multiply (2√5-√2)(√5+2√8) and then find its conjuate pair
Answers
Answer:
Conjugation is an automorphism of the field C . Don’t worry: it’s not a difficult concept.
A field is any set equipped with operations of addition and multiplication where the usual rules of algebra apply, including that any nonzero element has an inverse.
If F is a field, than an automorphism of F is a bijective map α:F→F such that
α(a+b)=α(a)+α(b),α(ab)=α(a)α(b)
for every a,b∈F .
The rational numbers form a field, which has no automorphism beside the identity. The same holds for the real numbers: the reason is that
a real number is positive if and only if it is a nonzero square; an automorphism maps squares into squares (and conversely, because its inverse is an automorphism as well);
any rational numbers must be mapped to itself;
an automorphism of the real numbers preserves inequalities (consequence of 1);
any real number is the supremum of an increasing sequence of rational numbers.
Each of the facts above admits a proof that’s not difficult, but it would take us far from our objective.
To the contrary, the complex numbers have lots of automorphisms. However, there is just one that maps the real numbers to themselves and it is the usual conjugation. There’s a “philosophical” reason why the complex numbers have such an automorphism: in a “platonic world” there is a unique model of the complex numbers and we grasp it by choosing one of the square roots of −1 . What would happen if we chose the “wrong one”? (In quotes, because there is no right or wrong one). Nothing at all would happen: we’d get a field that’s isomorphic to the other one. We cannot distinguish them through pure algebra. Hence the conjugation must be an automorphism.
There are many other field inside the complex numbers. If you do the same construction of the complex numbers starting instead with Q instead of R and add to it 5–√ , you get a field, usually denoted by Q(5–√) . Its elements are of the form a+b5–√ , with a,b∈Q .
But what would happen if we chose “the other square root of 5 ”? Nothing at all, we’d get the same field: no platonic world is necessary, because we can do it inside (a fixed model of) the complex numbers. So the map that fixes every rational number and sends 5–√ to −5–√ must be an automorphism of Q(5–√) . And it is the unique nonidentity automorphism.
So we call a−b5–√ the conjugate of a+b5–√ when talking about Q(5–√) .
The properties of this conjugation are very similar to the complex number conjugation: if we denote it by α , we see that when r=a+b5–√∈Q(5–√) , then
a=r+α(r)2,b=r−α(r)25–√
Do you see? In the complex numbers, if z=x+yi , then
x=z+z¯2,y=z−z¯2i
It’s just the same! But with α and 5–√ instead of i in the denominator.
The concept of field automorphism is a central one in modern Galois theory. Galois didn’t have available the machinery of abstract algebra (fields and extensions thereof), but his work has been reinterpreted in terms of automorphisms and has become a much more elegant and powerful theory.
Step-by-step explanation:
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