Algebra 2 textbooks that incorrectly claim that all solutions of polynomial equations can be found
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Over the years I have occasionally encountered a number of Algebra 2 textbooks that make an incorrect (or at very least extremely misleading) claim along the lines that "all solutions of a polynomial equation can be found using a combination of the Rational Roots Theorem, Synthetic Division, and the Quadratic Formula." For example, the following image is from a McGraw-Hill Algebra 2 ebook:
I found a similar claim in the 2007 edition of Holt Algbra 2 (section 6-5, p. 441). While I haven't done an exhaustive curriculum search, I suspect this is fairly widespread. Even more to the point, I expect that many Algebra 2 teachers tell their students this, even if it is not written in the textbook.
It is of course true that the problems that students encounter in the course can all be handled by these methods, but that is because they have been carefully curated for that purpose. More generally, while the Fundamental Theorem of Algebra guarantees the existence of solutions, there are certainly polynomials of degree 3 and 4 that are irreducible over $mathbbQ$, and finding zeros for such polynomials requires methods far beyond what is normally taught in high school. For degree 5 and higher, the Abel-Ruffini Theorem shows that it is in general not possible at all to find the irrational roots algebraically, and no "exact form" even exists, although the roots can be approximated to arbitrary precision using numerical methods.
Even when a text or teacher does not explicitly make the false claim that all polynomials can be solved using just a few methods, failing to state that in fact the opposite is true (while only presenting examples in which those methods work) seems to me to be at least a sin of omission, and I imagine most students come away believing that any polynomial equation can be solved.
Quite apart from the fact that we shouldn't be teaching students things that are actually false, the fact that some polynomial equations just can't be solved, no matter how clever you are, and that mathematics is capable of proving the impossibility of something, seems like an important piece of meta-knowledge to me. I put it in the same category as "you can't trisect an angle using only a compass and unmarked straightedge" -- we don't expect students to understand the proof, but the result seems worth knowing, if for no other reason than that it makes the constructions you can do seem more worthwhile.
My questions, after all this:
- How pervasive is this error in textbooks?
- Are there any high school textbooks that explicitly acknowledge that the methods included in the text are not adequate to solve all 3rd and 4th degree polynomial equations, and that in higher degrees that are no general methods at all?
- Has anybody ever studied this as an issue of teacher knowledge?
algebra textbooks teachers solving-polynomials
asked Nov 18 at 20:17
mweiss
12.1k12671
|
show 11 more comments
up vote
16
down vote
favorite
Over the years I have occasionally encountered a number of Algebra 2 textbooks that make an incorrect (or at very least extremely misleading) claim along the lines that "all solutions of a polynomial equation can be found using a combination of the Rational Roots Theorem, Synthetic Division, and the Quadratic Formula." For example, the following image is from a McGraw-Hill Algebra 2 ebook:
I found a similar claim in the 2007 edition of Holt Algbra 2 (section 6-5, p. 441). While I haven't done an exhaustive curriculum search, I suspect this is fairly widespread. Even more to the point, I expect that many Algebra 2 teachers tell their students this, even if it is not written in the textbook.
It is of course true that the problems that students encounter in the course can all be handled by these methods, but that is because they have been carefully curated for that purpose. More generally, while the Fundamental Theorem of Algebra guarantees the existence of solutions, there are certainly polynomials of degree 3 and 4 that are irreducible over $mathbbQ$, and finding zeros for such polynomials requires methods far beyond what is normally taught in high school. For degree 5 and higher, the Abel-Ruffini Theorem shows that it is in general not possible at all to find the irrational roots algebraically, and no "exact form" even exists, although the roots can be approximated to arbitrary precision using numerical methods.
Even when a text or teacher does not explicitly make the false claim that all polynomials can be solved using just a few methods, failing to state that in fact the opposite is true (while only presenting examples in which those methods work) seems to me to be at least a sin of omission, and I imagine most students come away believing that any polynomial equation can be solved.
Quite apart from the fact that we shouldn't be teaching students things that are actually false, the fact that some polynomial equations just can't be solved, no matter how clever you are, and that mathematics is capable of proving the impossibility of something, seems like an important piece of meta-knowledge to me. I put it in the same category as "you can't trisect an angle using only a compass and unmarked straightedge" -- we don't expect students to understand the proof, but the result seems worth knowing, if for no other reason than that it makes the constructions you can do seem more worthwhile.
My questions, after all this:
- How pervasive is this error in textbooks?
- Are there any high school textbooks that explicitly acknowledge that the methods included in the text are not adequate to solve all 3rd and 4th degree polynomial equations, and that in higher degrees that are no general methods at all?
- Has anybody ever studied this as an issue of teacher knowledge?
algebra textbooks teachers solving-polynomials
asked Nov 18 at 20:17
mweiss
12.1k12671
2
@JoelReyesNoche It was in the chapter on higher-degree polynomials, after covering the Rational Roots Theorem and the Factor & Remainder Theorems. Definitely not restricted to the quadratic case.
– mweiss
Nov 19 at 2:14
1
We are cool. ;-)
– guest
Nov 19 at 3:40
4
@alephzero I disagree entirely. In the context of high school Algebra 1, let alone Algebra 2, the solutions of $x^2-2x-2=0$ are $1 pm sqrt3$, not 2.73205 and -0.73205. We do teach both completing the square and the Quadratic Formula, after all.
– mweiss
Nov 19 at 14:10
3
@alephzero not in any high school math course or text that I ever saw. If the course wants to talk about numerical approximations, it'll explicitly state so.
– Carl Witthoft
Nov 19 at 14:35
2
Not to pile on, but I'll even go further: Even for a linear equation like $7x=3$ we expect an exact solution like $x=3/7$, not a decimal approximation like $0.42587$.
– mweiss
Nov 19 at 16:32
|
show 11 more comments
up vote
16
down vote
favorite
up vote
16
down vote
favorite
Over the years I have occasionally encountered a number of Algebra 2 textbooks that make an incorrect (or at very least extremely misleading) claim along the lines that "all solutions of a polynomial equation can be found using a combination of the Rational Roots Theorem, Synthetic Division, and the Quadratic Formula." For example, the following image is from a McGraw-Hill Algebra 2 ebook:
I found a similar claim in the 2007 edition of Holt Algbra 2 (section 6-5, p. 441). While I haven't done an exhaustive curriculum search, I suspect this is fairly widespread. Even more to the point, I expect that many Algebra 2 teachers tell their students this, even if it is not written in the textbook.
It is of course true that the problems that students encounter in the course can all be handled by these methods, but that is because they have been carefully curated for that purpose. More generally, while the Fundamental Theorem of Algebra guarantees the existence of solutions, there are certainly polynomials of degree 3 and 4 that are irreducible over $mathbbQ$, and finding zeros for such polynomials requires methods far beyond what is normally taught in high school. For degree 5 and higher, the Abel-Ruffini Theorem shows that it is in general not possible at all to find the irrational roots algebraically, and no "exact form" even exists, although the roots can be approximated to arbitrary precision using numerical methods.
Even when a text or teacher does not explicitly make the false claim that all polynomials can be solved using just a few methods, failing to state that in fact the opposite is true (while only presenting examples in which those methods work) seems to me to be at least a sin of omission, and I imagine most students come away believing that any polynomial equation can be solved.
Quite apart from the fact that we shouldn't be teaching students things that are actually false, the fact that some polynomial equations just can't be solved, no matter how clever you are, and that mathematics is capable of proving the impossibility of something, seems like an important piece of meta-knowledge to me. I put it in the same category as "you can't trisect an angle using only a compass and unmarked straightedge" -- we don't expect students to understand the proof, but the result seems worth knowing, if for no other reason than that it makes the constructions you can do seem more worthwhile.
My questions, after all this:
- How pervasive is this error in textbooks?
- Are there any high school textbooks that explicitly acknowledge that the methods included in the text are not adequate to solve all 3rd and 4th degree polynomial equations, and that in higher degrees that are no general methods at all?
- Has anybody ever studied this as an issue of teacher knowledge?
algebra textbooks teachers solving-polynomials
asked Nov 18 at 20:17
mweiss
12.1k12671
Over the years I have occasionally encountered a number of Algebra 2 textbooks that make an incorrect (or at very least extremely misleading) claim along the lines that "all solutions of a polynomial equation can be found using a combination of the Rational Roots Theorem, Synthetic Division, and the Quadratic Formula." For example, the following image is from a McGraw-Hill Algebra 2 ebook:
I found a similar claim in the 2007 edition of Holt Algbra 2 (section 6-5, p. 441). While I haven't done an exhaustive curriculum search, I suspect this is fairly widespread. Even more to the point, I expect that many Algebra 2 teachers tell their students this, even if it is not written in the textbook.
It is of course true that the problems that students encounter in the course can all be handled by these methods, but that is because they have been carefully curated for that purpose. More generally, while the Fundamental Theorem of Algebra guarantees the existence of solutions, there are certainly polynomials of degree 3 and 4 that are irreducible over $mathbbQ$, and finding zeros for such polynomials requires methods far beyond what is normally taught in high school. For degree 5 and higher, the Abel-Ruffini Theorem shows that it is in general not possible at all to find the irrational roots algebraically, and no "exact form" even exists, although the roots can be approximated to arbitrary precision using numerical methods.
Even when a text or teacher does not explicitly make the false claim that all polynomials can be solved using just a few methods, failing to state that in fact the opposite is true (while only presenting examples in which those methods work) seems to me to be at least a sin of omission, and I imagine most students come away believing that any polynomial equation can be solved.
Quite apart from the fact that we shouldn't be teaching students things that are actually false, the fact that some polynomial equations just can't be solved, no matter how clever you are, and that mathematics is capable of proving the impossibility of something, seems like an important piece of meta-knowledge to me. I put it in the same category as "you can't trisect an angle using only a compass and unmarked straightedge" -- we don't expect students to understand the proof, but the result seems worth knowing, if for no other reason than that it makes the constructions you can do seem more worthwhile.
My questions, after all this:
- How pervasive is this error in textbooks?
- Are there any high school textbooks that explicitly acknowledge that the methods included in the text are not adequate to solve all 3rd and 4th degree polynomial equations, and that in higher degrees that are no general methods at all?
- Has anybody ever studied this as an issue of teacher knowledge?
algebra textbooks teachers solving-polynomials
algebra textbooks teachers solving-polynomials
asked Nov 18 at 20:17
mweiss
12.1k12671
asked Nov 18 at 20:17
mweiss
12.1k12671
edited Nov 18 at 20:28
asked Nov 18 at 20:17
mweiss
12.1k12671
asked Nov 18 at 20:17
mweiss
12.1k12671
asked Nov 18 at 20:17
mweiss
12.1k12671
12.1k12671
2
@JoelReyesNoche It was in the chapter on higher-degree polynomials, after covering the Rational Roots Theorem and the Factor & Remainder Theorems. Definitely not restricted to the quadratic case.
– mweiss
Nov 19 at 2:14
1
We are cool. ;-)
– guest
Nov 19 at 3:40
4
@alephzero I disagree entirely. In the context of high school Algebra 1, let alone Algebra 2, the solutions of $x^2-2x-2=0$ are $1 pm sqrt3$, not 2.73205 and -0.73205. We do teach both completing the square and the Quadratic Formula, after all.
– mweiss
Nov 19 at 14:10
3
@alephzero not in any high school math course or text that I ever saw. If the course wants to talk about numerical approximations, it'll explicitly state so.
– Carl Witthoft
Nov 19 at 14:35
2
Not to pile on, but I'll even go further: Even for a linear equation like $7x=3$ we expect an exact solution like $x=3/7$, not a decimal approximation like $0.42587$.
– mweiss
Nov 19 at 16:32
|
show 11 more comments
2
@JoelReyesNoche It was in the chapter on higher-degree polynomials, after covering the Rational Roots Theorem and the Factor & Remainder Theorems. Definitely not restricted to the quadratic case.
– mweiss
Nov 19 at 2:14
1
We are cool. ;-)
– guest
Nov 19 at 3:40
4
@alephzero I disagree entirely. In the context of high school Algebra 1, let alone Algebra 2, the solutions of $x^2-2x-2=0$ are $1 pm sqrt3$, not 2.73205 and -0.73205. We do teach both completing the square and the Quadratic Formula, after all.
– mweiss
Nov 19 at 14:10
3
@alephzero not in any high school math course or text that I ever saw. If the course wants to talk about numerical approximations, it'll explicitly state so.
– Carl Witthoft
Nov 19 at 14:35
2
Not to pile on, but I'll even go further: Even for a linear equation like $7x=3$ we expect an exact solution like $x=3/7$, not a decimal approximation like $0.42587$.
– mweiss
Nov 19 at 16:32
2
2
@JoelReyesNoche It was in the chapter on higher-degree polynomials, after covering the Rational Roots Theorem and the Factor & Remainder Theorems. Definitely not restricted to the quadratic case.
– mweiss
Nov 19 at 2:14
@JoelReyesNoche It was in the chapter on higher-degree polynomials, after covering the Rational Roots Theorem and the Factor & Remainder Theorems. Definitely not restricted to the quadratic case.
– mweiss
Nov 19 at 2:14
1
1
We are cool. ;-)
– guest
Nov 19 at 3:40
We are cool. ;-)
– guest
Nov 19 at 3:40
4
4
@alephzero I disagree entirely. In the context of high school Algebra 1, let alone Algebra 2, the solutions of $x^2-2x-2=0$ are $1 pm sqrt3$, not 2.73205 and -0.73205. We do teach both completing the square and the Quadratic Formula, after all.
– mweiss
Nov 19 at 14:10
@alephzero I disagree entirely. In the context of high school Algebra 1, let alone Algebra 2, the solutions of $x^2-2x-2=0$ are $1 pm sqrt3$, not 2.73205 and -0.73205. We do teach both completing the square and the Quadratic Formula, after all.
– mweiss
Nov 19 at 14:10
3
3
@alephzero not in any high school math course or text that I ever saw. If the course wants to talk about numerical approximations, it'll explicitly state so.
– Carl Witthoft
Nov 19 at 14:35
@alephzero not in any high school math course or text that I ever saw. If the course wants to talk about numerical approximations, it'll explicitly state so.
– Carl Witthoft
Nov 19 at 14:35
2
2
Not to pile on, but I'll even go further: Even for a linear equation like $7x=3$ we expect an exact solution like $x=3/7$, not a decimal approximation like $0.42587$.
– mweiss
Nov 19 at 16:32
Not to pile on, but I'll even go further: Even for a linear equation like $7x=3$ we expect an exact solution like $x=3/7$, not a decimal approximation like $0.42587$.
– mweiss
Nov 19 at 16:32
|
show 11 more comments
2 Answers
2
active
oldest
votes
up vote
9
down vote
At the moment, I can answer bullet point two:
Are there any high school textbooks that explicitly acknowledge that the methods included in the text are not adequate to solve all 3rd and 4th degree polynomial equations, and that in higher degrees that are no general methods at all?
Yes, you can find this on p. 267 of CME Project's (2009) Algebra 2 text. See the final paragraph below:
(Separately, Cardano and Tartaglia are mentioned on p. 205.)
In case you have a copy that may be lying around, here is a photo of the cover of the textbook:
Side-note: In the second paragraph pictured above, it is claimed that "the first rigorous proof of the [fundamental] theorem [of algebra] was given by Gauss in his doctoral thesis in 1799." This assertion sometimes comes under scrutiny; for more on a gap and its potential fix, cf:
Basu, S., & Velleman, D. J. (2017). On Gauss's first proof of the fundamental theorem of algebra. The American Mathematical Monthly, 124(8), 688-694. Link (arXiv).
answered Nov 19 at 0:17
Benjamin Dickman
15.9k22892
2
As my professor said, "The Fundamental Theorem of Algebra is neither fundamental nor a theorem of algebra." Not sure if it was from a book.
– Matt Samuel
Nov 19 at 3:10
Per echochamber.me/viewtopic.php?t=107992#p3540137, it seems to be adapted from a remark in Walter Noll's "Finite-Dimensional Spaces" (1987), but it also seems to be a twist on Voltaire's joke "This body which called itself and which still calls itself the Holy Roman Empire was in no way holy, nor Roman, nor an empire."
– mweiss
Nov 19 at 16:48
@mweiss: Funny, I learned that Holy Roman Empire quote from a sketch of "Coffee Talk with Linda Richman" on SNL: en.wikipedia.org/wiki/Coffee_Talk#Discussion_topics
– Brendan W. Sullivan
2 days ago
add a comment |
up vote
2
down vote
I think every textbook I've used for pre-calculus (at community college level) has the Fundamental Theorem of Algebra in it, so I add the context that, "Yeah, we know there are n solutions to an nth degree polynomial, but we also know that for degree 5 and higher, there will be cases where we can't find those solutions. Math is weird sometimes."
This post made me realize that we could probably quantify that, and say most equations of degree 5 and higher won't have algebraic solutions.
edited Nov 19 at 17:50
Jasper
2,462716
answered Nov 19 at 6:36
Sue VanHattum♦
9,21711962
Well really, "essentially all" is much more to the point rather than "most".
– DRF
Nov 19 at 10:34
@DRF and at Sue: is there any way to make your claims about "most" and "essentially all", respectively, polynomials of degree larger than 4 not having algebraic solutions rigorous? All I can tell is that this cannot be a cardinality argument, since there are infinitely many equations that do have algebraic solutions (e.g. the minimal polynomials of algebraic expressions) and there are only countably many polynomials
– Bananach
Nov 19 at 12:06
@Bananach - why "only countably many polynomials"? The coefficients don't have to be integers or rationals! $x^2 + pi x + e$ is a perfectly good polynomial so far as I know.
– alephzero
Nov 19 at 12:24
@alephzero ah okay, I thought we're talking about polynomials with integer coefficients. Of course, if you allow transendental coefficients, you wouldn't expect algebraic solutions
– Bananach
Nov 19 at 12:58
1
@Bananach: I've seen quite a few papers (some by Paul Erdős, I believe) that address the kinds of questions you're asking, but I don't have any specific references on-hand right now. However, there are a lot of links and references to this sort of thing in the Mathematics Stack Exchange question/answer Natural density of solvable quintics. Note that for $n geq 5,$ having the Galois group equal to the symmetric group $S_n$ (maximum possible) is a way of expressing a stronger result than the roots not being expressible in terms of radicals.
– Dave L Renfro
Nov 19 at 21:32
|
show 3 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
At the moment, I can answer bullet point two:
Are there any high school textbooks that explicitly acknowledge that the methods included in the text are not adequate to solve all 3rd and 4th degree polynomial equations, and that in higher degrees that are no general methods at all?
Yes, you can find this on p. 267 of CME Project's (2009) Algebra 2 text. See the final paragraph below:
(Separately, Cardano and Tartaglia are mentioned on p. 205.)
In case you have a copy that may be lying around, here is a photo of the cover of the textbook:
Side-note: In the second paragraph pictured above, it is claimed that "the first rigorous proof of the [fundamental] theorem [of algebra] was given by Gauss in his doctoral thesis in 1799." This assertion sometimes comes under scrutiny; for more on a gap and its potential fix, cf:
Basu, S., & Velleman, D. J. (2017). On Gauss's first proof of the fundamental theorem of algebra. The American Mathematical Monthly, 124(8), 688-694. Link (arXiv).
answered Nov 19 at 0:17
Benjamin Dickman
15.9k22892
2
As my professor said, "The Fundamental Theorem of Algebra is neither fundamental nor a theorem of algebra." Not sure if it was from a book.
– Matt Samuel
Nov 19 at 3:10
Per echochamber.me/viewtopic.php?t=107992#p3540137, it seems to be adapted from a remark in Walter Noll's "Finite-Dimensional Spaces" (1987), but it also seems to be a twist on Voltaire's joke "This body which called itself and which still calls itself the Holy Roman Empire was in no way holy, nor Roman, nor an empire."
– mweiss
Nov 19 at 16:48
@mweiss: Funny, I learned that Holy Roman Empire quote from a sketch of "Coffee Talk with Linda Richman" on SNL: en.wikipedia.org/wiki/Coffee_Talk#Discussion_topics
– Brendan W. Sullivan
2 days ago
add a comment |
up vote
9
down vote
At the moment, I can answer bullet point two:
Are there any high school textbooks that explicitly acknowledge that the methods included in the text are not adequate to solve all 3rd and 4th degree polynomial equations, and that in higher degrees that are no general methods at all?
Yes, you can find this on p. 267 of CME Project's (2009) Algebra 2 text. See the final paragraph below:
(Separately, Cardano and Tartaglia are mentioned on p. 205.)
In case you have a copy that may be lying around, here is a photo of the cover of the textbook:
Side-note: In the second paragraph pictured above, it is claimed that "the first rigorous proof of the [fundamental] theorem [of algebra] was given by Gauss in his doctoral thesis in 1799." This assertion sometimes comes under scrutiny; for more on a gap and its potential fix, cf:
Basu, S., & Velleman, D. J. (2017). On Gauss's first proof of the fundamental theorem of algebra. The American Mathematical Monthly, 124(8), 688-694. Link (arXiv).
answered Nov 19 at 0:17
Benjamin Dickman
15.9k22892
2
As my professor said, "The Fundamental Theorem of Algebra is neither fundamental nor a theorem of algebra." Not sure if it was from a book.
– Matt Samuel
Nov 19 at 3:10
Per echochamber.me/viewtopic.php?t=107992#p3540137, it seems to be adapted from a remark in Walter Noll's "Finite-Dimensional Spaces" (1987), but it also seems to be a twist on Voltaire's joke "This body which called itself and which still calls itself the Holy Roman Empire was in no way holy, nor Roman, nor an empire."
– mweiss
Nov 19 at 16:48
@mweiss: Funny, I learned that Holy Roman Empire quote from a sketch of "Coffee Talk with Linda Richman" on SNL: en.wikipedia.org/wiki/Coffee_Talk#Discussion_topics
– Brendan W. Sullivan
2 days ago
add a comment |
up vote
9
down vote
up vote
9
down vote
At the moment, I can answer bullet point two:
Are there any high school textbooks that explicitly acknowledge that the methods included in the text are not adequate to solve all 3rd and 4th degree polynomial equations, and that in higher degrees that are no general methods at all?
Yes, you can find this on p. 267 of CME Project's (2009) Algebra 2 text. See the final paragraph below:
(Separately, Cardano and Tartaglia are mentioned on p. 205.)
In case you have a copy that may be lying around, here is a photo of the cover of the textbook:
Side-note: In the second paragraph pictured above, it is claimed that "the first rigorous proof of the [fundamental] theorem [of algebra] was given by Gauss in his doctoral thesis in 1799." This assertion sometimes comes under scrutiny; for more on a gap and its potential fix, cf:
Basu, S., & Velleman, D. J. (2017). On Gauss's first proof of the fundamental theorem of algebra. The American Mathematical Monthly, 124(8), 688-694. Link (arXiv).
answered Nov 19 at 0:17
Benjamin Dickman
15.9k22892
At the moment, I can answer bullet point two:
Are there any high school textbooks that explicitly acknowledge that the methods included in the text are not adequate to solve all 3rd and 4th degree polynomial equations, and that in higher degrees that are no general methods at all?
Yes, you can find this on p. 267 of CME Project's (2009) Algebra 2 text. See the final paragraph below:
(Separately, Cardano and Tartaglia are mentioned on p. 205.)
In case you have a copy that may be lying around, here is a photo of the cover of the textbook:
Side-note: In the second paragraph pictured above, it is claimed that "the first rigorous proof of the [fundamental] theorem [of algebra] was given by Gauss in his doctoral thesis in 1799." This assertion sometimes comes under scrutiny; for more on a gap and its potential fix, cf:
Basu, S., & Velleman, D. J. (2017). On Gauss's first proof of the fundamental theorem of algebra. The American Mathematical Monthly, 124(8), 688-694. Link (arXiv).
answered Nov 19 at 0:17
Benjamin Dickman
15.9k22892
answered Nov 19 at 0:17
Benjamin Dickman
15.9k22892
answered Nov 19 at 0:17
Benjamin Dickman
15.9k22892
answered Nov 19 at 0:17
Benjamin Dickman
15.9k22892
15.9k22892
2
As my professor said, "The Fundamental Theorem of Algebra is neither fundamental nor a theorem of algebra." Not sure if it was from a book.
– Matt Samuel
Nov 19 at 3:10
Per echochamber.me/viewtopic.php?t=107992#p3540137, it seems to be adapted from a remark in Walter Noll's "Finite-Dimensional Spaces" (1987), but it also seems to be a twist on Voltaire's joke "This body which called itself and which still calls itself the Holy Roman Empire was in no way holy, nor Roman, nor an empire."
– mweiss
Nov 19 at 16:48
@mweiss: Funny, I learned that Holy Roman Empire quote from a sketch of "Coffee Talk with Linda Richman" on SNL: en.wikipedia.org/wiki/Coffee_Talk#Discussion_topics
– Brendan W. Sullivan
2 days ago
add a comment |
2
As my professor said, "The Fundamental Theorem of Algebra is neither fundamental nor a theorem of algebra." Not sure if it was from a book.
– Matt Samuel
Nov 19 at 3:10
Per echochamber.me/viewtopic.php?t=107992#p3540137, it seems to be adapted from a remark in Walter Noll's "Finite-Dimensional Spaces" (1987), but it also seems to be a twist on Voltaire's joke "This body which called itself and which still calls itself the Holy Roman Empire was in no way holy, nor Roman, nor an empire."
– mweiss
Nov 19 at 16:48
@mweiss: Funny, I learned that Holy Roman Empire quote from a sketch of "Coffee Talk with Linda Richman" on SNL: en.wikipedia.org/wiki/Coffee_Talk#Discussion_topics
– Brendan W. Sullivan
2 days ago
2
2
As my professor said, "The Fundamental Theorem of Algebra is neither fundamental nor a theorem of algebra." Not sure if it was from a book.
– Matt Samuel
Nov 19 at 3:10
As my professor said, "The Fundamental Theorem of Algebra is neither fundamental nor a theorem of algebra." Not sure if it was from a book.
– Matt Samuel
Nov 19 at 3:10
Per echochamber.me/viewtopic.php?t=107992#p3540137, it seems to be adapted from a remark in Walter Noll's "Finite-Dimensional Spaces" (1987), but it also seems to be a twist on Voltaire's joke "This body which called itself and which still calls itself the Holy Roman Empire was in no way holy, nor Roman, nor an empire."
– mweiss
Nov 19 at 16:48
Per echochamber.me/viewtopic.php?t=107992#p3540137, it seems to be adapted from a remark in Walter Noll's "Finite-Dimensional Spaces" (1987), but it also seems to be a twist on Voltaire's joke "This body which called itself and which still calls itself the Holy Roman Empire was in no way holy, nor Roman, nor an empire."
– mweiss
Nov 19 at 16:48
@mweiss: Funny, I learned that Holy Roman Empire quote from a sketch of "Coffee Talk with Linda Richman" on SNL: en.wikipedia.org/wiki/Coffee_Talk#Discussion_topics
– Brendan W. Sullivan
2 days ago
@mweiss: Funny, I learned that Holy Roman Empire quote from a sketch of "Coffee Talk with Linda Richman" on SNL: en.wikipedia.org/wiki/Coffee_Talk#Discussion_topics
– Brendan W. Sullivan
2 days ago
add a comment |
up vote
2
down vote
I think every textbook I've used for pre-calculus (at community college level) has the Fundamental Theorem of Algebra in it, so I add the context that, "Yeah, we know there are n solutions to an nth degree polynomial, but we also know that for degree 5 and higher, there will be cases where we can't find those solutions. Math is weird sometimes."
This post made me realize that we could probably quantify that, and say most equations of degree 5 and higher won't have algebraic solutions.
edited Nov 19 at 17:50
Jasper
2,462716
answered Nov 19 at 6:36
Sue VanHattum♦
9,21711962
Well really, "essentially all" is much more to the point rather than "most".
– DRF
Nov 19 at 10:34
@DRF and at Sue: is there any way to make your claims about "most" and "essentially all", respectively, polynomials of degree larger than 4 not having algebraic solutions rigorous? All I can tell is that this cannot be a cardinality argument, since there are infinitely many equations that do have algebraic solutions (e.g. the minimal polynomials of algebraic expressions) and there are only countably many polynomials
– Bananach
Nov 19 at 12:06
@Bananach - why "only countably many polynomials"? The coefficients don't have to be integers or rationals! $x^2 + pi x + e$ is a perfectly good polynomial so far as I know.
– alephzero
Nov 19 at 12:24
@alephzero ah okay, I thought we're talking about polynomials with integer coefficients. Of course, if you allow transendental coefficients, you wouldn't expect algebraic solutions
– Bananach
Nov 19 at 12:58
1
@Bananach: I've seen quite a few papers (some by Paul Erdős, I believe) that address the kinds of questions you're asking, but I don't have any specific references on-hand right now. However, there are a lot of links and references to this sort of thing in the Mathematics Stack Exchange question/answer Natural density of solvable quintics. Note that for $n geq 5,$ having the Galois group equal to the symmetric group $S_n$ (maximum possible) is a way of expressing a stronger result than the roots not being expressible in terms of radicals.
– Dave L Renfro
Nov 19 at 21:32
|
show 3 more comments
up vote
2
down vote
I think every textbook I've used for pre-calculus (at community college level) has the Fundamental Theorem of Algebra in it, so I add the context that, "Yeah, we know there are n solutions to an nth degree polynomial, but we also know that for degree 5 and higher, there will be cases where we can't find those solutions. Math is weird sometimes."
This post made me realize that we could probably quantify that, and say most equations of degree 5 and higher won't have algebraic solutions.
edited Nov 19 at 17:50
Jasper
2,462716
answered Nov 19 at 6:36
Sue VanHattum♦
9,21711962
Well really, "essentially all" is much more to the point rather than "most".
– DRF
Nov 19 at 10:34
@DRF and at Sue: is there any way to make your claims about "most" and "essentially all", respectively, polynomials of degree larger than 4 not having algebraic solutions rigorous? All I can tell is that this cannot be a cardinality argument, since there are infinitely many equations that do have algebraic solutions (e.g. the minimal polynomials of algebraic expressions) and there are only countably many polynomials
– Bananach
Nov 19 at 12:06
@Bananach - why "only countably many polynomials"? The coefficients don't have to be integers or rationals! $x^2 + pi x + e$ is a perfectly good polynomial so far as I know.
– alephzero
Nov 19 at 12:24
@alephzero ah okay, I thought we're talking about polynomials with integer coefficients. Of course, if you allow transendental coefficients, you wouldn't expect algebraic solutions
– Bananach
Nov 19 at 12:58
1
@Bananach: I've seen quite a few papers (some by Paul Erdős, I believe) that address the kinds of questions you're asking, but I don't have any specific references on-hand right now. However, there are a lot of links and references to this sort of thing in the Mathematics Stack Exchange question/answer Natural density of solvable quintics. Note that for $n geq 5,$ having the Galois group equal to the symmetric group $S_n$ (maximum possible) is a way of expressing a stronger result than the roots not being expressible in terms of radicals.
– Dave L Renfro
Nov 19 at 21:32
|
show 3 more comments
up vote
2
down vote
up vote
2
down vote
I think every textbook I've used for pre-calculus (at community college level) has the Fundamental Theorem of Algebra in it, so I add the context that, "Yeah, we know there are n solutions to an nth degree polynomial, but we also know that for degree 5 and higher, there will be cases where we can't find those solutions. Math is weird sometimes."
This post made me realize that we could probably quantify that, and say most equations of degree 5 and higher won't have algebraic solutions.
edited Nov 19 at 17:50
Jasper
2,462716
answered Nov 19 at 6:36
Sue VanHattum♦
9,21711962
I think every textbook I've used for pre-calculus (at community college level) has the Fundamental Theorem of Algebra in it, so I add the context that, "Yeah, we know there are n solutions to an nth degree polynomial, but we also know that for degree 5 and higher, there will be cases where we can't find those solutions. Math is weird sometimes."
This post made me realize that we could probably quantify that, and say most equations of degree 5 and higher won't have algebraic solutions.
edited Nov 19 at 17:50
Jasper
2,462716
answered Nov 19 at 6:36
Sue VanHattum♦
9,21711962
edited Nov 19 at 17:50
Jasper
2,462716
edited Nov 19 at 17:50
Jasper
2,462716
edited Nov 19 at 17:50
Jasper
2,462716
2,462716
answered Nov 19 at 6:36
Sue VanHattum♦
9,21711962
answered Nov 19 at 6:36
Sue VanHattum♦
9,21711962
answered Nov 19 at 6:36
Sue VanHattum♦
9,21711962
9,21711962
Well really, "essentially all" is much more to the point rather than "most".
– DRF
Nov 19 at 10:34
@DRF and at Sue: is there any way to make your claims about "most" and "essentially all", respectively, polynomials of degree larger than 4 not having algebraic solutions rigorous? All I can tell is that this cannot be a cardinality argument, since there are infinitely many equations that do have algebraic solutions (e.g. the minimal polynomials of algebraic expressions) and there are only countably many polynomials
– Bananach
Nov 19 at 12:06
@Bananach - why "only countably many polynomials"? The coefficients don't have to be integers or rationals! $x^2 + pi x + e$ is a perfectly good polynomial so far as I know.
– alephzero
Nov 19 at 12:24
@alephzero ah okay, I thought we're talking about polynomials with integer coefficients. Of course, if you allow transendental coefficients, you wouldn't expect algebraic solutions
– Bananach
Nov 19 at 12:58
1
@Bananach: I've seen quite a few papers (some by Paul Erdős, I believe) that address the kinds of questions you're asking, but I don't have any specific references on-hand right now. However, there are a lot of links and references to this sort of thing in the Mathematics Stack Exchange question/answer Natural density of solvable quintics. Note that for $n geq 5,$ having the Galois group equal to the symmetric group $S_n$ (maximum possible) is a way of expressing a stronger result than the roots not being expressible in terms of radicals.
– Dave L Renfro
Nov 19 at 21:32
|
show 3 more comments
Well really, "essentially all" is much more to the point rather than "most".
– DRF
Nov 19 at 10:34
@DRF and at Sue: is there any way to make your claims about "most" and "essentially all", respectively, polynomials of degree larger than 4 not having algebraic solutions rigorous? All I can tell is that this cannot be a cardinality argument, since there are infinitely many equations that do have algebraic solutions (e.g. the minimal polynomials of algebraic expressions) and there are only countably many polynomials
– Bananach
Nov 19 at 12:06
@Bananach - why "only countably many polynomials"? The coefficients don't have to be integers or rationals! $x^2 + pi x + e$ is a perfectly good polynomial so far as I know.
– alephzero
Nov 19 at 12:24
@alephzero ah okay, I thought we're talking about polynomials with integer coefficients. Of course, if you allow transendental coefficients, you wouldn't expect algebraic solutions
– Bananach
Nov 19 at 12:58
1
@Bananach: I've seen quite a few papers (some by Paul Erdős, I believe) that address the kinds of questions you're asking, but I don't have any specific references on-hand right now. However, there are a lot of links and references to this sort of thing in the Mathematics Stack Exchange question/answer Natural density of solvable quintics. Note that for $n geq 5,$ having the Galois group equal to the symmetric group $S_n$ (maximum possible) is a way of expressing a stronger result than the roots not being expressible in terms of radicals.
– Dave L Renfro
Nov 19 at 21:32
Well really, "essentially all" is much more to the point rather than "most".
– DRF
Nov 19 at 10:34
Well really, "essentially all" is much more to the point rather than "most".
– DRF
Nov 19 at 10:34
@DRF and at Sue: is there any way to make your claims about "most" and "essentially all", respectively, polynomials of degree larger than 4 not having algebraic solutions rigorous? All I can tell is that this cannot be a cardinality argument, since there are infinitely many equations that do have algebraic solutions (e.g. the minimal polynomials of algebraic expressions) and there are only countably many polynomials
– Bananach
Nov 19 at 12:06
@DRF and at Sue: is there any way to make your claims about "most" and "essentially all", respectively, polynomials of degree larger than 4 not having algebraic solutions rigorous? All I can tell is that this cannot be a cardinality argument, since there are infinitely many equations that do have algebraic solutions (e.g. the minimal polynomials of algebraic expressions) and there are only countably many polynomials
– Bananach
Nov 19 at 12:06
@Bananach - why "only countably many polynomials"? The coefficients don't have to be integers or rationals! $x^2 + pi x + e$ is a perfectly good polynomial so far as I know.
– alephzero
Nov 19 at 12:24
@Bananach - why "only countably many polynomials"? The coefficients don't have to be integers or rationals! $x^2 + pi x + e$ is a perfectly good polynomial so far as I know.
– alephzero
Nov 19 at 12:24
@alephzero ah okay, I thought we're talking about polynomials with integer coefficients. Of course, if you allow transendental coefficients, you wouldn't expect algebraic solutions
– Bananach
Nov 19 at 12:58
@alephzero ah okay, I thought we're talking about polynomials with integer coefficients. Of course, if you allow transendental coefficients, you wouldn't expect algebraic solutions
– Bananach
Nov 19 at 12:58
1
1
@Bananach: I've seen quite a few papers (some by Paul Erdős, I believe) that address the kinds of questions you're asking, but I don't have any specific references on-hand right now. However, there are a lot of links and references to this sort of thing in the Mathematics Stack Exchange question/answer Natural density of solvable quintics. Note that for $n geq 5,$ having the Galois group equal to the symmetric group $S_n$ (maximum possible) is a way of expressing a stronger result than the roots not being expressible in terms of radicals.
– Dave L Renfro
Nov 19 at 21:32
@Bananach: I've seen quite a few papers (some by Paul Erdős, I believe) that address the kinds of questions you're asking, but I don't have any specific references on-hand right now. However, there are a lot of links and references to this sort of thing in the Mathematics Stack Exchange question/answer Natural density of solvable quintics. Note that for $n geq 5,$ having the Galois group equal to the symmetric group $S_n$ (maximum possible) is a way of expressing a stronger result than the roots not being expressible in terms of radicals.
– Dave L Renfro
Nov 19 at 21:32
|
show 3 more comments
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@JoelReyesNoche It was in the chapter on higher-degree polynomials, after covering the Rational Roots Theorem and the Factor & Remainder Theorems. Definitely not restricted to the quadratic case.
– mweiss
Nov 19 at 2:14
1
We are cool. ;-)
– guest
Nov 19 at 3:40
4
@alephzero I disagree entirely. In the context of high school Algebra 1, let alone Algebra 2, the solutions of $x^2-2x-2=0$ are $1 pm sqrt3$, not 2.73205 and -0.73205. We do teach both completing the square and the Quadratic Formula, after all.
– mweiss
Nov 19 at 14:10
3
@alephzero not in any high school math course or text that I ever saw. If the course wants to talk about numerical approximations, it'll explicitly state so.
– Carl Witthoft
Nov 19 at 14:35
2
Not to pile on, but I'll even go further: Even for a linear equation like $7x=3$ we expect an exact solution like $x=3/7$, not a decimal approximation like $0.42587$.
– mweiss
Nov 19 at 16:32