Equating rational and irrational parts: proof and worked problems

The basic fact

Let $a,b$ be rational numbers and let $\alpha$ be irrational. Then

\boxed{a+b\alpha=0\quad\Longrightarrow\quad a=b=0.}

Equivalently, if $a+b\alpha=c+d\alpha$ with all four coefficients rational, then $a=c$ and $b=d$ . In linear-algebra language, $1$ and $\alpha$ are linearly independent over the rational numbers.

Proof

If $b\ne0$ , rearrangement gives $\alpha=-a/b$ , a rational number. This contradicts the irrationality of $\alpha$ . Hence $b=0$ , and the original equation then gives $a=0$ .

The rationality assumptions matter. With arbitrary real coefficients the conclusion fails: $-\sqrt2+1\cdot\sqrt2=0$ , for example, does not have both coefficients zero.

How to use it in a written solution

State which expressions are rational and which number is irrational. For instance, suppose rational $x,y$ satisfy

(x+y+1)+(3x+2y)\sqrt2=0.

Both $x+y+1$ and $3x+2y$ are rational, while $\sqrt2$ is irrational. Therefore

x+y+1=0,\qquad3x+2y=0.

Solving gives

\boxed{(x,y)=(2,-3).}

Substitution verifies that the pair satisfies the original equation.

Problem 1: one equation determining two rational unknowns

Problem. Find rational $x,y$ such that

(1+\sqrt2)x+(1-\sqrt2)y=3+\sqrt2.

Collect the rational and irrational parts:

(x+y-3)+(x-y-1)\sqrt2=0.

Since both coefficients are rational,

x+y=3,\qquad x-y=1.

Consequently

\boxed{(x,y)=(2,1).}

There is no contradiction in obtaining two conditions from one equation: requiring the unknowns to be rational is an additional restriction.

Problem 2: rule out rational roots without solving explicitly

Problem. Show that

x^2-(1+\sqrt3)x+2\sqrt3=0

has no rational root.

Suppose a rational number $r$ were a root. Rearrange its equation as

(r^2-r)+(2-r)\sqrt3=0.

The two coefficients are rational and $\sqrt3$ is irrational, so

r^2-r=0,\qquad2-r=0.

The second equation requires $r=2$ , which does not satisfy the first. Hence there is no rational root.

In this particular example, the discriminant is $4-6\sqrt3<0$ , so the equation in fact has no real roots at all. The argument above still illustrates a useful method for equations with irrational coefficients; it does not depend on simplifying a quadratic-formula radical.

For polynomials with rational coefficients, the rational root theorem provides a different starting point.

Problem 3: why cube root of two cannot satisfy a rational quadratic

Let $\alpha=\sqrt[3]2$ .

Prove that $\alpha$ is irrational.
Prove that $\alpha$ is not a root of any quadratic polynomial with rational coefficients and a nonzero leading coefficient.

Part 1: irrationality

Suppose $\alpha=n/m$ with positive coprime integers $m,n$ . Cubing gives

n^3=2m^3.

Thus $n$ is even. Write $n=2k$ . Then $m^3=4k^3$ , so $m$ is even as well. This contradicts the assumption that $m,n$ are coprime. Therefore $\sqrt[3]2$ is irrational.

Part 2: reduce a hypothetical quadratic to a linear relation

Assume for contradiction that rational $a,b,c$ , with $a\ne0$ , satisfy

a\alpha^2+b\alpha+c=0.

We also know $\alpha^3-2=0$ . Polynomial division gives the identity

\alpha^3-2=(a\alpha^2+b\alpha+c)\left(\frac\alpha a-\frac b{a^2}\right)+\left(\frac{b^2}{a^2}-\frac ca\right)\alpha+\frac{bc}{a^2}-2.

The first term on the right is zero. Therefore

\left(\frac{b^2}{a^2}-\frac ca\right)\alpha+\left(\frac{bc}{a^2}-2\right)=0.

Both coefficients are rational, and $\alpha$ is irrational by part 1. The basic fact proved above yields

c=\frac{b^2}a,\qquad\frac{bc}{a^2}=2.

Substitute the first equation into the second:

\left(\frac ba\right)^3=2.

Since $b/a$ is real, this implies $b/a=\sqrt[3]2$ . The left side is rational and the right side irrational, a contradiction. No such quadratic exists.

Why not simply divide by the coefficient of alpha?

The coefficient $b^2/a^2-c/a$ might be zero. Dividing by it without a case distinction would leave a gap in the proof. Equating the rational and irrational parts handles the zero-coefficient case directly and safely.

Nor may we automatically equate the coefficients of $\alpha$ and $\alpha^2$ at the start merely because both are irrational. Two irrational numbers need not be linearly independent over the rationals. The reduction above establishes the stronger claim that is actually needed here.

What to remember

Use the method only after checking rationality of the coefficients. For more complicated powers of an irrational number, use its defining polynomial to reduce the degree. Avoid dividing by an expression before establishing that it is nonzero.