Questions about equation roots

[nmego12345]

At the beginning of the summer, I was studying a precalculus course, in which I was taught that whenever a polynomial equation has a root in the form a + sqrt(b)c or a + ib, then another root would be its conjugate, I took it for granted for that time and I thought it was intuitive.

Later on, now, summer ended, I have grown much more familiar with mathematics, was revisiting that course and tried to prove that

But I soon discovered that this theory is false
For example we may have

This is a polynomial : (x-2)(x+1-sqrt(3) = 0

While playing around I discovered that this theory works only if the coefficients of the polynomial expression are rational, yeah irrationals doesn't wiorkNothing wrong with it, but I just disproved a theorem written in a good math course, so I wonder was I right or not

 

[member 587159]

At the beginning of the summer, I was studying a precalculus course, in which I was taught that whenever a polynomial equation has a root in the form a + sqrt(b)c or a + ib, then another root would be its conjugate, I took it for granted for that time and I thought it was intuitive.

Later on, now, summer ended, I have grown much more familiar with mathematics, was revisiting that course and tried to prove that

But I soon discovered that this theory is false
For example we may have

This is a polynomial : (x-2)(x+1-sqrt(3) = 0

While playing around I discovered that this theory works only if the coefficients of the polynomial expression are rational, yeah irrationals doesn't wiorkNothing wrong with it, but I just disproved a theorem written in a good math course, so I wonder was I right or not

What you write does not make sense to me. Can you formally list the theorem that you wanted to proof/disproof?

 

[pwsnafu]

At the beginning of the summer, I was studying a precalculus course, in which I was taught that whenever a polynomial equation has a root in the form a + sqrt(b)c or a + ib, then another root would be its conjugate, I took it for granted for that time and I thought it was intuitive.

This is false.
Given a polynomial with real coefficients and $z \in \mathbb{C}\setminus\mathbb{R}$ is a root then $\bar{z}$ is a root.

 

[fresh_42]

At the beginning of the summer, I was studying a precalculus course, in which I was taught that whenever a polynomial equation has a root in the form a + sqrt(b)c or a + ib, then another root would be its conjugate, I took it for granted for that time and I thought it was intuitive.

Later on, now, summer ended, I have grown much more familiar with mathematics, was revisiting that course and tried to prove that

But I soon discovered that this theory is false
For example we may have

This is a polynomial : (x-2)(x+1-sqrt(3) = 0

While playing around I discovered that this theory works only if the coefficients of the polynomial expression are rational, yeah irrationals doesn't wiorkNothing wrong with it, but I just disproved a theorem written in a good math course, so I wonder was I right or not

The point is, that you haven't read, memorized, quoted or whatever the theorem carefully enough.

If $p$ is an irreducible polynomial in $\mathbb{F}[x]$, and $a$ a root of $p$ in a field extension $\mathbb{G} \supset \mathbb{F}$ then $\sigma(a)$ is also a root for all $\sigma \in Aut_\mathbb{F}(\mathbb{G})$.
The latter is the group of automorphisms of $\mathbb{G}$ that leave elements of $\mathbb{F}$ fixed.

In your example, you have $\mathbb{F} = \mathbb{Q}$ and $\mathbb{G}=\mathbb{Q}[\sqrt{b}]$, where $b$ is not a square in $\mathbb{Q}$ and $\sigma(\sqrt{b}) = -\sqrt{b}$. This is an automorphism of $\mathbb{Q}[\sqrt{b}]$ of order $2$ that leaves $\mathbb{Q}$ fixed and conjugates the extension element $\sqrt{b}$.

In the other example, you have $\mathbb{F} = \mathbb{R}$ and $\mathbb{G} = \mathbb{R} [ \sqrt{-1} ] = \mathbb{R} [ i ] = \mathbb{C}$.
Here $b=-1$ is not a square in $\mathbb{R}$ and $\sigma(\sqrt{b}) = \sigma(\sqrt{-1}) = \sigma(i) =- i$. This is an automorphism of $\mathbb{C}$ of order $2$ that leaves $\mathbb{R}$ fixed and conjugates in $\mathbb{C}$.

So your observations have all been correct. Only your memory of the theorem was a bit too sloppy formulated.

Edit: In these cases $\sigma$, the conjugation, and the identity are the only automorphisms. Therefore the requirement that $p$ is irreducible, i.e. cannot be split into non-trivial factors. Otherwise you could attach any zeros as you want, which makes the overall situation a bit messy.

 

[nmego12345]

What you write does not make sense to me. Can you formally list the theorem that you wanted to proof/disproof?

If a number a has a conjugate b, and a is a root of the polynomial function f(x) then b must be a root of f(x)

This is false.
Given a polynomial with real coefficients and $z \in \mathbb{C}\setminus\mathbb{R}$ is a root then $\bar{z}$ is a root.

Sure, but I just proven you're wrong, try (x + 1- sqrt(3))(x-2), How can you explain that, furthermore, you've just stated the theorem without any proof, just "you're wrong"

Sorry if i was a little harsh, thanks for taking your time to help

If $p$ is an irreducible polynomial in $\mathbb{F}[x]$, and $a$ a root of $p$ in a field extension $\mathbb{G} \supset \mathbb{F}$ then $\sigma(a)$ is also a root for all $\sigma \in Aut_\mathbb{F}(\mathbb{G})$.
The latter is the group of automorphisms of $\mathbb{G}$ that leave elements of $\mathbb{F}$ fixed.
y.

I'm convinced but do you have a proof of that?

 

[nmego12345]

What you write does not make sense to me. Can you formally list the theorem that you wanted to proof/disproof?

If a number a has a conjugate b, and a is a root of the polynomial function f(x) then b must be a root of f(x)

This is false.
Given a polynomial with real coefficients and $z \in \mathbb{C}\setminus\mathbb{R}$ is a root then $\bar{z}$ is a root.

Sure, but I just proven you're wrong, try (x + 1- sqrt(3))(x-2), How can you explain that, furthermore, you've just stated the theorem without any proof, just "you're wrong"

Sorry if i was a little harsh, thanks for taking your time to help

If $p$ is an irreducible polynomial in $\mathbb{F}[x]$, and $a$ a root of $p$ in a field extension $\mathbb{G} \supset \mathbb{F}$ then $\sigma(a)$ is also a root for all $\sigma \in Aut_\mathbb{F}(\mathbb{G})$.
The latter is the group of automorphisms of $\mathbb{G}$ that leave elements of $\mathbb{F}$ fixed.
y.

I'm convinced but do you have a proof of that?

 

[nmego12345]

What you write does not make sense to me. Can you formally list the theorem that you wanted to proof/disproof?

If a number a has a conjugate b, and a is a root of the polynomial function f(x) then b must be a root of f(x)

This is false.
Given a polynomial with real coefficients and $z \in \mathbb{C}\setminus\mathbb{R}$ is a root then $\bar{z}$ is a root.

Sure, but I just proven you're wrong, try (x + 1- sqrt(3))(x-2), How can you explain that?

If $p$ is an irreducible polynomial in $\mathbb{F}[x]$, and $a$ a root of $p$ in a field extension $\mathbb{G} \supset \mathbb{F}$ then $\sigma(a)$ is also a root for all $\sigma \in Aut_\mathbb{F}(\mathbb{G})$.
The latter is the group of automorphisms of $\mathbb{G}$ that leave elements of $\mathbb{F}$ fixed.
y.

I'm convinced but do you have a proof of that?

 

[fresh_42]

I'm convinced but do you have a proof of that?

Google Galois Theory and field extensions.
For a textbook reference, see e.g.
https://www.amazon.com/dp/0387406247/?tag=pfamazon01-20

 

[PeroK]

This is a polynomial : (x-2)(x+1-sqrt(3) = 0

While playing around I discovered that this theory works only if the coefficients of the polynomial expression are rational, yeah irrationals doesn't wiorkNothing wrong with it, but I just disproved a theorem written in a good math course, so I wonder was I right or not

The solutions to your polynomial equation would appear to be $2$ and $-1 + \sqrt{3}$.

Given that these roots are real, then they equal their complex conjugates and thus, trivially, the theorem holds.

Have I misunderstood what you mean?

 

[pwsnafu]

Sure, but I just proven you're wrong, try (x + 1- sqrt(3))(x-2), How can you explain that, furthermore, you've just stated the theorem without any proof, just "you're wrong"

You haven't proved me wrong at all, I'm saying you are misremembering the theorem you were taught. Notice I quoted the part where you state the theorem, not your analysis of it.

 

[symbolipoint]

$(x+1-sqrt(3))(x-2)$ is NOT a polynomial but the fact might or might not help you.

 

[pwsnafu]

$(x+1-sqrt(3))(x-2)$ is NOT a polynomial but the fact might or might not help you.

Why do you say that?

 

[symbolipoint]

I should really recheck the textbook information for discussions about what expressions are and what are not polynomials, but a sum of terms which contains a constant which is irrational would not be a polynomial; a sum of terms which contains any irrational coefficient is not a polynomial. I may be wrong on one of those and could use some refinement in knowing how "polynomial" is defined. Tell I I'm correct or incorrect and give support - and I'll do a quick check on Wikipedia right now.

 

[symbolipoint]

Why do you say that?

After a quick Wiki check, I seem supported, but true polynomial or not, this about a definition is likely not important to nmego's question. I simply noted the mention of the expression as a polynomial when in fact it is not a polynomial.

 

[PeroK]

I should really recheck the textbook information for discussions about what expressions are and what are not polynomials, but a sum of terms which contains a constant which is irrational would not be a polynomial; a sum of terms which contains any irrational coefficient is not a polynomial. I may be wrong on one of those and could use some refinement in knowing how "polynomial" is defined. Tell I I'm correct or incorrect and give support - and I'll do a quick check on Wikipedia right now.

There are polynomials with rational coefficients and polynomials with real coefficients (for which the roots are either real or in complex conjugate pairs) and polynomials with complex coefficients (for which, trivially, the roots do not necessarily come in conjugate pairs).

There was something I read a while ago, where a large number of students thought that the solution to the quadratic equation:

$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

only applied when $a, b, c$ were whole numbers. Whereas, of course, $a, b, c$ can be any real or complex numbers.

 

[PeroK]

After a quick Wiki check, I seem supported, but true polynomial or not, this about a definition is likely not important to nmego's question. I simply noted the mention of the expression as a polynomial when in fact it is not a polynomial.

This is really not right at all. Almost all the examples of polynomials have whole numbers as coefficients, but the coefficients can be any real or complex numbers.

In any case, what do you say?

$ax^2 + bx + c = 0$

Has solutions:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Does this only apply when $a, b, c$ are integer/rational numbers? Or, can $a, b, c$ can be any real or complex numbers?

 

[pwsnafu]

After a quick Wiki check, I seem supported, but true polynomial or not, this about a definition is likely not important to nmego's question. I simply noted the mention of the expression as a polynomial when in fact it is not a polynomial.

No, polynomials exist over any ring, that includes reals and complex numbers. The algebraic definition of polynomial is as follows:

Let $R$ be a ring, and consider a subset of $R^{\mathbb{N}}$ defined as $\{ p \in R^{\mathbb{N}} : \exists N \in \mathbb{N} , \forall i > N \, p_i = 0 \}$. That is consider all sequences with only finite number of non-zero terms. This is clearly a $R$-module. We turn it into a ring by defining $c_n = \sum_{i=0}^n a_i \, b_{n-i}$. It is easy to see this satisfies distributivity and associativity. We then define $x = (0,1,0,0,0,\ldots)$ and then we can re-write any of our sequences in terms of monomials because $x^n$ now forms a basis.

 

[micromass]

$(x+1-sqrt(3))(x-2)$ is NOT a polynomial but the fact might or might not help you.

What? That function is about as polynomial as polynomials get!