Subject:MathPrice: Bought3

See below. (It is not a simple answer.) (I welcome contributions from others to correct or expand any points I make below.)

There are several appropriate answers depending on the background of the person asking. The answer is different for binomials in which one or both terms involve square roots as opposed to general elements of an abstract field extension. (Which I could not explain without reviewing field theory.)

For students working with the integers, rationals, reals and complex numbers here is the idea.

Suppose that we are working in the integers or the rational numbers and we add to the set of numbers a zero of the polynomial ##x^2-2## (a solution to the equation ##x^2-2=0##).

We say we have adjoined a zero of the polynomial. In this example, we'll denote it ##sqrt2##.

We want to keep all of the standard properties of the base set, so we also include ##-sqrt2##. If we are working in the rational numbers, we also need the reciprocal, ##1/sqrt2## and it's opposite ##-1/sqrt2##.

Finally we need to have every number that can be made by adding, subtracting and multiplying (in the rationals we also need dividing) members of the base set and the new element ##sqrt2##

Expression involving ##asqrt2 + bsqrt2## (a, and b are in the base set -- integers or rational numbers) can be written as ##(a+b)sqrt2## where ##a+b## is also in the base set.

But expressions like ##3+5sqrt2## cannot be written as a member of the base set alone, nor as a member of the base set times ##sqrt2##

Every member of the extended set can be written as ##a+bsqrt2## where ##a## and ##b## are in the base set.

The conjugate of ##a+bsqrt2## is the member of the extended set we multiply by, so that the product is in the base set.

Since ##(3+5sqrt2)(3-5sqrt2) = 9-50 = -41##, we say that

in this system, ##(3+5sqrt2)## and ##(3-5sqrt2)## are conjugates of each other.

Simlarly if ##sqrtc## is not an integer or a rational number, then

##a+bsqrtc## and ##a-bsqrtc## are conjugates of each other.

If we adjoin two square roots, the general situation is more complicated. Adjoining ##sqrt3## and ##sqrt5## leads to a set in which every element can be written

##a+bsqrt3+csqrt5+dsqrt15##.

These things also have conjugates. That is: another expression of the same form that, when we multiply gets us a memebr of the original small set (integer or rational.)

The most common kind of expression to see in early classes is ##bsqrt3+csqrt5## which has ##bsqrt3-csqrt5## (and also ##-bsqrt3+csqrt5##) as conjugates.

By extension of terminology, it is common to say that the **algebraic** expressions ##2+7sqrtx## and ##2-7sqrtx## are conjugates. The product of these expression does not involve a square root of the variable ##x##. (the product of ##2+7sqrtx## and ##-2+7sqrtx## also has no ##sqrtx##)

**Complex Conjugates**

When we adjoin a zero of ##x^2+1## we cannot keep the order properties of the original numbers system, but we can the other important algebraic properties.

In this case every member of the extended set can be written ##a+bsqrt(-1)## where ##a## and ##b## are real numbers.

The product of ##a+bsqrt(-1)## and ##a-bsqrt(-1)## is simply a real number (it does not involve ##sqrt-1##) so these are conjugates of each other.