Proving Equality of Rational Functions with Two Polynomials

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Homework Help Overview

The discussion revolves around proving the equality of two rational functions formed by polynomials, specifically under the condition that their greatest common divisor is one. The original poster seeks to establish that if the degrees of the numerators are less than the degrees of the respective denominators, the only solution to the equality is when both numerators are zero.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the degrees of the polynomials and the conditions under which one polynomial can divide another. There is discussion about the consequences of the gcd being one and how it relates to the equality of the rational functions.

Discussion Status

Participants are actively engaging with the problem, questioning the assumptions about polynomial degrees, particularly regarding the zero polynomial. Some guidance has been offered regarding the implications of the degree constraints, but no consensus has been reached on the final proof structure.

Contextual Notes

There is a noted ambiguity regarding the definition of the degree of the zero polynomial, which may affect the reasoning about the equality of the functions.

silvermane
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Homework Statement


We have two polynomials F(x) and g(x) over a field, and suppose that we know gcd(f,g)=1.
Consider two rational functions b(x)/f(x) and c(x)/g(x) in which we have deg(b)<deg(f) and deg(c)<deg(g). Prove that b(x)/f(x) = c(x)/g(x) can only be true when b(x) = c(x) = 0


The Attempt at a Solution


I really need help starting this proof.
So far, I have that f(x) won't divide b(x) since deg(b)<deg(f), and g(x) won't divide c(x) since deg(c)<deg(g). Because of this, we cross multiply to obtain:

b(x)*g(x) = c(x)*f(x)

But then it gets hazy here.
I know that f(x) and g(x) won't cancel out since their gcd=1, and b(x) must equal c(x) for them to cancel out. But then we'll be left with g(x) = f(x), so b(x) must equal c(x) and both terms must be equal to 0.

Is that all I have to do/say? It seems too simple.
Thanks for your help in advance! It means a lot! :blushing:
 
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silvermane said:

b(x)*g(x) = c(x)*f(x)


g(x) divides the LHS so it must divide the RHS. Since gcd(g,f) = 1, this means g(x) must divide c(x). By the same reasoning, f(x) must divide b(x). But you have some constraints on the degrees of these polynomials. What do the constraints imply?
 
By the way, the question may not make any sense as worded. If b(x) is the zero polynomial, what is deg(b)? Some people would say deg(b) is undefined in that case, others would set it to -\infty. What is the definition in this context?
 
jbunniii said:
g(x) divides the LHS so it must divide the RHS. Since gcd(g,f) = 1, this means g(x) must divide c(x). By the same reasoning, f(x) must divide b(x). But you have some constraints on the degrees of these polynomials. What do the constraints imply?

The constraints imply that f(x) can't divide b(x) and g(x) can't divide c(x). So then we would have that they must be 0 for the two sides to be equal?
 
silvermane said:
The constraints imply that f(x) can't divide b(x) and g(x) can't divide c(x). So then we would have that they must be 0 for the two sides to be equal?

Or, putting it more pedantically, f(x) cannot divide a polynomial of lower degree unless that polynomial is zero. [This assumes "degree" is defined for the zero polynomial.] Therefore b(x) must be the zero polynomial, and similarly for c(x).
 
jbunniii said:
Or, putting it more pedantically, f(x) cannot divide a polynomial of lower degree unless that polynomial is zero. [This assumes "degree" is defined for the zero polynomial.] Therefore b(x) must be the zero polynomial, and similarly for c(x).

Ah, that makes sense. Thanks so much for your help :smile:
 

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