# Can a polynomial have an irrational coefficient?

• barryj
In summary, the conversation discusses whether a polynomial can have irrational coefficients and the concept of "polynomials over a field." It also touches on a historical perspective of polynomial coefficients and the use of LaTeX in mathematical expressions. The final conclusion is that, unless specified, polynomials are assumed to have coefficients in the field of real numbers.
barryj
Homework Statement
can a polynomial have an irrational cooeficient?
Relevant Equations
is this is poloynomial? y = x^2 + sqrt(5)x + 1
Is this a polynomial? y = x^2 + sqrt(5)x + 1
I was told NO, the coefficients had to be rational numbers. I this true?

It seem to me this is an OK polynomial.
I can graph it and use the quad formula to find the roots? so why or why not?

It is a polynomial over the reals. The only time it would not be called a polynomial is if the coefficients were specifically specified as restricted to rational numbers. That is when you would specify that it is a "polynomial over the rationals". 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##. (see http://mathonline.wikidot.com/polynomials-over-a-field )

PS. It is even common to have polynomials with complex coefficients. They would be called "polynomials over the complex numbers".

Delta2 and scottdave
barryj said:
Homework Statement:: can a polynomial have an irrational cooeficient?
Relevant Equations:: is this is poloynomial? y = x^2 + sqrt(5)x + 1

Is this a polynomial? y = x^2 + sqrt(5)x + 1
I was told NO, the coefficients had to be rational numbers. I this true?

It seem to me this is an OK polynomial.
I can graph it and use the quad formula to find the roots? so why or why not?
Yes, in modern mathematics the coefficients may be any real number. In Medieval times, the coefficients had to be rational. Here's the history of it:

https://www.cambridge.org/core/jour...ance-algebra/38391278E789631131A9F9BFC4DC8B0B

Orodruin, Delta2, scottdave and 2 others
I had a homework problem that asked , given these roots, 3i, -3i, and (sqrt)5 , what is the smallest degree polynomial could I have. I said degree 3, a cubic. It can be found by multiplying the three factors...
(x+3i)(x-3i)(x-(sqrt)5) = x^3 -(sqrt)5x^2 + 9x - 9(sqrt)5

This polynomial has a coefficient of (sqrt)5 and the teacher said it was wrong??

Yes? I know, I need to learn to use Text.

I agree with your answer. The only reason I can imagine that the teacher would object is if he specified somewhere that the polynomial was over the rationals.

Thank you for confirming what I thought was correct.
This is an Algebra 2 class.
I do not think the term "polynomial over rationals" has been mentioned.

barryj said:
(x+3i)(x-3i)(x-(sqrt)5) = x^3 -(sqrt)5x^2 + 9x - 9(sqrt)5
In LaTeX this is ##(x + 3i)(x - 3i)(x - \sqrt 5) = x^3 - \sqrt 5 x^2 + 9x - 9\sqrt 5##

The raw, unrendered script looks like this:
##(x + 3i)(x - 3i)(x - \sqrt 5) = x^3 - \sqrt 5 x^2 + 9x - 9\sqrt 5##
barryj said:
This polynomial has a coefficient of (sqrt)5 and the teacher said it was wrong??
One coefficient is ##-\sqrt 5## and another is ##-9\sqrt 5##. I don't know why your teacher said it was wrong.
barryj said:
Yes? I know, I need to learn to use Text.
That's Tex, not Text. Take a look at our LaTeX Guide -- the link is in the lower left corner.

barryj said:
This is an Algebra 2 class.
I do not think the term "polynomial over rationals" has been mentioned.
I think that the concept of "polynomials over rationals" (or over any other field) is a subject for Abstract Algebra and would not be mentioned in Algebra 2.

barryj said:
Thank you for confirming what I thought was correct.
This is an Algebra 2 class.
I do not think the term "polynomial over rationals" has been mentioned.
I believe it's quite common among maths students (and possibly high-school maths teachers) that coefficients such as ##a, b, c## represent whole numbers. And, quite erroneously, they believe that, for example:
$$x^2 + 2\pi x + 1 = 0$$ is not a valid quadratic equation.

PS they allow rational coefficients because you can multiply by a suitable factor to make all coefficients whole numbers.

Last edited:
FactChecker
I just remembered something from 40 years ago! When I was a maths student, we were given the question to find a quadratic equation with ##\pi## as a root. This stumped many in the class, because they were only trying whole numbers for the coefficients.

In those pre-Internet days we had never heard of a "transcendental" number.

And, I think I saw something a few years ago about a study that found that many maths students throught that ##a, b, c## must be whole numbers.

Orodruin, FactChecker and Hall
PeroK said:
And, I think I saw something a few years ago about a study that found that many maths students throught that ##a, b, c## must be whole numbers.
Do you remember what level the students were at? I would think that it must be a beginning level.

FactChecker said:
Do you remember what level the students were at? I would think that it must be a beginning level.
This would have been first year.

FactChecker said:
I agree with your answer. The only reason I can imagine that the teacher would object is if he specified somewhere that the polynomial was over the rationals.

"Minimum degree of a polnomial with $n$ distinct roots" is going to be $n$ unless "with rational coefficients" or similar is specified, in which case other rationalizing factors may be necessary and the problem is no longer trivial.

PeroK said:
I believe it's quite common among maths students (and possibly high-school maths teachers) that coefficients such as ##a, b, c## represent whole numbers. And, quite erroneously, they believe that, for example:
$$x^2 + 2\pi x + 1 = 0$$ is not a valid quadratic equation.

PS they allow rational coefficients because you can multiply by a suitable factor to make all coefficients whole numbers.
That's not a function, that is a relation. And I don't see how you can multiply this by anything to make all coefficients rational, algebraic, or whole numbers.

Edit: I misinterpreted you last part, but most of my last point I think is right... Maybe?

valenumr said:
That's not a function, that is a relation. And I don't see how you can multiply this by anything to make all coefficients rational, algebraic, or whole numbers.

Edit: I misinterpreted you last part, but most of my last point I think is right... Maybe?
Sorry, I don't understand what you mean.

PeroK said:
Sorry, I don't understand what you mean.
I mean how can one multiply through to make a non whole number go away?

valenumr said:
I mean how can one multiply through to make a non whole number go away?
Multiply by the LCM of the denominators of the coefficients. Or, any common multiple thereof.

PeroK said:
Multiply by the LCM of the denominators of the coefficients. Or, any common multiple thereof.
But that may work for certain cases of expressions, but not functions... I guess that was my initial thought. Not that I disagree with the concept.

valenumr said:
But that may work for certain cases of expressions, but not functions... I guess that was my initial thought. Not that I disagree with the concept.
Shoot, what I'm thinking is that you can write a (say) quadratic equation or relation with any real coefficients. For all cases, I don't see a relation that would lead to all coefficients necissarily being whole numbers.

PeroK said:
Multiply by the LCM of the denominators of the coefficients. Or, any common multiple thereof.
I guess you are speaking about rational numbers...

valenumr said:
I guess you are speaking about rational numbers...
Yes. Polynomials with rational coefficients are essentially the same as Polynomials with integer coefficients.

valenumr
PeroK said:
Yes. Polynomials with rational coefficients are essentially the same as Polynomials with integer coefficients.
I got off track from your specific example... No reason that can't be a polynomial either.

valenumr said:
That's not a function, that is a relation.
What @PeroK posted (post #9) was an equation; namely ##x^2 + 2\pi x + 1 = 0##, so your comment above is a bit bewildering.

Mark44 said:
What @PeroK posted (post #9) was an equation; namely ##x^2 + 2\pi x + 1 = 0##, so your comment above is a bit bewildering.
Well I guess my line of thinking was that it's a bit different multiplying both sides by a factor when one side equals zero.

valenumr said:
Well I guess my line of thinking was that it's a bit different multiplying both sides by a factor when one side equals zero.
That registers high on my bewilderment scale!

Tom.G and Mark44
valenumr said:
Well I guess my line of thinking was that it's a bit different multiplying both sides by a factor when one side equals zero.
No, it's not different. If some expression = 0, then k times that expression = k * 0 = 0.

PeroK said:
That registers high on my bewilderment scale!
Why? A quadratic relation has at most two exact solutions. A quadratic function has infinite many "solitions".

fresh_42
Mark44 said:
No, it's not different. If some expression = 0, then k times that expression = k * 0 = 0.
The difference in treating a relation vs an edit: equation function

valenumr said:
The difference in treating a relation vs an edit: equation function
And sorry, I'm using the wrong terminology. I should say equality, not relation.

valenumr said:
The difference in treating a relation vs an equation.
This is irrelevant here. @PeroK explicitly wrote an equation and described it as such. There was no mention of either a function or relation.

What PeroK was talking about was the difference between an equation such as ##x^2 + \frac 3 2 x + 5 = 0## on the one hand, vs. his example of ##x^2 + 2\pi x + 1 = 0##.
The first equation can be converted to the equivalent equation ##2x^3 + 3x + 10 = 0## by multiplying both sides of the equation by 2.

The second equation cannot be converted to a quadratic equation with integer coefficients.

PeroK
valenumr said:
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.

PeroK said:
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.

Last edited:
valenumr
Hall said:
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.

valenumr said:
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?

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).

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