Mixing
Is there 'Algebraic number' which cannot display with Arithmetic operation and root
$begingroup$
Let $a + bi$ be an algebraic number.
Then there is polynomial which coefficients are rational number and one of root is $a+bi$.
I think..
$$x = a + bi$$
we can subtract $c_1$ (which is rational number) from both sides.
$$x-c_1=a-c_1+bi$$
and we can power both side.
$$(x-c_1)^n=(a-c_1+bi)^n$$
and repeat we can get polynomial which coefficients are rational number and one of root is $a+bi$.
Therefore, I think there is no 'Algebraic number' which cannot be displayed with arithmetic operations and roots.
algebraic-numbers
edited Feb 3 at 13:06
Matt Samuel
39.3k63870
asked Feb 3 at 2:44
math mathmath math
61
$endgroup$
add a comment |
$begingroup$
Let $a + bi$ be an algebraic number.
Then there is polynomial which coefficients are rational number and one of root is $a+bi$.
I think..
$$x = a + bi$$
we can subtract $c_1$ (which is rational number) from both sides.
$$x-c_1=a-c_1+bi$$
and we can power both side.
$$(x-c_1)^n=(a-c_1+bi)^n$$
and repeat we can get polynomial which coefficients are rational number and one of root is $a+bi$.
Therefore, I think there is no 'Algebraic number' which cannot be displayed with arithmetic operations and roots.
algebraic-numbers
edited Feb 3 at 13:06
Matt Samuel
39.3k63870
asked Feb 3 at 2:44
math mathmath math
61
$endgroup$
3
$begingroup$
There certainly are such numbers.
$endgroup$
– Matt Samuel
Feb 3 at 2:59
$begingroup$
should you explain about that or show me example?
$endgroup$
– math math
Feb 3 at 3:35
1
$begingroup$
The solutions of $x^5-4x+2=0$ cannot be expressed in terms of radicals.
$endgroup$
– Matt Samuel
Feb 3 at 3:37
2
$begingroup$
See "Abel-Ruffini Theorem" ..... in Wikipedia.
$endgroup$
– DanielWainfleet
Feb 3 at 3:50
add a comment |
$begingroup$
Let $a + bi$ be an algebraic number.
Then there is polynomial which coefficients are rational number and one of root is $a+bi$.
I think..
$$x = a + bi$$
we can subtract $c_1$ (which is rational number) from both sides.
$$x-c_1=a-c_1+bi$$
and we can power both side.
$$(x-c_1)^n=(a-c_1+bi)^n$$
and repeat we can get polynomial which coefficients are rational number and one of root is $a+bi$.
Therefore, I think there is no 'Algebraic number' which cannot be displayed with arithmetic operations and roots.
algebraic-numbers
edited Feb 3 at 13:06
Matt Samuel
39.3k63870
asked Feb 3 at 2:44
math mathmath math
61
$endgroup$
Let $a + bi$ be an algebraic number.
Then there is polynomial which coefficients are rational number and one of root is $a+bi$.
I think..
$$x = a + bi$$
we can subtract $c_1$ (which is rational number) from both sides.
$$x-c_1=a-c_1+bi$$
and we can power both side.
$$(x-c_1)^n=(a-c_1+bi)^n$$
and repeat we can get polynomial which coefficients are rational number and one of root is $a+bi$.
Therefore, I think there is no 'Algebraic number' which cannot be displayed with arithmetic operations and roots.
algebraic-numbers
algebraic-numbers
edited Feb 3 at 13:06
Matt Samuel
39.3k63870
asked Feb 3 at 2:44
math mathmath math
61
edited Feb 3 at 13:06
Matt Samuel
39.3k63870
asked Feb 3 at 2:44
math mathmath math
61
edited Feb 3 at 13:06
Matt Samuel
39.3k63870
edited Feb 3 at 13:06
Matt Samuel
39.3k63870
edited Feb 3 at 13:06
Matt Samuel
39.3k63870
39.3k63870
asked Feb 3 at 2:44
math mathmath math
61
asked Feb 3 at 2:44
math mathmath math
61
asked Feb 3 at 2:44
math mathmath math
61
61
3
$begingroup$
There certainly are such numbers.
$endgroup$
– Matt Samuel
Feb 3 at 2:59
$begingroup$
should you explain about that or show me example?
$endgroup$
– math math
Feb 3 at 3:35
1
$begingroup$
The solutions of $x^5-4x+2=0$ cannot be expressed in terms of radicals.
$endgroup$
– Matt Samuel
Feb 3 at 3:37
2
$begingroup$
See "Abel-Ruffini Theorem" ..... in Wikipedia.
$endgroup$
– DanielWainfleet
Feb 3 at 3:50
add a comment |
3
$begingroup$
There certainly are such numbers.
$endgroup$
– Matt Samuel
Feb 3 at 2:59
$begingroup$
should you explain about that or show me example?
$endgroup$
– math math
Feb 3 at 3:35
1
$begingroup$
The solutions of $x^5-4x+2=0$ cannot be expressed in terms of radicals.
$endgroup$
– Matt Samuel
Feb 3 at 3:37
2
$begingroup$
See "Abel-Ruffini Theorem" ..... in Wikipedia.
$endgroup$
– DanielWainfleet
Feb 3 at 3:50
3
3
$begingroup$
There certainly are such numbers.
$endgroup$
– Matt Samuel
Feb 3 at 2:59
$begingroup$
There certainly are such numbers.
$endgroup$
– Matt Samuel
Feb 3 at 2:59
$begingroup$
should you explain about that or show me example?
$endgroup$
– math math
Feb 3 at 3:35
$begingroup$
should you explain about that or show me example?
$endgroup$
– math math
Feb 3 at 3:35
1
1
$begingroup$
The solutions of $x^5-4x+2=0$ cannot be expressed in terms of radicals.
$endgroup$
– Matt Samuel
Feb 3 at 3:37
$begingroup$
The solutions of $x^5-4x+2=0$ cannot be expressed in terms of radicals.
$endgroup$
– Matt Samuel
Feb 3 at 3:37
2
2
$begingroup$
See "Abel-Ruffini Theorem" ..... in Wikipedia.
$endgroup$
– DanielWainfleet
Feb 3 at 3:50
$begingroup$
See "Abel-Ruffini Theorem" ..... in Wikipedia.
$endgroup$
– DanielWainfleet
Feb 3 at 3:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The algebraic numbers are divided into the explicit algebraic numbers and the implicit algebraic numbers.
The explicit algebraic numbers can be presented explicitly from the rational complex numbers by the arithmetic operations (addition, subtraction, multiplication, division, raising to integer powers, and $n$-th roots where $n$ is an integer). They are called solutions in radicals.
All solutions of algebraic equations of degree $leq$ 4 are solutions in radicals. There are equations of degree $ge$ 5 that have solutions that are not solutions in radicals. This is stated by Abel–Ruffini theorem.
The algebraic equation $x^5-x+1$ for example doesn't have solutions in radicals.
answered Feb 3 at 14:50
IV_IV_
1,561525
$endgroup$
add a comment |
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$begingroup$
The algebraic numbers are divided into the explicit algebraic numbers and the implicit algebraic numbers.
The explicit algebraic numbers can be presented explicitly from the rational complex numbers by the arithmetic operations (addition, subtraction, multiplication, division, raising to integer powers, and $n$-th roots where $n$ is an integer). They are called solutions in radicals.
All solutions of algebraic equations of degree $leq$ 4 are solutions in radicals. There are equations of degree $ge$ 5 that have solutions that are not solutions in radicals. This is stated by Abel–Ruffini theorem.
The algebraic equation $x^5-x+1$ for example doesn't have solutions in radicals.
answered Feb 3 at 14:50
IV_IV_
1,561525
$endgroup$
add a comment |
$begingroup$
The algebraic numbers are divided into the explicit algebraic numbers and the implicit algebraic numbers.
The explicit algebraic numbers can be presented explicitly from the rational complex numbers by the arithmetic operations (addition, subtraction, multiplication, division, raising to integer powers, and $n$-th roots where $n$ is an integer). They are called solutions in radicals.
All solutions of algebraic equations of degree $leq$ 4 are solutions in radicals. There are equations of degree $ge$ 5 that have solutions that are not solutions in radicals. This is stated by Abel–Ruffini theorem.
The algebraic equation $x^5-x+1$ for example doesn't have solutions in radicals.
answered Feb 3 at 14:50
IV_IV_
1,561525
$endgroup$
add a comment |
$begingroup$
The algebraic numbers are divided into the explicit algebraic numbers and the implicit algebraic numbers.
The explicit algebraic numbers can be presented explicitly from the rational complex numbers by the arithmetic operations (addition, subtraction, multiplication, division, raising to integer powers, and $n$-th roots where $n$ is an integer). They are called solutions in radicals.
All solutions of algebraic equations of degree $leq$ 4 are solutions in radicals. There are equations of degree $ge$ 5 that have solutions that are not solutions in radicals. This is stated by Abel–Ruffini theorem.
The algebraic equation $x^5-x+1$ for example doesn't have solutions in radicals.
answered Feb 3 at 14:50
IV_IV_
1,561525
$endgroup$
The algebraic numbers are divided into the explicit algebraic numbers and the implicit algebraic numbers.
The explicit algebraic numbers can be presented explicitly from the rational complex numbers by the arithmetic operations (addition, subtraction, multiplication, division, raising to integer powers, and $n$-th roots where $n$ is an integer). They are called solutions in radicals.
All solutions of algebraic equations of degree $leq$ 4 are solutions in radicals. There are equations of degree $ge$ 5 that have solutions that are not solutions in radicals. This is stated by Abel–Ruffini theorem.
The algebraic equation $x^5-x+1$ for example doesn't have solutions in radicals.
answered Feb 3 at 14:50
IV_IV_
1,561525
answered Feb 3 at 14:50
IV_IV_
1,561525
answered Feb 3 at 14:50
IV_IV_
1,561525
answered Feb 3 at 14:50
IV_IV_
1,561525
1,561525
add a comment |
add a comment |
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$begingroup$
There certainly are such numbers.
$endgroup$
– Matt Samuel
Feb 3 at 2:59
$begingroup$
should you explain about that or show me example?
$endgroup$
– math math
Feb 3 at 3:35
1
$begingroup$
The solutions of $x^5-4x+2=0$ cannot be expressed in terms of radicals.
$endgroup$
– Matt Samuel
Feb 3 at 3:37
2
$begingroup$
See "Abel-Ruffini Theorem" ..... in Wikipedia.
$endgroup$
– DanielWainfleet
Feb 3 at 3:50