Question on the logic of rational roots theorem

In summary, the conversation discusses a contradiction in the rational roots theorem and the existence of rational roots in a given function. The speaker initially assumes that the function has no rational roots based on the theorem, but is confused when they graph the function and find one root. They then question whether this root could be irrational, but realizes that the conjugate of an irrational root is always a root, which would result in two roots. However, they are corrected by others in the conversation, clarifying that irrational and complex roots are not the same and that the theorem only applies to rational roots. The speaker then understands their mistake and thanks everyone for their help.
  • #1
dnt
238
0
this is not a homework question, but rather I feel like there is a contradiction in the theorem and just want clarity. I know the theorem is correct so I am looking for help in where the mistake is in my logic.

take f(x) = x^3 + x^2 - 4x- 7

the rational roots theorem says if there are any rational roots they must be in the set: plus or minus 7 and 1. none of which work using synthetic division. so the logic would dictate there should be no rational roots.

however, when you graph it you can see there is one root. but if this root were irrational, I believe the conjugate is always a root - hence there should be two. so therefore it cannot be an irrational root? therefore it must be rational?

which contradicts my previous statement from the rational roots theorem stating there were no rational zeros in this function.

hence my confusion...can someone clarify where I made my mistake in my logic or which assumption I made that was wrong? thanks!
 
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  • #2
##\sqrt[3]{3}## is the only real root of ##x^3-3##. ##\sqrt[3]{3}## is not rational.
 
  • #3
but if this root were irrational, I believe the conjugate is always a root

The conjugate? Are you confusing irrational roots with complex roots?
 
  • #4
If a polynomial equation, having only real coefficients, has a non-real complex number as root, it must also have the complex conjugate of that number as root.

But "irrational" does NOT mean "complex"!
 
  • #5
Having checked that [itex]f(1),f(-1),f(7),f(-7)[/itex] are all [itex]\neq 0[/itex], you've indeed shown (by the rational roots theorem) that [itex]f[/itex] has no rational roots. That is, are no [itex]a\in\mathbb Z, b\in\mathbb N[/itex] such that [itex]f\left(\frac{a}{b}\right)=0[/itex]. As others have pointed out, this result doesn't at all involve talking about complex conjugation (given that [itex]\mathbb Q\subseteq\mathbb R[/itex]).
 
  • #6
dnt said:
but if this root were irrational, I believe the conjugate is always a root - hence there should be two. so therefore it cannot be an irrational root? therefore it must be rational?

"The conjugate is a root" statement refers to complex conjugation. You are mixing that up with conjugation from rationalizing the denominator.
 
  • #7
i see - thanks for the help everyone! makes sense. i did get complex confused with irrational.
 

1. What is the rational roots theorem?

The rational roots theorem, also known as the rational zeros theorem, is a mathematical theorem that helps to determine the possible rational roots of a polynomial equation. It states that if a polynomial has integer coefficients, then any rational root of the polynomial must be of the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.

2. How is the rational roots theorem used?

The rational roots theorem is used to find the possible rational roots of a polynomial equation. This can be helpful in solving the equation or in finding the factors of the polynomial. By narrowing down the possible rational roots, it can make solving the equation more efficient and manageable.

3. What is the difference between rational and irrational roots?

Rational roots are numbers that can be expressed as a ratio of two integers, while irrational roots cannot be expressed in this way. Rational roots can be written as terminating or repeating decimals, while irrational roots are non-repeating and non-terminating decimals.

4. Can the rational roots theorem be used for all polynomial equations?

No, the rational roots theorem can only be used for polynomial equations with integer coefficients. If the coefficients are not integers, then the theorem does not apply and other methods must be used to find the roots.

5. Is the rational roots theorem a guarantee for finding all the roots of a polynomial equation?

No, the rational roots theorem only provides a list of possible rational roots. It does not guarantee that all the roots of the polynomial equation will be rational. There may be irrational or complex roots that also satisfy the equation, which the theorem does not account for.

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