Mixing
Definition of algebraic integer
$begingroup$
Can the definition of algebraic Integer
The roots of polynomials, such as $x^3 + bx^2 + cx + d = 0$, with integer (or rational) coefficients.
be accurately paraphrased as
Any real or complex number $A + iB$ where $A$ and $B$ are integers.
If not, what belongs to my definition that does not belong to the usual one?
algebraic-number-theory
edited Jul 20 '15 at 21:14
coldnumber
3,3021820
asked Jul 17 '15 at 10:21
xatabayxatabay
94110
$endgroup$
add a comment |
$begingroup$
Can the definition of algebraic Integer
The roots of polynomials, such as $x^3 + bx^2 + cx + d = 0$, with integer (or rational) coefficients.
be accurately paraphrased as
Any real or complex number $A + iB$ where $A$ and $B$ are integers.
If not, what belongs to my definition that does not belong to the usual one?
algebraic-number-theory
edited Jul 20 '15 at 21:14
coldnumber
3,3021820
asked Jul 17 '15 at 10:21
xatabayxatabay
94110
$endgroup$
$begingroup$
What about $sqrt 2$? It is root of $x^2 - 2$. Your numbers are gaussian integers
$endgroup$
– Michael Galuza
Jul 17 '15 at 10:27
2
$begingroup$
Your definition of algebraic integer isn't even correct. An algebraic integer is a complex number which is the root of a monic polynomial with integer coefficients.
$endgroup$
– lokodiz
Jul 17 '15 at 10:34
$begingroup$
You got it the other way around. The numbers you mention are all algebraic integers, but not all algebraic integers take that form. For example think about the roots of unity.
$endgroup$
– A.P.
Jul 17 '15 at 12:05
add a comment |
$begingroup$
Can the definition of algebraic Integer
The roots of polynomials, such as $x^3 + bx^2 + cx + d = 0$, with integer (or rational) coefficients.
be accurately paraphrased as
Any real or complex number $A + iB$ where $A$ and $B$ are integers.
If not, what belongs to my definition that does not belong to the usual one?
algebraic-number-theory
edited Jul 20 '15 at 21:14
coldnumber
3,3021820
asked Jul 17 '15 at 10:21
xatabayxatabay
94110
$endgroup$
Can the definition of algebraic Integer
The roots of polynomials, such as $x^3 + bx^2 + cx + d = 0$, with integer (or rational) coefficients.
be accurately paraphrased as
Any real or complex number $A + iB$ where $A$ and $B$ are integers.
If not, what belongs to my definition that does not belong to the usual one?
algebraic-number-theory
algebraic-number-theory
edited Jul 20 '15 at 21:14
coldnumber
3,3021820
asked Jul 17 '15 at 10:21
xatabayxatabay
94110
edited Jul 20 '15 at 21:14
coldnumber
3,3021820
asked Jul 17 '15 at 10:21
xatabayxatabay
94110
edited Jul 20 '15 at 21:14
coldnumber
3,3021820
edited Jul 20 '15 at 21:14
coldnumber
3,3021820
edited Jul 20 '15 at 21:14
coldnumber
3,3021820
3,3021820
asked Jul 17 '15 at 10:21
xatabayxatabay
94110
asked Jul 17 '15 at 10:21
xatabayxatabay
94110
asked Jul 17 '15 at 10:21
xatabayxatabay
94110
94110
$begingroup$
What about $sqrt 2$? It is root of $x^2 - 2$. Your numbers are gaussian integers
$endgroup$
– Michael Galuza
Jul 17 '15 at 10:27
2
$begingroup$
Your definition of algebraic integer isn't even correct. An algebraic integer is a complex number which is the root of a monic polynomial with integer coefficients.
$endgroup$
– lokodiz
Jul 17 '15 at 10:34
$begingroup$
You got it the other way around. The numbers you mention are all algebraic integers, but not all algebraic integers take that form. For example think about the roots of unity.
$endgroup$
– A.P.
Jul 17 '15 at 12:05
add a comment |
$begingroup$
What about $sqrt 2$? It is root of $x^2 - 2$. Your numbers are gaussian integers
$endgroup$
– Michael Galuza
Jul 17 '15 at 10:27
2
$begingroup$
Your definition of algebraic integer isn't even correct. An algebraic integer is a complex number which is the root of a monic polynomial with integer coefficients.
$endgroup$
– lokodiz
Jul 17 '15 at 10:34
$begingroup$
You got it the other way around. The numbers you mention are all algebraic integers, but not all algebraic integers take that form. For example think about the roots of unity.
$endgroup$
– A.P.
Jul 17 '15 at 12:05
$begingroup$
What about $sqrt 2$? It is root of $x^2 - 2$. Your numbers are gaussian integers
$endgroup$
– Michael Galuza
Jul 17 '15 at 10:27
$begingroup$
What about $sqrt 2$? It is root of $x^2 - 2$. Your numbers are gaussian integers
$endgroup$
– Michael Galuza
Jul 17 '15 at 10:27
2
2
$begingroup$
Your definition of algebraic integer isn't even correct. An algebraic integer is a complex number which is the root of a monic polynomial with integer coefficients.
$endgroup$
– lokodiz
Jul 17 '15 at 10:34
$begingroup$
Your definition of algebraic integer isn't even correct. An algebraic integer is a complex number which is the root of a monic polynomial with integer coefficients.
$endgroup$
– lokodiz
Jul 17 '15 at 10:34
$begingroup$
You got it the other way around. The numbers you mention are all algebraic integers, but not all algebraic integers take that form. For example think about the roots of unity.
$endgroup$
– A.P.
Jul 17 '15 at 12:05
$begingroup$
You got it the other way around. The numbers you mention are all algebraic integers, but not all algebraic integers take that form. For example think about the roots of unity.
$endgroup$
– A.P.
Jul 17 '15 at 12:05
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If by "i" you mean $i = sqrt{-1}$, then your "paraphrase" is lacking. It would be like saying only beagles are dogs, thus ignoring terriers, colliers, retrievers, etc. Your "such as" seemed to indicate knowledge of algebraic integers of degree $3$, such as $1 + root 3 of 2$, which is a little difficult to represent as $A + iB$.
I recommend that you play around with Wolfram Alpha, asking it questions like:
is 1/2 + sqrt(-7)/2 an algebraic integer?is sqrt(-7)/2 an algebraic integer?is 3^(1/5) an algebraic integer?is pi an algebraic integer?- etc.
$endgroup$
add a comment |
$begingroup$
You're getting your Gaussian integers mixed up with your algebraic integers. All Gaussian integers are algebraic integers, but not all algebraic integers are Gaussian integers.
The polynomials are of crucial importance to determining what is or what isn't an algebraic integer. Take a look at this number: $$frac{28 + root 3 of {10} + 19(root 3 of {10})^2}{3}$$ With an approximate decimal value of $39.4482$, this is clearly not your typical integer. But it is a solution to $x^3 - 28x^2 + 198x - 25626 = 0$ (so then $b = 28$, $c = 198$ and $d = -25626$).
Hey, where's $a$? It's there, it's equal to $a = 1$, so you can just leave it out. Which brings me to my next point: all algebraic integers are algebraic numbers, but not all algebraic numbers are algebraic integers. This is not an algebraic integer, but it is an algebraic number: $$frac{root 3 of {11}}{3}$$ The relevant polynomial is $27x^3 - 11 = 0$. We have $b = 0$, $c = 0$, $d = -11$ but $a = 27$.
At the risk of sounding monotonous, the root of the polynomial is important, and trying to leave it out of a definition is just not worth it.
But the nice thing about Gaussian integers is that you can very easily guess the relevant polynomial. If $a + bi$ is a Gaussian integer, then the polynomial is probably is $x^2 - 2a + (a^2 + b^2)$.
$endgroup$
add a comment |
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2 Answers
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oldest
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2 Answers
2
active
oldest
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oldest
votes
$begingroup$
If by "i" you mean $i = sqrt{-1}$, then your "paraphrase" is lacking. It would be like saying only beagles are dogs, thus ignoring terriers, colliers, retrievers, etc. Your "such as" seemed to indicate knowledge of algebraic integers of degree $3$, such as $1 + root 3 of 2$, which is a little difficult to represent as $A + iB$.
I recommend that you play around with Wolfram Alpha, asking it questions like:
is 1/2 + sqrt(-7)/2 an algebraic integer?is sqrt(-7)/2 an algebraic integer?is 3^(1/5) an algebraic integer?is pi an algebraic integer?- etc.
$endgroup$
add a comment |
$begingroup$
If by "i" you mean $i = sqrt{-1}$, then your "paraphrase" is lacking. It would be like saying only beagles are dogs, thus ignoring terriers, colliers, retrievers, etc. Your "such as" seemed to indicate knowledge of algebraic integers of degree $3$, such as $1 + root 3 of 2$, which is a little difficult to represent as $A + iB$.
I recommend that you play around with Wolfram Alpha, asking it questions like:
is 1/2 + sqrt(-7)/2 an algebraic integer?is sqrt(-7)/2 an algebraic integer?is 3^(1/5) an algebraic integer?is pi an algebraic integer?- etc.
$endgroup$
add a comment |
$begingroup$
If by "i" you mean $i = sqrt{-1}$, then your "paraphrase" is lacking. It would be like saying only beagles are dogs, thus ignoring terriers, colliers, retrievers, etc. Your "such as" seemed to indicate knowledge of algebraic integers of degree $3$, such as $1 + root 3 of 2$, which is a little difficult to represent as $A + iB$.
I recommend that you play around with Wolfram Alpha, asking it questions like:
is 1/2 + sqrt(-7)/2 an algebraic integer?is sqrt(-7)/2 an algebraic integer?is 3^(1/5) an algebraic integer?is pi an algebraic integer?- etc.
$endgroup$
If by "i" you mean $i = sqrt{-1}$, then your "paraphrase" is lacking. It would be like saying only beagles are dogs, thus ignoring terriers, colliers, retrievers, etc. Your "such as" seemed to indicate knowledge of algebraic integers of degree $3$, such as $1 + root 3 of 2$, which is a little difficult to represent as $A + iB$.
I recommend that you play around with Wolfram Alpha, asking it questions like:
is 1/2 + sqrt(-7)/2 an algebraic integer?is sqrt(-7)/2 an algebraic integer?is 3^(1/5) an algebraic integer?is pi an algebraic integer?- etc.
answered Jul 20 '15 at 13:58
user153918
add a comment |
add a comment |
$begingroup$
You're getting your Gaussian integers mixed up with your algebraic integers. All Gaussian integers are algebraic integers, but not all algebraic integers are Gaussian integers.
The polynomials are of crucial importance to determining what is or what isn't an algebraic integer. Take a look at this number: $$frac{28 + root 3 of {10} + 19(root 3 of {10})^2}{3}$$ With an approximate decimal value of $39.4482$, this is clearly not your typical integer. But it is a solution to $x^3 - 28x^2 + 198x - 25626 = 0$ (so then $b = 28$, $c = 198$ and $d = -25626$).
Hey, where's $a$? It's there, it's equal to $a = 1$, so you can just leave it out. Which brings me to my next point: all algebraic integers are algebraic numbers, but not all algebraic numbers are algebraic integers. This is not an algebraic integer, but it is an algebraic number: $$frac{root 3 of {11}}{3}$$ The relevant polynomial is $27x^3 - 11 = 0$. We have $b = 0$, $c = 0$, $d = -11$ but $a = 27$.
At the risk of sounding monotonous, the root of the polynomial is important, and trying to leave it out of a definition is just not worth it.
But the nice thing about Gaussian integers is that you can very easily guess the relevant polynomial. If $a + bi$ is a Gaussian integer, then the polynomial is probably is $x^2 - 2a + (a^2 + b^2)$.
$endgroup$
add a comment |
$begingroup$
You're getting your Gaussian integers mixed up with your algebraic integers. All Gaussian integers are algebraic integers, but not all algebraic integers are Gaussian integers.
The polynomials are of crucial importance to determining what is or what isn't an algebraic integer. Take a look at this number: $$frac{28 + root 3 of {10} + 19(root 3 of {10})^2}{3}$$ With an approximate decimal value of $39.4482$, this is clearly not your typical integer. But it is a solution to $x^3 - 28x^2 + 198x - 25626 = 0$ (so then $b = 28$, $c = 198$ and $d = -25626$).
Hey, where's $a$? It's there, it's equal to $a = 1$, so you can just leave it out. Which brings me to my next point: all algebraic integers are algebraic numbers, but not all algebraic numbers are algebraic integers. This is not an algebraic integer, but it is an algebraic number: $$frac{root 3 of {11}}{3}$$ The relevant polynomial is $27x^3 - 11 = 0$. We have $b = 0$, $c = 0$, $d = -11$ but $a = 27$.
At the risk of sounding monotonous, the root of the polynomial is important, and trying to leave it out of a definition is just not worth it.
But the nice thing about Gaussian integers is that you can very easily guess the relevant polynomial. If $a + bi$ is a Gaussian integer, then the polynomial is probably is $x^2 - 2a + (a^2 + b^2)$.
$endgroup$
add a comment |
$begingroup$
You're getting your Gaussian integers mixed up with your algebraic integers. All Gaussian integers are algebraic integers, but not all algebraic integers are Gaussian integers.
The polynomials are of crucial importance to determining what is or what isn't an algebraic integer. Take a look at this number: $$frac{28 + root 3 of {10} + 19(root 3 of {10})^2}{3}$$ With an approximate decimal value of $39.4482$, this is clearly not your typical integer. But it is a solution to $x^3 - 28x^2 + 198x - 25626 = 0$ (so then $b = 28$, $c = 198$ and $d = -25626$).
Hey, where's $a$? It's there, it's equal to $a = 1$, so you can just leave it out. Which brings me to my next point: all algebraic integers are algebraic numbers, but not all algebraic numbers are algebraic integers. This is not an algebraic integer, but it is an algebraic number: $$frac{root 3 of {11}}{3}$$ The relevant polynomial is $27x^3 - 11 = 0$. We have $b = 0$, $c = 0$, $d = -11$ but $a = 27$.
At the risk of sounding monotonous, the root of the polynomial is important, and trying to leave it out of a definition is just not worth it.
But the nice thing about Gaussian integers is that you can very easily guess the relevant polynomial. If $a + bi$ is a Gaussian integer, then the polynomial is probably is $x^2 - 2a + (a^2 + b^2)$.
$endgroup$
You're getting your Gaussian integers mixed up with your algebraic integers. All Gaussian integers are algebraic integers, but not all algebraic integers are Gaussian integers.
The polynomials are of crucial importance to determining what is or what isn't an algebraic integer. Take a look at this number: $$frac{28 + root 3 of {10} + 19(root 3 of {10})^2}{3}$$ With an approximate decimal value of $39.4482$, this is clearly not your typical integer. But it is a solution to $x^3 - 28x^2 + 198x - 25626 = 0$ (so then $b = 28$, $c = 198$ and $d = -25626$).
Hey, where's $a$? It's there, it's equal to $a = 1$, so you can just leave it out. Which brings me to my next point: all algebraic integers are algebraic numbers, but not all algebraic numbers are algebraic integers. This is not an algebraic integer, but it is an algebraic number: $$frac{root 3 of {11}}{3}$$ The relevant polynomial is $27x^3 - 11 = 0$. We have $b = 0$, $c = 0$, $d = -11$ but $a = 27$.
At the risk of sounding monotonous, the root of the polynomial is important, and trying to leave it out of a definition is just not worth it.
But the nice thing about Gaussian integers is that you can very easily guess the relevant polynomial. If $a + bi$ is a Gaussian integer, then the polynomial is probably is $x^2 - 2a + (a^2 + b^2)$.
edited Jul 21 '15 at 14:13
answered Jul 20 '15 at 20:58
user155234
add a comment |
add a comment |
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$begingroup$
What about $sqrt 2$? It is root of $x^2 - 2$. Your numbers are gaussian integers
$endgroup$
– Michael Galuza
Jul 17 '15 at 10:27
2
$begingroup$
Your definition of algebraic integer isn't even correct. An algebraic integer is a complex number which is the root of a monic polynomial with integer coefficients.
$endgroup$
– lokodiz
Jul 17 '15 at 10:34
$begingroup$
You got it the other way around. The numbers you mention are all algebraic integers, but not all algebraic integers take that form. For example think about the roots of unity.
$endgroup$
– A.P.
Jul 17 '15 at 12:05