Can a polynomial have an irrational coefficient?

[PeroK]

Why? A quadratic relation has at most two exact solutions. A quadratic function has infinite many "solitions".

And that's off the scale! I have no idea what you are talking about.

 

[Hall]

And that's off the scale! I have no idea what you are talking about.

I think Valenumr has this notion that a quadratic function is
$$
f(x) = x^2 +2\pi x +1$$
And what you wrote was
$$
x^2 + 2\pi x +1=0$$
And so, he is saying that you can, in some cases get integral coefficients in the second type of formss, but in the first form anything that you will multiply on RHS will get multiplied to f(x) also.

I think we should tell him that we are interested in finding the root of f(x), and hence setting it to zero is the most natural and rational thing to do. The function f(x) does remain as it is, with non integral coefficients but its root may coincide with the one whose coefficients are integral.

 

[PeroK]

I think Valenumr has this notion that a quadratic function is
$$
f(x) = x^2 +2\pi x +1$$
And what you wrote was
$$
x^2 + 2\pi x +1=0$$
And so, he is saying that you can, in some cases get integral coefficients in the second type of formss, but in the first form anything that you will multiply on RHS will get multiplied to f(x) also.

I think we should tell him that we are interested in finding the root of f(x), and hence setting it to zero is the most natural and rational thing to do. The function f(x) does remain as it is, with non integral coefficients but its root may coincide with the one whose coefficients are integral.

That might make some sense.

$f(x) = 0$ iff $kf(x) = 0$ where $k \ne 0$. So, the roots of a function are not changed if you multiply by a non-zero constant.

 

[WWGD]

And sorry, I'm using the wrong terminology. I should say equality, not relation.

Not sure what you mean . Do you mean $\{p(x) \}$ is an infinite set for nonconstant polynomials?

 

[epenguin]

The teacher had maybe some confused memory of term and concept "algebraic number"?

"An algebraic number is a number that is a root of a non-zero polynomial in one variable with integer (or, equivalently, rational) coefficients." (Wiki).

 

[Prof B]

In general, given a field, $F$, you can talk about "polynomials over the field $F$" when the coefficients are required to be in the field $F$

The integers are not a field. But the integers are a ring, and you can talk about polynomials over any ring. If you don't know which ring from context, it is a good idea to ask. One reason teachers ask questions like the OP's is to find out whether you know which ring.

 

[FactChecker]

The integers are not a field. But the integers are a ring, and you can talk about polynomials over any ring. If you don't know which ring from context, it is a good idea to ask. One reason teachers ask questions like the OP's is to find out whether you know which ring.

Good point. I stand corrected.