# Can a polynomial have an irrational coefficient?

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?

## Answers and Replies

FactChecker
Gold Member
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".

scottdave
PeroK
Homework Helper
Gold Member
2021 Award
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

scottdave, hutchphd and FactChecker
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.

FactChecker
Gold Member
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.

Mark44
Mentor
(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##
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.
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.

FactChecker
Gold Member
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.

PeroK
Homework Helper
Gold Member
2021 Award
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
PeroK
Homework Helper
Gold Member
2021 Award
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.

FactChecker and Hall
FactChecker
Gold Member
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.

PeroK
Homework Helper
Gold Member
2021 Award
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.

pasmith
Homework Helper
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.

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?

PeroK
Homework Helper
Gold Member
2021 Award
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.

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

PeroK
Homework Helper
Gold Member
2021 Award
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.

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.

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.

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

PeroK
Homework Helper
Gold Member
2021 Award
I guess you are speaking about rational numbers...
Yes. Polynomials with rational coefficients are essentially the same as Polynomials with integer coefficients.

valenumr
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.

Mark44
Mentor
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.

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.

PeroK