Question

Define \( \oplus \text{ on } \mathbb{R} X \mathbb{R} \) by setting \( (a,b) \oplus (c,d)=(ac-bd, ad+bc) \)
Show that \( ( \mathbb{R} X \mathbb{R} , \oplus ) \) is an algebraic system

Answer 1

ohhhhh The definition is An algebraic system is an ordered pair (AO) , where A is a set, called the underlying set of the algebraic system, and O is a set, called the operator set, of finitary operations on A .

Answer 2

so I guess we gotta prove that its non empty and it has a binary operation?

Answer 3

what do we have here
http://planetmath.org/AlgebraicSystem.html

Answer 4

hmmmmm

Answer 5

The thing is I just checked in my book and there were 2 definitions of algebraic system which is is quite baffling

Answer 6

The first one is similar to the one stated in planet math

Answer 7

the second one imo is really the def of a group which is a kind of algebraic structure

Answer 8

like if its an algebraic system then it has the following 4 properties.
commutativity, associativity, an identity element and each element has an inverse

Answer 9

ohh right and its obviously closed under that binary operation

Answer 10

haha there are way too many diff definitions in my book. i dont get it y they dont state it all at once

Answer 11

This is what someone responded but I dont think its correct
-->The identity is (1, 0), since (a, b)*(1, 0)=(a(1)-b(0), a(0)+b(1))=(a, b)=(1, 0)*(a, b). But now we have a contradiction; there is no inverse for the element (0, 0), since (a, b)*(0, 0) = (0, 0) ≠ (1, 0) So, G does not form a group under *.
Since the question says show then it must be an algebraic structure

Answer 12

is it clear to you what this is?

Answer 13

Oops!! sorry ... i totally missed. how long did OS froze?

Answer 14

you want a hint?

Answer 15

yaaaaaaaa

Answer 16

Like I am not sure exactly what an algebraic structure is

Answer 17

here is my hint:
let \(z_1=a+bi,z_2=c+di\) be two complex numbers. what is \(z_1\times z_2\)?

Answer 18

i just read your post above.
one does not expect to find an inverse for the zero element, but you should be able to find an inverse for anything else

Answer 19

An algebraic system is an ordered pair (A,O) , where A is a set, called the underlying set of the algebraic system, and O is a set, called the operator set, of finitary operations on A .

Answer 20

I bet RxR is A

Answer 21

for O, we have ⊕

Answer 22

right

Answer 23

it isn't difficult to prove that it's closed and commutative.

Answer 24

(1, 0) is the identity element
inverse ... yep here is the trouble (0,0) does not have inverse.

Answer 25

but what was satellite's hint

Answer 26

How does that come into play?

Answer 27

let's try this Q before
is (R, *) a group?

Answer 28

yes

Answer 29

did you multiply \((a+bi)(c+di)\) ?

Answer 30

i guess yes
http://in.answers.yahoo.com/question/index?qid=20091104013411AA3mtNV
from planet math:
"A prototypical example of an algebraic system is a group, which consists of the underlying set G , and a set O consisting of three operators: a constant e called the multiplicative identity, a unary operator called the multiplicative inverse, and a binary operator called the multiplication."

Answer 31

\((\mathbb{R},\times)\) is not a group because zero has no inverse. you do not expect 0 (the identity under addition) to have a multiplicative inverse

Answer 32

So then what abt this example?

Answer 33

so what do we say here? on this particular example ... i'm not sure either ...algebra has never been my thing.

Answer 34

the zero element here, namely \((0,0)\) has no inverse, you would not expect it to
my hint was to get you to see that this is the beginning of the construction of complex numbers from the real numbers

Answer 35

perhaps not
http://en.wikipedia.org/wiki/Group_(mathematics)#Definition
says "each element must have multiplicative inverse"

Answer 36

you start by defining a set \(\mathbb{R}\times \mathbb{R}\)
@experimentX that is for a group, this is not a group

Answer 37

why ... does algebraic system require further criteria?

Answer 38

Like what is an algebraic structure
what are the properties of an algebraic structure?

Answer 39

you define addition for \(\mathbb{R}\times \mathbb{R}\) to be \((a,b)+(c,d)=(a+c,b+d)\) and multiplication to be defined as above namely
\((a,b), (c,d)=(ac-bc,ad+bc)\)
then you show \((0,1)\times (0,1)=(-1,0)\)
you can then show that this is a field, and finally you define \((0,1)=i\) and rewrite the ordered pair \((a,b)\) as \(a+bi\)

Answer 40

@swissgirl work directly from the definition, whatever definition you have
i think there is not really much to show
you are not expected to show that this is a group (because it isn't) only that you have a set and a binary operation

Answer 41

ok here is what you wrote as the definition
The definition is An algebraic system is an ordered pair (AO) , where A is a set, called the underlying set of the algebraic system, and O is a set, called the operator set, of finitary operations on A
so in this example
\[A=\mathbb{R}\times \mathbb{R}\] and \(O=\oplus\)

Answer 42

since \(\mathbb{R}\times \mathbb{R}\) is a set, and since \(\oplus\) is a binary operation (takes two elements of the set and send it to one element in the set) then by definition this is an algebraic structure
there is really no more to be done

Answer 43

but what were u showing earlier

Answer 44

but what's the difference between algebraic structure and algebraic system?

Answer 45

I think its the same thing

Answer 46

I googled and got this pdf with Galois stuff ...

Answer 47

me? i was showing that this is how on constructs complex numbers out of real numbers, by starting with ordered pairs
but that is not what the question asks, so you can ignore that part