# When solving a linear system for x and y, am i in a group? ring? field?

1. May 3, 2012

### ksmith630

Hi everyone, I'm currently taking an abstract Algebra course and need a little guidance with an analysis of solving a system of linear equations.

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 in a system? Thanks in advance for any insight!

2. May 3, 2012

### micromass

When solving a linear system of equations, you usually want to live in a vector space or a module over a suitable field or ring.

You can also work in an arbitrary ring or field, but that is a special occasion as working in a vector space or module.

3. May 3, 2012

### HallsofIvy

One way of solving that particular system would be to add the two equations to get
(x+ y)+ (x- y)= 2x= 5+ 1= 6 and then divide by 2. In order to add the two equations, you have to be working in at least a group but in order to divide by 2 you must we working in a ring or vector space.

4. May 3, 2012

### ksmith630

Thanks for the responses! So i guess i'd have to say i'm solving the system in a ring?
This is what i have so far (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

(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!

5. May 3, 2012

### micromass

Why does 1/2 exist in your ring???

6. May 3, 2012

### ksmith630

I'm guessing it shouldn't; is this an indicator that this isn't in a ring? What should i be looking for to determine this? Should indicate reals rather than integers? Maybe integral domain instead?

7. May 3, 2012

### DonAntonio

Well, $\,\,\mathbb Z\,\,$ is an integral domain and yet 1/2 does't exist in it. Whenever you want to make some operation

in a ring you must make sure it is a valid one there. For example, you can use 1/2 in the ring $\,\,\mathbb Z/15\mathbb Z\,\,$ ...

DonAntonio

8. May 3, 2012

### ksmith630

So am i in a ring of reals?

9. May 4, 2012

### HallsofIvy

The real numbers is a field, not a ring. (Essentially, a field is a ring in which every non-zero member has a multiplicative inverse.)

10. May 4, 2012

### DonAntonio

You tell us! Unless you're completely free to choose, the problem or the teacher imposes the conditions. You have to suit yourself

to those conditions.

BTW, the set of real numbers with the usual operations is not only " a ring ", but in fact a field. I think you need to read a little more

about general (college and up) algebra and/or on algebraic structures as it iss clear you are not very sure what's going on here.

DonAntonio

11. May 4, 2012

### micromass

So, can you tell us which properties you need for solving equations?? First tell us the properties needed and then you can find out which structure better suits your needs.

12. May 4, 2012

### ksmith630

Actually we weren't given any instructions on whether to use properties from groups, rings, fields or integral domains. We must pick the structure that best represents the method of elimination and again for the method of substitution. Did I solve the systeem correctly above for each of these methods? If so, then the structure must maintain addition and multiplication/division in the real numbers. So I'm assuming it cannot be in a group, and since it has reals it must be a field. If this is wrong then what other indicators should I be looking for? There is absolutely no mention of systems of linear equations in any of the chapters on groups, rings, fields, integral domains of our text- so I have no idea.

13. May 4, 2012

### micromass

Right, so you can solve linear systems in a field. So, a field is certainly a correct answer.
But can you go more general?? Vector spaces also work...

14. May 4, 2012

### DonAntonio

....

15. May 4, 2012

### ksmith630

Well I know I can't do it in a group since I need two operations, and I know fractions don't exist in a ring, so what else can I pick from? I want to use the properties of real numbers in addition and mult, but to cite properties I need the algebraic structure first.

16. May 4, 2012

### DonAntonio

....

17. May 4, 2012

### micromass

I don't see anybody claiming that your coefficients and your x,y,z need to be in the same space. I think it's an unnecessary assumption that they do.

Vector spaces are precisely there for being able to solve linear systems of equations. I can solve any linear system of equation while working in a vector space. Furthermore, even if the coefficients and the x,y,z are in the same space, then I'm still working in a vector space, a one-dimensional one.

18. May 4, 2012

### ksmith630

huh??

19. May 4, 2012

### DonAntonio

If you don't quote it is impossible to know whom you're addressing.

DonAntonio

20. May 4, 2012

### DonAntonio

From the example the OP presented I think it was clear that he/she was struggling with the idea that of what alg. structure her/his equations'

coefficients were to be taken from, and thus what operations and elements could be used. This isn't answered by working with vector spaces

as one can't multiply vectors in general vector spaces.

Vectors spaces provide, among other things, an adequate alg. frame in which we can work with linear equations, but the basic operations with vectors

usually have a field (or division ring) as base alg. structure from where the operations are taken. This is what I meant.

DonAntonio