Proving the Field Property of a Group of Numbers

[transgalactic]

i got this group of numbers
http://img205.imageshack.us/img205/5291/20039746rf4.gif

prove that its a field

??

 

[HallsofIvy]

A field must be closed under addition and multiplication and every non-zero number must have a multiplicative inverse.

Is $(a+ b\sqrt{5})+ (c+ d\sqrt{5})$ in that same set?
Is $(a+ b\sqrt{5})(c+ d\sqrt{5})$ in that same set?

Is $$\frac{1}{a+b\sqrt{5}}$$ in that same set? (Hint: rationalize the denominator.)

 

[transgalactic]

the problem is that i remember these proofs as
Q(x)+Q(y)=Q(x+y)

Q(x)*Q(y)=Q(x*y)
but here the input the root of 5 instead of sum variable

??

i don't know how to construct these equations because there is no variable input?

 

[Mark44]

Halls has shown you what you need to do. It seems pretty clear to me.

 

[transgalactic]

i clearly remember building these equations:
Q(x)+Q(y)=Q(x+y)

Q(x)*Q(y)=Q(x*y)

but in my function there is no variable in the input

??

 

[sutupidmath]

the problem is that i remember these proofs as
Q(x)+Q(y)=Q(x+y)

Q(x)*Q(y)=Q(x*y)
but here the input the root of 5 instead of sum variable

??

What is a field, you probbably already know. However, In order to show that it is a ring, alll we need to show is that it is closed under subtraction, and under multiplication. Like Halls already told you.

What you are doing there, is basically a hommomorphism, that is the perservation of the operation, but we are not dealing with mapps between two fields in here.

So let x, and y be any two elements on that set, then obviously they will have the form

$x=(a+ b\sqrt{5}); y= (c+ d\sqrt{5})$

so

$x-y=(a+ b\sqrt{5})- (c+ d\sqrt{5})=(a-c)+(b-d)\sqrt{5}$ so $$x-y \in Q(\sqrt{5})$$

now also

[tex]xy=(a+ b\sqrt{5})(c+ d\sqrt{5})=ac+ad\sqrt{5}+bc\sqrt{5}+bd5=(ac+bd5)+(ad+bc)\sqrt{5}[/itex] which again, for obvious reasons is in $$Q(\sqrt{5})$$

Now,in order for that to be a field, like Halls already stated,we need to show that every el. in that set has an inverse.

let x be in that set. so it will obviously have the form $$=x(a+ b\sqrt{5})$$ let
s suppose that there is another el, y, such that

$$xy=1 => y=\frac{1}{x}=\frac{1}{a+b\sqrt{5}}=\frac{a-b\sqrt{5}}{a^2-5b^2}$$ SO, the same question still holds, is this in the same set?

I'll give u one more hint: rewrite the last expression as

$$\frac{a}{a^2-5b^2}+\frac{-b}{a^2-5b^2}\sqrt{5}$$

 

[sutupidmath]

??

i don't know how to construct these equations because there is no variable input?

you don't need to construct these, because you are not dealing with a mapping between two rings or fields, and thus you are not asked to show whether that kind of mapping is a homomorphism. because, those expressions are used when we want to show that a particular mapping say between two rings is a homomorphism.

 

[sutupidmath]

A field must be closed under addition

you meant under subtraction here, right?

 

[transgalactic]

yes if we split it:

a/(a^2-5b^2) -(b/a^2-5b^2)*5^(1/2)

is that correct?

 

[sutupidmath]

yes if we split it:

a/(a^2-5b^2) -(b/a^2-5b^2)*5^(1/2)

is that correct?

Well, now the problem breaks down into showing whether
$$\frac{a}{a^2-5b^2}\in Q; and\frac{-b}{a^2-5b^2}\in Q ?$$

You can either prove this one too, or take it for granted. since a, b are rational nr. then by adding, subtracting, multiplying, and dividing them, we will still end up having rational numbers.

 

[transgalactic]

hallsofIvy said to prove that that inverse of the formula is of the same set
and
a/(a^2-5b^2) -(b/a^2-5b^2)*5^(1/2) => a/(a^2-5b^2) +(-b/a^2+5b^2)*5^(1/2)

its of the same set now(it looks like the original formula)?

 

[transgalactic]

how to prove that 0 or 1 are members
??

 

[HallsofIvy]

If you want to write 0 or 1 as "$a+ b\sqrt{5}$", what would a and b be equal to?

 

[sutupidmath]

how to prove that 0 or 1 are members
??

$$0=0+0\sqrt{5}, 1=1+0\sqrt{5}$$ fair enough!

P.S this is what Halls already said!...lol...

 

[transgalactic]

how to recognize a ring proof
and field prove
i can't see the difference between them?

 

[NoMoreExams]

Well fields are a generalization of rings ...http://en.wikipedia.org/wiki/Field_(mathematics)#Rings_vs._fields so the proofs might go a similar route

 

[transgalactic]

what is the form of a field?

what is the form of a ring?

(how each of their formula looks?)

 

[HallsofIvy]

how to recognize a ring proof
and field prove
i can't see the difference between them?

What are the definitions of "field" and "ring"?

 

[transgalactic]

ring is like x=(a+b,c,g+1)
and i need to
prove that it stays the same type by addition
and multiplication by scalar
and existence of zero.

field is like a formula
f(7)=a+7*b

is this correct

 

[NoMoreExams]

ring is like x=(a+b,c,g+1)
and i need to
prove that it stays the same type by addition
and multiplication by scalar
and existence of zero.

field is like a formula
f(7)=a+7*b

is this correct

Read the wiki article I linked, what you wrote is not very mathematical in nature i.e. a field is definitely not "like a formula f(7) = a + 7*b". These are pretty advanced mathematical ideas and you should have at the very least the definition of each ingrained in your brain before you go out and try to prove things about them.

 

[HallsofIvy]

ring is like x=(a+b,c,g+1)
and i need to
prove that it stays the same type by addition
and multiplication by scalar
and existence of zero.

field is like a formula
f(7)=a+7*b

is this correct

"ring is like" is not a DEFINITION. It is very important to learn the precise words of mathematical definitions. Definitions in mathematics are "working" definitions- you use the precise words of the definitions in proofs and problems.

A "ring" is a set of objects together with two binary operations, + and *, such that:
1) The set with the operation + is an abelian group (so you need to know the definition of that!).
2) The associative law is true for *, a*(b*c)= (a*b)*c and the distributive law a*(b+c)= a*b+ a*c is true.

A "field" is a set of objects together with two binary operations, + and *, such that:
1) The set with the operation + is an abelian group.
2) The associative law is true for * and the distributive law is true.
3) There exist an "identity" for *: a*1= 1*a= a for all a in the set.
4) Every element of the set except the additive identity has a multiplicative inverse.


(3) and (4) are, of course, the difference between a ring and a field. The set of all two by two matrices over the real numbers is an example of a ring (with identity) that is not a field. Some non-zero matrices do not have inverses.

 

[transgalactic]

regarding what you said about the field:
i remember you said that in order to prove a field
i need to prove that the sum and multiplication is of the same
structure.
and that the inverse is of the same structure
and that 0 exist

is that the tests i need to perform
?

 

[NoMoreExams]

Halls just outlined what you need to prove. Aren't you missing a few points?

 

[sutupidmath]

"ring is like" is not a DEFINITION. It is very important to learn the precise words of mathematical definitions. Definitions in mathematics are "working" definitions- you use the precise words of the definitions in proofs and problems.

A "ring" is a set of objects together with two binary operations, + and *, such that:
1) The set with the operation + is an abelian group (so you need to know the definition of that!).
2) The associative law is true for *, a*(b*c)= (a*b)*c and the distributive law a*(b+c)= a*b+ a*c is true.

Should we also explicitly mention that it is closed under the second operation * ? i.e fo every two el. a,b from the ring R, a*b is also in the ring.

This is somehow clear since the associative law is true, but the book from which i learned these things stated it explicitly as well. So, i think it is important to state it, especially since when you want to prove that something is a subring of a ring R, then all u need to prove is that it is closed under subtraction, and multiplication.

 

[sutupidmath]

regarding what you said about the field:
i remember you said that in order to prove a field
i need to prove that the sum and multiplication is of the same
structure.
and that the inverse is of the same structure
and that 0 exist

is that the tests i need to perform
?

I know for which post you are talking about, but this isn't exactly the same situation, since in that case we only needed to prove that Q is a subring. So, we already knew that it is a set of a ring. I really think you should go back and read these things, because if you are having trouble with the definition of things, i cannot immagine how you can even think about starting to prove stuff.