Some roots are irrational. This page proves that is irrational, shows how to tell that a polynomial has an irrational root, and builds polynomials with rational coefficients from irrational roots by pairing each root with its conjugate.
Problem 10.13.1. Prove that is irrational.
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Let be coprime such that , that is to say, can be represented rationally. Thus
so is even by definition. But if ,, then
ah! This cannot be, since we assumed were coprime! Thus we have a contradiction, and there is no rational representation of .
Problem 10.13.2. Prove that if , then (given that ).
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Let and , with and . Then
and since , their products are in , so .
Remark. Similar to complex conjugate notation (), we have radical conjugate notation: . (This is non-standard, however.)
Problem 10.13.3. Let . Show that there are no rational roots and that there exists at least one irrational root.
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By the Rational Root Theorem, if rational roots exist then .
and , so there exists a real root there. But as we just showed, this root cannot be rational.
Problem 10.13.4. Find a polynomial with rational coefficients that has and as roots.
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Clearly such a polynomial must be of degree at least :
The first pair gives
and the second pair gives . So
so .
Problem 10.13.5. Prove there exists such that for all and there is a root with .
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First try, with the quadratic formula:
There is a factor of to divide out, which leaves something of the form . Matching, , and (that is, ). This cannot be, since we need for all .
Instead, multiply out conjugates:
so
Its factors are , so .
Problem 10.13.6. Let . Find all roots and factor.
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By the Rational Root Theorem,
Since , we need , so . So the root cannot be in , and we now have
. An educated guess, since , is :
so there exists a root between and . Next, , so . seems possible:
so , but it must be irrational.
Let’s try , tracking signs in synthetic division:
So .
Now let . Then
(By the quadratic formula, , with .)
So
From ,; from ,.
Exercises
Exercise 10.13.7. Let .
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Note and , so there is a root between and . But , thus the root is irrational.
Exercise 10.13.8. is a root of .
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So try : ✗. Also and . So there is a root and another between and .
For : let , so , which gives
So those are the two roots for :
and the first factor gives
(Side scratch: the quadratic formula ;,.)