Page 1 of 2 12 LastLast
Results 1 to 10 of 11

Thread: Solving Linear Equations: group? ring? field? in abstract algebra

  1. #1

    Solving Linear Equations: group? ring? field? in abstract algebra

    Hi everyone, I'm currently taking an abstract Algebra course and need a little guidance with a problem.

    We are given two linear equations and need to solve for x and y using the method of "substitution" and again using "elimination". However, we must provide theorems and properties explaining each step. I have no problem solving the system, i'm just having difficulty citing the properties used.

    Here is the system, which yields one solution (3, 2).

    x + y = 5
    x - y = 1

    We first need to determine the appropriate algebraic structure in which we are solving the system: either a group, ring, field, or integral domain.
    We were told that for solving something like x + 2 = 5 (with one variable) we would be in a "group" <Z,+> where + is the normal binary op of addition in Z.

    Our text defines a "group" as having only one binary operation, while defining a "ring" as having two binary operations (addition and multiplication). We are then given the definitions of "field" and "integral domain" which seem to be special cases of a ring... So when solving the system x + y = 5 and x - y = 1 for x and y, would i then be in a "ring" since we have mult and addition? Would i be in the ring <Z,+,x> with the usual ops of addition and mult in Z? Or would i be in the reals? Or am i not even in a ring but rather a field or integral domain?

    Next, we need to determine which properties and theorems i'll be using to solve the system. But since i can't determine whether we're in a group, ring, field or integral domain, i don't know which properties to cite.

    We were told that when asked to solve for x in something like x + 2 = 5 we would use the theorem: For elements a and b in a group <G,*> if we are given that a=b, then for any c in G that a*c=b*c.
    which when applied to solving for x, would let us do (x+2)+(-2)=5+(-2) which results in x+(2+(-2))=5+(-2) since addition is associative in G, which gives us x+0=3 since the additive ID in the group Z is 0, which gives us the answer x=3 using the definition of additive inverse and the definition of + in Z.

    But how do i write something like this for solving the system of two linear equations with two variables x and y: x + y = 5 and x - y = 1?

    I want to use the addition property of equality: (If a=b and c=d then a+c=b+d) but i'm not sure if this is of a group, ring, field, etc...

    Can anyone tell me if i'm going in the right direction? Am i OK by saying i'd be in the ring <Z,+,x> with the usual ops of addition and mult in Z when solving for x and y here?
    Thanks for the help!

  2. #2
    Here's my work in actually solving the system (note that i have to do this for both methods below).

    For the "elimination" method:

    x+y=5
    x-y=1

    1(1x+1y)=(5)1 Multiply the top equation (both sides) by 1
    -1(1x11y)=(1)(-1) Multiply the bottom equation (both sides) by -1

    note 1+(-1)=0

    add eqns together

    (1x-1x)+(1y+1y)=5-1
    (101)x+(1+1)y=5-1
    1+(-1)x+(1+1)y=5-1
    0+(1+1)y=5-1
    2y=4

    divide both sides by 2...
    2y/2=4/2
    y=2

    plug into eqn x+y=5
    1x+1(2)=5
    1x=5-2 subtract 2 from both sides
    1x=3
    (1/1)(1)x=(3)(1/1) mult both sides by (1/1)
    x=3

    So solution is (3,2)

    ********************************************

    By the "substitution" method:

    x+y=5
    x-y=1

    solve for y in eq

    1y=5-1x Subtract 1x from both sides
    y=(5-1x) Divide both sides by 1
    y=5-1x

    substitute

    1x+(-1)(5-1x)=1
    1x-1(5)-1(-1)x=1 Distribute -1 to 5-1x
    1x-5+1x=1

    1x-5+1x+5=1+5 Add 5 to both sides
    1x+1x=6
    2x=6

    (1/2)(2/1)x=(6/1)(1/2) Mult both sides by (1/2)
    x=3

    Substitute into eq

    1(3)-1y=1
    3-1y=1
    -1y=1-3 Subtract both sides by 3
    -1y=-2
    (1/(-1))(-1)y=(-2/1)(1/(-1)) Mult both sides by 1/(-1)
    y=-2/(-1)
    y=2

    So solution is yet again (3,2)

    **********************************************

    Part of what we need to do is indicate which algebraic structure we're solving it in. Is it safe to write: "When solving the system, I will work in the ring <Z,+,x> with the standard definitions of mult and addition in Z." ?

    I then need to list any theorems or properties i used to solve the system. Can i then cite properties of groups, like: "For elements a and b in a group <G,*> if we are given that a=b, then for any c in G that a*c=b*c"? Or do i need to replace <G,*> with something else?

    I have something called the "property of equality": "If a=b and c=d then a+c=b+d)" but do i write that this is for elements a, b, c, d in a "Group" or in a "ring"?

    Thanks again for the help everyone!

  3. #3
    Full Member
    Join Date
    Aug 2011
    Posts
    933
    In any field, you may do elimination like you do in a linear algebra course. Just make sure not to multiply or divide by the field's characteristic (in the case of a finite field).

    In rings you cannot assume the existence of multiplicative inverses. You can still do gaussian elimination to an extent. Think of what happens in Z/4Z though. The system

    2x+y=0
    x+2y=0

    Has multiple solutions: (1,2), (2,1), (0,0)

    About your other question: It is standard to define a (unitary) ring as an abelian group under an operation +, whose elements also form a monoid under an operation * (and such that * interacts with + as we would expect).

    So, to me, simply writing '+' indicates you are talking about the commutative group in which the ring sits. The group elements are the ring elements.
    Last edited by daon2; 05-03-2012 at 07:17 PM.

  4. #4
    Did i solve the system correctly above for both the "elimination" and "substitution" methods? If so, this means i need to use the fractions, right? in this case would i then be working in a group with real numbers vs integers? Or a field? ring?

  5. #5
    Full Member
    Join Date
    Aug 2011
    Posts
    933
    If you are working solely over the integers, you may "pretend" you are solving the system over the field of real numbers (your ring is a subset of this field). If your solution consists of integers, then your answer should be fine. The solution set inherits all the same properties from the reals (or complex numbers) in there being either a unique solution, infinitely many, or none at all.

  6. #6
    Quote Originally Posted by daon2 View Post
    If you are working solely over the integers, you may "pretend" you are solving the system over the field of real numbers (your ring is a subset of this field). If your solution consists of integers, then your answer should be fine. The solution set inherits all the same properties from the reals (or complex numbers) in there being either a unique solution, infinitely many, or none at all.
    Well both equations have integer coeficients, and the solutions are also integers. But when solving the system, I have to add integers and mult/divide integers and fractions. So I guess I have to be working in the field of real numbers?

    For both the elimination and substitution methods, I have to begin by writing:

    To solve this system, I will be working in the (group, ring, field, integral domain) represented by <?,?,?...?> with the usual operations of (?) and (?) in the (reals, integers. Etc).

    Does this make any sense?

  7. #7
    What about the field of rationals? Why isn't this OK?

  8. #8
    Full Member
    Join Date
    Aug 2011
    Posts
    933
    Quote Originally Posted by ksmith630 View Post
    Well both equations have integer coeficients, and the solutions are also integers. But when solving the system, I have to add integers and mult/divide integers and fractions. So I guess I have to be working in the field of real numbers?

    For both the elimination and substitution methods, I have to begin by writing:

    To solve this system, I will be working in the (group, ring, field, integral domain) represented by <?,?,?...?> with the usual operations of (?) and (?) in the (reals, integers. Etc).

    Does this make any sense?
    You actually do not need to do any division if the solutions are integral. Sometimes it might be faster.

    x + y = 5
    x - y = 1

    Add the two equations together. You'll get:


    2x = 6 = 2*3.


    Since the integers form a domain, you know that for a nonzero, a*b = a*c => b=c.


    So 2x = 2*3 => x=3.

  9. #9
    Quote Originally Posted by daon2 View Post
    You actually do not need to do any division if the solutions are integral. Sometimes it might be faster.

    x + y = 5
    x - y = 1

    Add the two equations together. You'll get:


    2x = 6 = 2*3.


    Since the integers form a domain, you know that for a nonzero, a*b = a*c => b=c.


    So 2x = 2*3 => x=3.
    That makes perfect sense to me but i think i need to do it specifically for the "elimination" and "substitution" methods of solving linear eqns. Thus, i'd need to have rationals, right? Did i solve the systems correctly above for each method?

  10. #10
    Full Member
    Join Date
    Aug 2011
    Posts
    933
    Quote Originally Posted by ksmith630 View Post
    That makes perfect sense to me but i think i need to do it specifically for the "elimination" and "substitution" methods of solving linear eqns. Thus, i'd need to have rationals, right? Did i solve the systems correctly above for each method?
    I think you provided too much detail, if there is such a thing. But that is for your professor to decide, not me. As I said, you don't need rational numbers. But I'm not certain what your definition of "elimination method" is. I would say the above is also the elimination method. The "RREF" form cannot be attained in the same way through standard elementary row operations over (only) the ring of integers.

    2x=6 implies x=3 whether or not you are allowed to use rational scalars in a row replacement.

    Edit: For one thing, multiplying by "1/1" is silly. x=1x, it is the definition of 1.
    Last edited by daon2; 05-05-2012 at 01:15 AM.

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •