mapping makes sense (as in well defined and preserves algebraic structure)? In both cases, $alpha$ gets mapped to $alpha$ and we declare that the conjugate of $alpha$ is itself.
– Frenzy Li
Jul 30 at 15:19
6
@FrenzyLi In fact, that's what Daniel Mroz is hinting at with his part on field extensions. Conjugation is an automorphism of the bigger field, leaving the smaller field fixed; an element of the Galois group.
– Babelfish
Jul 30 at 16:11
add a comment |Â
up vote
42
down vote
accepted
up vote
42
down vote
accepted
Careful! These are two different notions of conjugate.
First we have the complex conjugate, given by $overlinea+bi = a-bi$. Then, since we can write a real number $x$ as $x+0i$, the complex conjugate of a real number is itself.
There is also a second idea of a rational conjugate, where as in your example, if $a,b$ are rational and $d$ is squarefree, the conjugate of $a+bsqrtd$ is $a-bsqrtd$.
There is a connection between these two ideas. In general, given a field extension $E/F$, take an algebraic element $alpha$ of $E$, and let $m(x)$ be it's minimal polynomial over $F$. Then we call the other roots of $m$ in $E$ the conjugates of $alpha$.
In the case of the extensions $mathbbC/mathbbR$ and $mathbbQ(sqrtd)/ mathbbQ$ this agrees with the above.
edited Jul 30 at 19:59
Community♦
1
answered Jul 30 at 12:22
Daniel Mroz
851314
Careful! These are two different notions of conjugate.
First we have the complex conjugate, given by $overlinea+bi = a-bi$. Then, since we can write a real number $x$ as $x+0i$, the complex conjugate of a real number is itself.
There is also a second idea of a rational conjugate, where as in your example, if $a,b$ are rational and $d$ is squarefree, the conjugate of $a+bsqrtd$ is $a-bsqrtd$.
There is a connection between these two ideas. In general, given a field extension $E/F$, take an algebraic element $alpha$ of $E$, and let $m(x)$ be it's minimal polynomial over $F$. Then we call the other roots of $m$ in $E$ the conjugates of $alpha$.
In the case of the extensions $mathbbC/mathbbR$ and $mathbbQ(sqrtd)/ mathbbQ$ this agrees with the above.
edited Jul 30 at 19:59
Community♦
1
answered Jul 30 at 12:22
Daniel Mroz
851314
edited Jul 30 at 19:59
Community♦
1
edited Jul 30 at 19:59
Community♦
1
edited Jul 30 at 19:59
Community♦
1
1
answered Jul 30 at 12:22
Daniel Mroz
851314
answered Jul 30 at 12:22
Daniel Mroz
851314
answered Jul 30 at 12:22
Daniel Mroz
851314
851314
I can see that those are two different kinds of conjugates, but by treating $i$ and $sqrtd$ as (mere) indeterminate variables, don't we get a unified notion of conjugation, as in mapping from $alpha+beta y$ to $alpha-beta y$ when this mapping makes sense (as in well defined and preserves algebraic structure)? In both cases, $alpha$ gets mapped to $alpha$ and we declare that the conjugate of $alpha$ is itself.
– Frenzy Li
Jul 30 at 15:19
6
@FrenzyLi In fact, that's what Daniel Mroz is hinting at with his part on field extensions. Conjugation is an automorphism of the bigger field, leaving the smaller field fixed; an element of the Galois group.
– Babelfish
Jul 30 at 16:11
add a comment |Â
I can see that those are two different kinds of conjugates, but by treating $i$ and $sqrtd$ as (mere) indeterminate variables, don't we get a unified notion of conjugation, as in mapping from $alpha+beta y$ to $alpha-beta y$ when this mapping makes sense (as in well defined and preserves algebraic structure)? In both cases, $alpha$ gets mapped to $alpha$ and we declare that the conjugate of $alpha$ is itself.
– Frenzy Li
Jul 30 at 15:19
6
@FrenzyLi In fact, that's what Daniel Mroz is hinting at with his part on field extensions. Conjugation is an automorphism of the bigger field, leaving the smaller field fixed; an element of the Galois group.
– Babelfish
Jul 30 at 16:11
I can see that those are two different kinds of conjugates, but by treating $i$ and $sqrtd$ as (mere) indeterminate variables, don't we get a unified notion of conjugation, as in mapping from $alpha+beta y$ to $alpha-beta y$ when this mapping makes sense (as in well defined and preserves algebraic structure)? In both cases, $alpha$ gets mapped to $alpha$ and we declare that the conjugate of $alpha$ is itself.
– Frenzy Li
Jul 30 at 15:19
I can see that those are two different kinds of conjugates, but by treating $i$ and $sqrtd$ as (mere) indeterminate variables, don't we get a unified notion of conjugation, as in mapping from $alpha+beta y$ to $alpha-beta y$ when this mapping makes sense (as in well defined and preserves algebraic structure)? In both cases, $alpha$ gets mapped to $alpha$ and we declare that the conjugate of $alpha$ is itself.
– Frenzy Li
Jul 30 at 15:19
6
6
@FrenzyLi In fact, that's what Daniel Mroz is hinting at with his part on field extensions. Conjugation is an automorphism of the bigger field, leaving the smaller field fixed; an element of the Galois group.
– Babelfish
Jul 30 at 16:11
@FrenzyLi In fact, that's what Daniel Mroz is hinting at with his part on field extensions. Conjugation is an automorphism of the bigger field, leaving the smaller field fixed; an element of the Galois group.
– Babelfish
Jul 30 at 16:11
add a comment |Â
up vote
6
down vote
The notion of a conjugate comes up when you have a naturally defined function $c$ from a set to itself such that for all $x$ you have $c(c(x)) = x$. Then $c(x)$ is the conjugate of $x$.
That happens naturally when the numbers you are thinking about are of the form $a+bsqrtd$ for some $d$. Then you define the conjugating function by switching the sign of $b$.
In this case $2$ will be its own conjugate.
The conjugate of $1 + sqrt2$ is tricky. It's sometimes $1 - sqrt2$ but it's just $1 + sqrt2$ (itself) in the complex numbers, since it's a real number.
answered Jul 30 at 12:14
Ethan Bolker
35.7k54199
Perhaps outside the scope of the question, but I would say a conjugate must be with respect to some subset, i.e. $c(x)=x$ for every $x$ in the sub-set as well as $c(c(x))=x$ for every $x$ in the whole set. Assuming only members of the sub-set satisfy the first condition, we have $c(x cdot c(x))=c(x)cdot c(c(x))=xcdot c(x)$ which means $xcdot c(x)$ is always in our "special subset" (in other words - a conjugate of $L$ over $K$ is a non-trivial involution in $mathrmAut(L/K)$)
– Itai
Jul 30 at 13:27
I think you should clarify "numbers of the form a + b sqrt (d)", it's evidently not well defined when considering real numbers, or when d is a perfect square. Also the "It's sometimes..." sentence could be clarified, it's not like it is non-deterministic, depending on the context and object considered it will be clear what notion of conjugation one is referring to.
– Maxim
Jul 31 at 9:30
add a comment |Â
up vote
6
down vote
The notion of a conjugate comes up when you have a naturally defined function $c$ from a set to itself such that for all $x$ you have $c(c(x)) = x$. Then $c(x)$ is the conjugate of $x$.
That happens naturally when the numbers you are thinking about are of the form $a+bsqrtd$ for some $d$. Then you define the conjugating function by switching the sign of $b$.
In this case $2$ will be its own conjugate.
The conjugate of $1 + sqrt2$ is tricky. It's sometimes $1 - sqrt2$ but it's just $1 + sqrt2$ (itself) in the complex numbers, since it's a real number.
answered Jul 30 at 12:14
Ethan Bolker
35.7k54199
Perhaps outside the scope of the question, but I would say a conjugate must be with respect to some subset, i.e. $c(x)=x$ for every $x$ in the sub-set as well as $c(c(x))=x$ for every $x$ in the whole set. Assuming only members of the sub-set satisfy the first condition, we have $c(x cdot c(x))=c(x)cdot c(c(x))=xcdot c(x)$ which means $xcdot c(x)$ is always in our "special subset" (in other words - a conjugate of $L$ over $K$ is a non-trivial involution in $mathrmAut(L/K)$)
– Itai
Jul 30 at 13:27
I think you should clarify "numbers of the form a + b sqrt (d)", it's evidently not well defined when considering real numbers, or when d is a perfect square. Also the "It's sometimes..." sentence could be clarified, it's not like it is non-deterministic, depending on the context and object considered it will be clear what notion of conjugation one is referring to.
– Maxim
Jul 31 at 9:30
add a comment |Â
up vote
6
down vote
up vote
6
down vote
The notion of a conjugate comes up when you have a naturally defined function $c$ from a set to itself such that for all $x$ you have $c(c(x)) = x$. Then $c(x)$ is the conjugate of $x$.
That happens naturally when the numbers you are thinking about are of the form $a+bsqrtd$ for some $d$. Then you define the conjugating function by switching the sign of $b$.
In this case $2$ will be its own conjugate.
The conjugate of $1 + sqrt2$ is tricky. It's sometimes $1 - sqrt2$ but it's just $1 + sqrt2$ (itself) in the complex numbers, since it's a real number.
answered Jul 30 at 12:14
Ethan Bolker
35.7k54199
The notion of a conjugate comes up when you have a naturally defined function $c$ from a set to itself such that for all $x$ you have $c(c(x)) = x$. Then $c(x)$ is the conjugate of $x$.
That happens naturally when the numbers you are thinking about are of the form $a+bsqrtd$ for some $d$. Then you define the conjugating function by switching the sign of $b$.
In this case $2$ will be its own conjugate.
The conjugate of $1 + sqrt2$ is tricky. It's sometimes $1 - sqrt2$ but it's just $1 + sqrt2$ (itself) in the complex numbers, since it's a real number.
answered Jul 30 at 12:14
Ethan Bolker
35.7k54199
answered Jul 30 at 12:14
Ethan Bolker
35.7k54199
answered Jul 30 at 12:14
Ethan Bolker
35.7k54199
answered Jul 30 at 12:14
Ethan Bolker
35.7k54199
35.7k54199
Perhaps outside the scope of the question, but I would say a conjugate must be with respect to some subset, i.e. $c(x)=x$ for every $x$ in the sub-set as well as $c(c(x))=x$ for every $x$ in the whole set. Assuming only members of the sub-set satisfy the first condition, we have $c(x cdot c(x))=c(x)cdot c(c(x))=xcdot c(x)$ which means $xcdot c(x)$ is always in our "special subset" (in other words - a conjugate of $L$ over $K$ is a non-trivial involution in $mathrmAut(L/K)$)
– Itai
Jul 30 at 13:27
I think you should clarify "numbers of the form a + b sqrt (d)", it's evidently not well defined when considering real numbers, or when d is a perfect square. Also the "It's sometimes..." sentence could be clarified, it's not like it is non-deterministic, depending on the context and object considered it will be clear what notion of conjugation one is referring to.
– Maxim
Jul 31 at 9:30
add a comment |Â
Perhaps outside the scope of the question, but I would say a conjugate must be with respect to some subset, i.e. $c(x)=x$ for every $x$ in the sub-set as well as $c(c(x))=x$ for every $x$ in the whole set. Assuming only members of the sub-set satisfy the first condition, we have $c(x cdot c(x))=c(x)cdot c(c(x))=xcdot c(x)$ which means $xcdot c(x)$ is always in our "special subset" (in other words - a conjugate of $L$ over $K$ is a non-trivial involution in $mathrmAut(L/K)$)
– Itai
Jul 30 at 13:27
I think you should clarify "numbers of the form a + b sqrt (d)", it's evidently not well defined when considering real numbers, or when d is a perfect square. Also the "It's sometimes..." sentence could be clarified, it's not like it is non-deterministic, depending on the context and object considered it will be clear what notion of conjugation one is referring to.
– Maxim
Jul 31 at 9:30
Perhaps outside the scope of the question, but I would say a conjugate must be with respect to some subset, i.e. $c(x)=x$ for every $x$ in the sub-set as well as $c(c(x))=x$ for every $x$ in the whole set. Assuming only members of the sub-set satisfy the first condition, we have $c(x cdot c(x))=c(x)cdot c(c(x))=xcdot c(x)$ which means $xcdot c(x)$ is always in our "special subset" (in other words - a conjugate of $L$ over $K$ is a non-trivial involution in $mathrmAut(L/K)$)
– Itai
Jul 30 at 13:27
Perhaps outside the scope of the question, but I would say a conjugate must be with respect to some subset, i.e. $c(x)=x$ for every $x$ in the sub-set as well as $c(c(x))=x$ for every $x$ in the whole set. Assuming only members of the sub-set satisfy the first condition, we have $c(x cdot c(x))=c(x)cdot c(c(x))=xcdot c(x)$ which means $xcdot c(x)$ is always in our "special subset" (in other words - a conjugate of $L$ over $K$ is a non-trivial involution in $mathrmAut(L/K)$)
– Itai
Jul 30 at 13:27
I think you should clarify "numbers of the form a + b sqrt (d)", it's evidently not well defined when considering real numbers, or when d is a perfect square. Also the "It's sometimes..." sentence could be clarified, it's not like it is non-deterministic, depending on the context and object considered it will be clear what notion of conjugation one is referring to.
– Maxim
Jul 31 at 9:30
I think you should clarify "numbers of the form a + b sqrt (d)", it's evidently not well defined when considering real numbers, or when d is a perfect square. Also the "It's sometimes..." sentence could be clarified, it's not like it is non-deterministic, depending on the context and object considered it will be clear what notion of conjugation one is referring to.
– Maxim
Jul 31 at 9:30
add a comment |Â
up vote
2
down vote
Usually our definition of "conjugate" refers to complex numbers: the conjugate of $a+bi$ is $a-bi$. You could say "complex conjugate" be be extra specific.
Note that $1+sqrt2$ is a real number, so its conjugate is $1+sqrt2$.
A nice way of thinking about conjugates is how they are related in the complex plane (on an Argand diagram). Given a complex number, reflect it across the horizontal (real) axis to get its conjugate. Since $1$, $2$ and $1+sqrt2$ all lie on the real line, they are their own conjugate.
Getting a little bit more advanced, you can define conjugates in different fields. The complex numbers are an extension of the real numbers with the number $i$. We can write this "field extension" as $mathbbC=mathbbR(i)$.
Working instead in $mathbbQ(sqrt2)$ (the rationals extended with $sqrt2$), you can define the conjugate of $a+bsqrt2$ as $a-bsqrt2$. But don't worry so much about this if you are new to the topic!
answered Jul 30 at 12:16
Malkin
914421
add a comment |Â
up vote
2
down vote
Usually our definition of "conjugate" refers to complex numbers: the conjugate of $a+bi$ is $a-bi$. You could say "complex conjugate" be be extra specific.
Note that $1+sqrt2$ is a real number, so its conjugate is $1+sqrt2$.
A nice way of thinking about conjugates is how they are related in the complex plane (on an Argand diagram). Given a complex number, reflect it across the horizontal (real) axis to get its conjugate. Since $1$, $2$ and $1+sqrt2$ all lie on the real line, they are their own conjugate.
Getting a little bit more advanced, you can define conjugates in different fields. The complex numbers are an extension of the real numbers with the number $i$. We can write this "field extension" as $mathbbC=mathbbR(i)$.
Working instead in $mathbbQ(sqrt2)$ (the rationals extended with $sqrt2$), you can define the conjugate of $a+bsqrt2$ as $a-bsqrt2$. But don't worry so much about this if you are new to the topic!
answered Jul 30 at 12:16
Malkin
914421
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Usually our definition of "conjugate" refers to complex numbers: the conjugate of $a+bi$ is $a-bi$. You could say "complex conjugate" be be extra specific.
Note that $1+sqrt2$ is a real number, so its conjugate is $1+sqrt2$.
A nice way of thinking about conjugates is how they are related in the complex plane (on an Argand diagram). Given a complex number, reflect it across the horizontal (real) axis to get its conjugate. Since $1$, $2$ and $1+sqrt2$ all lie on the real line, they are their own conjugate.
Getting a little bit more advanced, you can define conjugates in different fields. The complex numbers are an extension of the real numbers with the number $i$. We can write this "field extension" as $mathbbC=mathbbR(i)$.
Working instead in $mathbbQ(sqrt2)$ (the rationals extended with $sqrt2$), you can define the conjugate of $a+bsqrt2$ as $a-bsqrt2$. But don't worry so much about this if you are new to the topic!
answered Jul 30 at 12:16
Malkin
914421
Usually our definition of "conjugate" refers to complex numbers: the conjugate of $a+bi$ is $a-bi$. You could say "complex conjugate" be be extra specific.
Note that $1+sqrt2$ is a real number, so its conjugate is $1+sqrt2$.
A nice way of thinking about conjugates is how they are related in the complex plane (on an Argand diagram). Given a complex number, reflect it across the horizontal (real) axis to get its conjugate. Since $1$, $2$ and $1+sqrt2$ all lie on the real line, they are their own conjugate.
Getting a little bit more advanced, you can define conjugates in different fields. The complex numbers are an extension of the real numbers with the number $i$. We can write this "field extension" as $mathbbC=mathbbR(i)$.
Working instead in $mathbbQ(sqrt2)$ (the rationals extended with $sqrt2$), you can define the conjugate of $a+bsqrt2$ as $a-bsqrt2$. But don't worry so much about this if you are new to the topic!
answered Jul 30 at 12:16
Malkin
914421
edited Jul 30 at 12:24
answered Jul 30 at 12:16
Malkin
914421
answered Jul 30 at 12:16
Malkin
914421
answered Jul 30 at 12:16
Malkin
914421
914421
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Since $mathbbR subset mathbbC$, any $a in mathbbR$ can be written in the following way: $a=a+icdot 0$.
– Iuli
Jul 30 at 12:10
14
There is complex conjugate for complex numbers, and there is also quadratic conjugate for algebraic numbers of the form $p+qsqrtd$ with $p,qinmathbbQ$ and $dinmathbbZ$ is not a square.
– Batominovski
Jul 30 at 12:11