Do Irrational Conjugates always come in pairs?
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5
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favorite
Please forgive me for any mistakes while asking this question.
While applying the quadratic formula to find out the roots, or using any other methods to find the roots, I observe that the roots of the equation
$y=x^2-3$ are $sqrt3$ and $-sqrt3$. These roots evaluate to be irrational, and the roots always happen to come in irrational conjugates. If I do this with any other polynomial equation, the irrational solutions almost always come in conjugates. However, I have found a few exceptions to this, and I do not know why this occurs. Take the equation $-x^3-2x^2+2$, where two of the solutions are complex solutions. After applying long division, why do I get the other answer to be an irrational solution? I thought that if there is one irrational solution, surely there must be another, but this is not the case in the equation provided. Why is this so?
roots graphing-functions quadratics
asked Jan 16 at 23:31
Yash Jain
1
 |Â
show 1 more comment
up vote
5
down vote
favorite
Please forgive me for any mistakes while asking this question.
While applying the quadratic formula to find out the roots, or using any other methods to find the roots, I observe that the roots of the equation
$y=x^2-3$ are $sqrt3$ and $-sqrt3$. These roots evaluate to be irrational, and the roots always happen to come in irrational conjugates. If I do this with any other polynomial equation, the irrational solutions almost always come in conjugates. However, I have found a few exceptions to this, and I do not know why this occurs. Take the equation $-x^3-2x^2+2$, where two of the solutions are complex solutions. After applying long division, why do I get the other answer to be an irrational solution? I thought that if there is one irrational solution, surely there must be another, but this is not the case in the equation provided. Why is this so?
roots graphing-functions quadratics
asked Jan 16 at 23:31
Yash Jain
1
2
It's complicated. The subject you want is called Galois theory.
– Qiaochu Yuan
Jan 16 at 23:34
I think I don't really understand the question. What is it exactly that you are wondering? You can produce a polynomial with any number of irrational roots you want, just choose whatever roots you want in $mathbbC$ and then multiply the corresponding linear factors
– Pedro
Jan 16 at 23:37
1
$x^3 + 3$ has one real irrational root, and two complex (also irrational) roots. However, complex roots (of polynomials with real or integer coefficients) always come in conjugate pairs.
– Doug M
Jan 16 at 23:39
1
I think the motivation is that if $p + sqrtq$ is a root of a quadratic (with rational coefficients) then $p - sqrtq$ is as well (in many "elementary" math classes, expressions of the form $a + b$ and $a - b$ are called conjugates, no matter what type of things $a$ and $b$ are). [...]
– pjs36
Jan 17 at 0:13
1
[...] A thought on why this doesn't generalize neatly: conjugates (in the above sense) are strongly motivated by quadratics, in particular by the difference of squares identity $(a + b)(a - b) = a^2 - b^2$. But if $a$ and $b$ are cube roots, instead of square roots, then $a^2 - b^2$ is probably not rational; e.g., $(1 - sqrt[3]5)(1 + sqrt[3]5) = 1 - sqrt[3]25$. Conjugates (in the above sense) lose their niceness, if $a$ and $b$ aren't the "right" type of thing. But I don't really know.
– pjs36
Jan 17 at 0:13
 |Â
show 1 more comment
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Please forgive me for any mistakes while asking this question.
While applying the quadratic formula to find out the roots, or using any other methods to find the roots, I observe that the roots of the equation
$y=x^2-3$ are $sqrt3$ and $-sqrt3$. These roots evaluate to be irrational, and the roots always happen to come in irrational conjugates. If I do this with any other polynomial equation, the irrational solutions almost always come in conjugates. However, I have found a few exceptions to this, and I do not know why this occurs. Take the equation $-x^3-2x^2+2$, where two of the solutions are complex solutions. After applying long division, why do I get the other answer to be an irrational solution? I thought that if there is one irrational solution, surely there must be another, but this is not the case in the equation provided. Why is this so?
roots graphing-functions quadratics
asked Jan 16 at 23:31
Yash Jain
1
Please forgive me for any mistakes while asking this question.
While applying the quadratic formula to find out the roots, or using any other methods to find the roots, I observe that the roots of the equation
$y=x^2-3$ are $sqrt3$ and $-sqrt3$. These roots evaluate to be irrational, and the roots always happen to come in irrational conjugates. If I do this with any other polynomial equation, the irrational solutions almost always come in conjugates. However, I have found a few exceptions to this, and I do not know why this occurs. Take the equation $-x^3-2x^2+2$, where two of the solutions are complex solutions. After applying long division, why do I get the other answer to be an irrational solution? I thought that if there is one irrational solution, surely there must be another, but this is not the case in the equation provided. Why is this so?
roots graphing-functions quadratics
roots graphing-functions quadratics
asked Jan 16 at 23:31
Yash Jain
1
asked Jan 16 at 23:31
Yash Jain
1
asked Jan 16 at 23:31
Yash Jain
1
asked Jan 16 at 23:31
Yash Jain
1
asked Jan 16 at 23:31
Yash Jain
1
1
2
It's complicated. The subject you want is called Galois theory.
– Qiaochu Yuan
Jan 16 at 23:34
I think I don't really understand the question. What is it exactly that you are wondering? You can produce a polynomial with any number of irrational roots you want, just choose whatever roots you want in $mathbbC$ and then multiply the corresponding linear factors
– Pedro
Jan 16 at 23:37
1
$x^3 + 3$ has one real irrational root, and two complex (also irrational) roots. However, complex roots (of polynomials with real or integer coefficients) always come in conjugate pairs.
– Doug M
Jan 16 at 23:39
1
I think the motivation is that if $p + sqrtq$ is a root of a quadratic (with rational coefficients) then $p - sqrtq$ is as well (in many "elementary" math classes, expressions of the form $a + b$ and $a - b$ are called conjugates, no matter what type of things $a$ and $b$ are). [...]
– pjs36
Jan 17 at 0:13
1
[...] A thought on why this doesn't generalize neatly: conjugates (in the above sense) are strongly motivated by quadratics, in particular by the difference of squares identity $(a + b)(a - b) = a^2 - b^2$. But if $a$ and $b$ are cube roots, instead of square roots, then $a^2 - b^2$ is probably not rational; e.g., $(1 - sqrt[3]5)(1 + sqrt[3]5) = 1 - sqrt[3]25$. Conjugates (in the above sense) lose their niceness, if $a$ and $b$ aren't the "right" type of thing. But I don't really know.
– pjs36
Jan 17 at 0:13
 |Â
show 1 more comment
2
It's complicated. The subject you want is called Galois theory.
– Qiaochu Yuan
Jan 16 at 23:34
I think I don't really understand the question. What is it exactly that you are wondering? You can produce a polynomial with any number of irrational roots you want, just choose whatever roots you want in $mathbbC$ and then multiply the corresponding linear factors
– Pedro
Jan 16 at 23:37
1
$x^3 + 3$ has one real irrational root, and two complex (also irrational) roots. However, complex roots (of polynomials with real or integer coefficients) always come in conjugate pairs.
– Doug M
Jan 16 at 23:39
1
I think the motivation is that if $p + sqrtq$ is a root of a quadratic (with rational coefficients) then $p - sqrtq$ is as well (in many "elementary" math classes, expressions of the form $a + b$ and $a - b$ are called conjugates, no matter what type of things $a$ and $b$ are). [...]
– pjs36
Jan 17 at 0:13
1
[...] A thought on why this doesn't generalize neatly: conjugates (in the above sense) are strongly motivated by quadratics, in particular by the difference of squares identity $(a + b)(a - b) = a^2 - b^2$. But if $a$ and $b$ are cube roots, instead of square roots, then $a^2 - b^2$ is probably not rational; e.g., $(1 - sqrt[3]5)(1 + sqrt[3]5) = 1 - sqrt[3]25$. Conjugates (in the above sense) lose their niceness, if $a$ and $b$ aren't the "right" type of thing. But I don't really know.
– pjs36
Jan 17 at 0:13
2
2
It's complicated. The subject you want is called Galois theory.
– Qiaochu Yuan
Jan 16 at 23:34
It's complicated. The subject you want is called Galois theory.
– Qiaochu Yuan
Jan 16 at 23:34
I think I don't really understand the question. What is it exactly that you are wondering? You can produce a polynomial with any number of irrational roots you want, just choose whatever roots you want in $mathbbC$ and then multiply the corresponding linear factors
– Pedro
Jan 16 at 23:37
I think I don't really understand the question. What is it exactly that you are wondering? You can produce a polynomial with any number of irrational roots you want, just choose whatever roots you want in $mathbbC$ and then multiply the corresponding linear factors
– Pedro
Jan 16 at 23:37
1
1
$x^3 + 3$ has one real irrational root, and two complex (also irrational) roots. However, complex roots (of polynomials with real or integer coefficients) always come in conjugate pairs.
– Doug M
Jan 16 at 23:39
$x^3 + 3$ has one real irrational root, and two complex (also irrational) roots. However, complex roots (of polynomials with real or integer coefficients) always come in conjugate pairs.
– Doug M
Jan 16 at 23:39
1
1
I think the motivation is that if $p + sqrtq$ is a root of a quadratic (with rational coefficients) then $p - sqrtq$ is as well (in many "elementary" math classes, expressions of the form $a + b$ and $a - b$ are called conjugates, no matter what type of things $a$ and $b$ are). [...]
– pjs36
Jan 17 at 0:13
I think the motivation is that if $p + sqrtq$ is a root of a quadratic (with rational coefficients) then $p - sqrtq$ is as well (in many "elementary" math classes, expressions of the form $a + b$ and $a - b$ are called conjugates, no matter what type of things $a$ and $b$ are). [...]
– pjs36
Jan 17 at 0:13
1
1
[...] A thought on why this doesn't generalize neatly: conjugates (in the above sense) are strongly motivated by quadratics, in particular by the difference of squares identity $(a + b)(a - b) = a^2 - b^2$. But if $a$ and $b$ are cube roots, instead of square roots, then $a^2 - b^2$ is probably not rational; e.g., $(1 - sqrt[3]5)(1 + sqrt[3]5) = 1 - sqrt[3]25$. Conjugates (in the above sense) lose their niceness, if $a$ and $b$ aren't the "right" type of thing. But I don't really know.
– pjs36
Jan 17 at 0:13
[...] A thought on why this doesn't generalize neatly: conjugates (in the above sense) are strongly motivated by quadratics, in particular by the difference of squares identity $(a + b)(a - b) = a^2 - b^2$. But if $a$ and $b$ are cube roots, instead of square roots, then $a^2 - b^2$ is probably not rational; e.g., $(1 - sqrt[3]5)(1 + sqrt[3]5) = 1 - sqrt[3]25$. Conjugates (in the above sense) lose their niceness, if $a$ and $b$ aren't the "right" type of thing. But I don't really know.
– pjs36
Jan 17 at 0:13
 |Â
show 1 more comment
3 Answers
3
active
oldest
votes
up vote
5
down vote
accepted
The product of all the roots of the polynomial is exactly the constant term of the polynomial, up to a factor of $pm 1$. So if your polynomial has all integer coefficients and at least one root not in $mathbbQ$, then it has to have at least one other root not in $mathbbQ$ - otherwise, we would multiply all the roots and have an irrational number equal to a rational one.
These roots can, of course, be complex; in this case, either the real part or the imaginary part must be irrational. This is exactly what you're observing in this case. There is one real, irrational root and a pair of complex (conjugate) roots that have irrational real and imaginary parts.
answered Jan 16 at 23:44
T. Bongers
21.4k54259
add a comment |Â
up vote
3
down vote
I think you are confusing irrational solutions with complex solutions.
Irrational solutions need not come in pairs. The equation
$$
x^3 - 2 = 0
$$
has three roots. One is the irrational real number $alpha = 2^1/3$.The other two are the complex conjugate pair
$$
alpha left(frac-1 pm i sqrt32 right).
$$
The complex roots of a polynomial with real coefficients always come in conjugate pairs.
answered Jan 16 at 23:39
Ethan Bolker
36.5k54299
Please provide the explanation to why the irrational solutions do not need to come in pairs, otherwise just give me the link to where you found the explanation, as I am currently arguing with my teacher about this.
– Yash Jain
Jan 16 at 23:54
1
I can't provide a better explanation for why the pairing of irrational real roots doesn't happen other than a counterexample, like the one in the answer. If you edit the question to provide an argument for why you or your teacher think real irrational roots should come in pairs then someone here may be able to point out the error.
– Ethan Bolker
Jan 17 at 0:06
add a comment |Â
up vote
0
down vote
Here is a real polynomial with two roots: one rational and one irrational:
$$
P(x) = (x+3)(x+sqrt3)
$$
They are not irrational conjugates, naturally.
edited Sep 4 at 7:47
Henrik
5,83771930
answered Sep 4 at 7:12
Zibo
1
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
The product of all the roots of the polynomial is exactly the constant term of the polynomial, up to a factor of $pm 1$. So if your polynomial has all integer coefficients and at least one root not in $mathbbQ$, then it has to have at least one other root not in $mathbbQ$ - otherwise, we would multiply all the roots and have an irrational number equal to a rational one.
These roots can, of course, be complex; in this case, either the real part or the imaginary part must be irrational. This is exactly what you're observing in this case. There is one real, irrational root and a pair of complex (conjugate) roots that have irrational real and imaginary parts.
answered Jan 16 at 23:44
T. Bongers
21.4k54259
add a comment |Â
up vote
5
down vote
accepted
The product of all the roots of the polynomial is exactly the constant term of the polynomial, up to a factor of $pm 1$. So if your polynomial has all integer coefficients and at least one root not in $mathbbQ$, then it has to have at least one other root not in $mathbbQ$ - otherwise, we would multiply all the roots and have an irrational number equal to a rational one.
These roots can, of course, be complex; in this case, either the real part or the imaginary part must be irrational. This is exactly what you're observing in this case. There is one real, irrational root and a pair of complex (conjugate) roots that have irrational real and imaginary parts.
answered Jan 16 at 23:44
T. Bongers
21.4k54259
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
The product of all the roots of the polynomial is exactly the constant term of the polynomial, up to a factor of $pm 1$. So if your polynomial has all integer coefficients and at least one root not in $mathbbQ$, then it has to have at least one other root not in $mathbbQ$ - otherwise, we would multiply all the roots and have an irrational number equal to a rational one.
These roots can, of course, be complex; in this case, either the real part or the imaginary part must be irrational. This is exactly what you're observing in this case. There is one real, irrational root and a pair of complex (conjugate) roots that have irrational real and imaginary parts.
answered Jan 16 at 23:44
T. Bongers
21.4k54259
The product of all the roots of the polynomial is exactly the constant term of the polynomial, up to a factor of $pm 1$. So if your polynomial has all integer coefficients and at least one root not in $mathbbQ$, then it has to have at least one other root not in $mathbbQ$ - otherwise, we would multiply all the roots and have an irrational number equal to a rational one.
These roots can, of course, be complex; in this case, either the real part or the imaginary part must be irrational. This is exactly what you're observing in this case. There is one real, irrational root and a pair of complex (conjugate) roots that have irrational real and imaginary parts.
answered Jan 16 at 23:44
T. Bongers
21.4k54259
answered Jan 16 at 23:44
T. Bongers
21.4k54259
answered Jan 16 at 23:44
T. Bongers
21.4k54259
answered Jan 16 at 23:44
T. Bongers
21.4k54259
21.4k54259
add a comment |Â
add a comment |Â
up vote
3
down vote
I think you are confusing irrational solutions with complex solutions.
Irrational solutions need not come in pairs. The equation
$$
x^3 - 2 = 0
$$
has three roots. One is the irrational real number $alpha = 2^1/3$.The other two are the complex conjugate pair
$$
alpha left(frac-1 pm i sqrt32 right).
$$
The complex roots of a polynomial with real coefficients always come in conjugate pairs.
answered Jan 16 at 23:39
Ethan Bolker
36.5k54299
Please provide the explanation to why the irrational solutions do not need to come in pairs, otherwise just give me the link to where you found the explanation, as I am currently arguing with my teacher about this.
– Yash Jain
Jan 16 at 23:54
1
I can't provide a better explanation for why the pairing of irrational real roots doesn't happen other than a counterexample, like the one in the answer. If you edit the question to provide an argument for why you or your teacher think real irrational roots should come in pairs then someone here may be able to point out the error.
– Ethan Bolker
Jan 17 at 0:06
add a comment |Â
up vote
3
down vote
I think you are confusing irrational solutions with complex solutions.
Irrational solutions need not come in pairs. The equation
$$
x^3 - 2 = 0
$$
has three roots. One is the irrational real number $alpha = 2^1/3$.The other two are the complex conjugate pair
$$
alpha left(frac-1 pm i sqrt32 right).
$$
The complex roots of a polynomial with real coefficients always come in conjugate pairs.
answered Jan 16 at 23:39
Ethan Bolker
36.5k54299
Please provide the explanation to why the irrational solutions do not need to come in pairs, otherwise just give me the link to where you found the explanation, as I am currently arguing with my teacher about this.
– Yash Jain
Jan 16 at 23:54
1
I can't provide a better explanation for why the pairing of irrational real roots doesn't happen other than a counterexample, like the one in the answer. If you edit the question to provide an argument for why you or your teacher think real irrational roots should come in pairs then someone here may be able to point out the error.
– Ethan Bolker
Jan 17 at 0:06
add a comment |Â
up vote
3
down vote
up vote
3
down vote
I think you are confusing irrational solutions with complex solutions.
Irrational solutions need not come in pairs. The equation
$$
x^3 - 2 = 0
$$
has three roots. One is the irrational real number $alpha = 2^1/3$.The other two are the complex conjugate pair
$$
alpha left(frac-1 pm i sqrt32 right).
$$
The complex roots of a polynomial with real coefficients always come in conjugate pairs.
answered Jan 16 at 23:39
Ethan Bolker
36.5k54299
I think you are confusing irrational solutions with complex solutions.
Irrational solutions need not come in pairs. The equation
$$
x^3 - 2 = 0
$$
has three roots. One is the irrational real number $alpha = 2^1/3$.The other two are the complex conjugate pair
$$
alpha left(frac-1 pm i sqrt32 right).
$$
The complex roots of a polynomial with real coefficients always come in conjugate pairs.
answered Jan 16 at 23:39
Ethan Bolker
36.5k54299
answered Jan 16 at 23:39
Ethan Bolker
36.5k54299
answered Jan 16 at 23:39
Ethan Bolker
36.5k54299
answered Jan 16 at 23:39
Ethan Bolker
36.5k54299
36.5k54299
Please provide the explanation to why the irrational solutions do not need to come in pairs, otherwise just give me the link to where you found the explanation, as I am currently arguing with my teacher about this.
– Yash Jain
Jan 16 at 23:54
1
I can't provide a better explanation for why the pairing of irrational real roots doesn't happen other than a counterexample, like the one in the answer. If you edit the question to provide an argument for why you or your teacher think real irrational roots should come in pairs then someone here may be able to point out the error.
– Ethan Bolker
Jan 17 at 0:06
add a comment |Â
Please provide the explanation to why the irrational solutions do not need to come in pairs, otherwise just give me the link to where you found the explanation, as I am currently arguing with my teacher about this.
– Yash Jain
Jan 16 at 23:54
1
I can't provide a better explanation for why the pairing of irrational real roots doesn't happen other than a counterexample, like the one in the answer. If you edit the question to provide an argument for why you or your teacher think real irrational roots should come in pairs then someone here may be able to point out the error.
– Ethan Bolker
Jan 17 at 0:06
Please provide the explanation to why the irrational solutions do not need to come in pairs, otherwise just give me the link to where you found the explanation, as I am currently arguing with my teacher about this.
– Yash Jain
Jan 16 at 23:54
Please provide the explanation to why the irrational solutions do not need to come in pairs, otherwise just give me the link to where you found the explanation, as I am currently arguing with my teacher about this.
– Yash Jain
Jan 16 at 23:54
1
1
I can't provide a better explanation for why the pairing of irrational real roots doesn't happen other than a counterexample, like the one in the answer. If you edit the question to provide an argument for why you or your teacher think real irrational roots should come in pairs then someone here may be able to point out the error.
– Ethan Bolker
Jan 17 at 0:06
I can't provide a better explanation for why the pairing of irrational real roots doesn't happen other than a counterexample, like the one in the answer. If you edit the question to provide an argument for why you or your teacher think real irrational roots should come in pairs then someone here may be able to point out the error.
– Ethan Bolker
Jan 17 at 0:06
add a comment |Â
up vote
0
down vote
Here is a real polynomial with two roots: one rational and one irrational:
$$
P(x) = (x+3)(x+sqrt3)
$$
They are not irrational conjugates, naturally.
edited Sep 4 at 7:47
Henrik
5,83771930
answered Sep 4 at 7:12
Zibo
1
add a comment |Â
up vote
0
down vote
Here is a real polynomial with two roots: one rational and one irrational:
$$
P(x) = (x+3)(x+sqrt3)
$$
They are not irrational conjugates, naturally.
edited Sep 4 at 7:47
Henrik
5,83771930
answered Sep 4 at 7:12
Zibo
1
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Here is a real polynomial with two roots: one rational and one irrational:
$$
P(x) = (x+3)(x+sqrt3)
$$
They are not irrational conjugates, naturally.
edited Sep 4 at 7:47
Henrik
5,83771930
answered Sep 4 at 7:12
Zibo
1
Here is a real polynomial with two roots: one rational and one irrational:
$$
P(x) = (x+3)(x+sqrt3)
$$
They are not irrational conjugates, naturally.
edited Sep 4 at 7:47
Henrik
5,83771930
answered Sep 4 at 7:12
Zibo
1
edited Sep 4 at 7:47
Henrik
5,83771930
edited Sep 4 at 7:47
Henrik
5,83771930
edited Sep 4 at 7:47
Henrik
5,83771930
5,83771930
answered Sep 4 at 7:12
Zibo
1
answered Sep 4 at 7:12
Zibo
1
answered Sep 4 at 7:12
Zibo
1
1
add a comment |Â
add a comment |Â
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2
It's complicated. The subject you want is called Galois theory.
– Qiaochu Yuan
Jan 16 at 23:34
I think I don't really understand the question. What is it exactly that you are wondering? You can produce a polynomial with any number of irrational roots you want, just choose whatever roots you want in $mathbbC$ and then multiply the corresponding linear factors
– Pedro
Jan 16 at 23:37
1
$x^3 + 3$ has one real irrational root, and two complex (also irrational) roots. However, complex roots (of polynomials with real or integer coefficients) always come in conjugate pairs.
– Doug M
Jan 16 at 23:39
1
I think the motivation is that if $p + sqrtq$ is a root of a quadratic (with rational coefficients) then $p - sqrtq$ is as well (in many "elementary" math classes, expressions of the form $a + b$ and $a - b$ are called conjugates, no matter what type of things $a$ and $b$ are). [...]
– pjs36
Jan 17 at 0:13
1
[...] A thought on why this doesn't generalize neatly: conjugates (in the above sense) are strongly motivated by quadratics, in particular by the difference of squares identity $(a + b)(a - b) = a^2 - b^2$. But if $a$ and $b$ are cube roots, instead of square roots, then $a^2 - b^2$ is probably not rational; e.g., $(1 - sqrt[3]5)(1 + sqrt[3]5) = 1 - sqrt[3]25$. Conjugates (in the above sense) lose their niceness, if $a$ and $b$ aren't the "right" type of thing. But I don't really know.
– pjs36
Jan 17 at 0:13