Confusion about the conjugate (square roots)

Question

In this Wikipedia article it says: "In mathematics, the conjugate of an expression of the form $a+b\sqrt{d}$ is $a-b\sqrt{d}$, provided that $\sqrt{d}$ does not appear in $a$ and $b$."

Why can't $\sqrt{d}$ appear in $a$ and $b$? In my textbook it is defied like Wikipedia did it except that it didn't say that $\sqrt{d}$ can't appear in $a$ and $b$.

Answer 1

In this case Wikipedia's explanation is quite poor.

Let's take a step back. Consider a polynomial $$ P(X)=X^d+a_1X^{d-1}+a_2X^{d-2}+\cdots+a_d $$ and assume:

• the coefficients $a_i$ are rational numbers;
• the polynomial is irreducible, meaning that it admits no decomposition $P(X)=A(X)B(X)$ where $A(X)$ and $B(X)$ are both non constant.

Then it is known (essentially by the Fundamental Theorem of Algebra) that $P(X)$ admits $d$ different roots $\alpha_1,...,\alpha_d$ in the complex numbers, none of which is a rational number. A root of a polynomial $P(X)$ is some number $\alpha$ such that $P(\alpha)=0$.

We say that two complex numbers $\alpha$ and $\beta$ are conjugated if they both appear among the roots of a polynomial as above.

Back to square roots.

When the degree of the polynomial is $2$, i.e. $$ P(X)=X^2+aX+b, $$ the roots are $\alpha_{1,2}=\frac12\left(-a\pm\sqrt{a^2-4b}\right)$ so that in fact one root is transformed into the other (conjugation) by switching sign to the square root.

The "radical not appearing in $a$" clause should be read "$a$ and $b$ are rational numbers".

A more general situation is that when the polynomia $P(X)$ is allowed to have coefficients in just any field, but let's keep it simple for the time being.

Answer 2

In front of a radical sign we have an implied $\pm$ sign.

If a radical appears in $a$ or $b$ then it would make for four instead of two results and there would be ambiguity.

They would arise as roots from a fourth order and not a quadratic equation with real coefficients.

It is better to say, imho, that :

"$a-b\sqrt{d}$, provided that the radical does not appear in $a$ and $b$ real numbers"