Question on the logic of rational roots theorem

  • Thread starter dnt
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  • #1
dnt
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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 Im 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!
 

Answers and Replies

  • #2
575
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##\sqrt[3]{3}## is the only real root of ##x^3-3##. ##\sqrt[3]{3}## is not rational.
 
  • #3
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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
HallsofIvy
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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
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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
pwsnafu
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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
dnt
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i see - thanks for the help everyone! makes sense. i did get complex confused with irrational.
 

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