Difference between revisions of "Rational Root Theorem"
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3. Prove that <math>\sqrt{2}</math> is irrational, using the Rational Root Theorem. | 3. Prove that <math>\sqrt{2}</math> is irrational, using the Rational Root Theorem. | ||
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===Answers=== | ===Answers=== |
Revision as of 12:46, 25 September 2020
Given a polynomial with integral coefficients,
. The Rational Root Theorem states that if
has a rational root
with
relatively prime positive integers,
is a divisor of
and
is a divisor of
.
As a consequence, every rational root of a monic polynomial with integral coefficients must be integral.
This gives us a relatively quick process to find all "nice" roots of a given polynomial, since given the coefficients we have only a finite number of rational numbers to check.
Contents
Proof
Given is a rational root of a polynomial
, where the
's are integers, we wish to show that
and
. Since
is a root,
Multiplying by
, we have:
Examining this in modulo
, we have
. As
and
are relatively prime,
. With the same logic, but with modulo
, we have
, which completes the proof.
Problems
Easy
1. Factor the polynomial .
Intermediate
2. Find all rational roots of the polynomial .
3. Prove that is irrational, using the Rational Root Theorem.
Answers
1.
2.
3. A polynomial with integer coefficients and has a root as must also have
as a root. The simplest polynomial is
which is
. We see that the only possible rational roots are
and
, and when substituted, none of these roots work.
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