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PostPosted: Fri, 4 May 2012 19:39:20 UTC 
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Hi everyone, i'm hoping to get a little confirmation as to whether or not i'm on the right track with something.

I've been given two linear equations and need to solve for x and y using the method of "substitution" and again using "elimination".
However, i must also:

1. State the algebraic structure in which to solve the system (ie: group, ring, field, integral domain).
2. Once i have chosen the algebraic structure, i must cite theorems/properties for EACH step of the solutions.


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

x + y = 5
x - y = 1

Here's my work for solving the system in the two methods:

***************************************************************
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)

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

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.

But since (as my work shows above for each method of solving) i now have two binary operations (and rational numbers) so i cannot solve the system in a "group" of integers.
So when solving the system x + y = 5 and x - y = 1 for x and y, would i then be in a field (of rationals) since i have two binary operations and fractions?

Once i know what structure i'm working in, i can then cite specific properties and theorems from that structure at each step of my work. Note i need to do this twice; once for the "elimination" method and once for the "substitution" method.

The start of my work needs to start with a line that looks like:

To solve this system of linear equations, i will be in the (group, ring, field, integral domain) represented by <Z,R,Q,+,x,?,?> where + and x are the ordinary binary ops of the (integers, real numbers, rational numbers).

What would this line look like?


Then once the structure is decided, i would cite a theorem from that structure that looks like:

For any a and b in a (group, ring, field, integral domain) <Z,R,Q,+,x,?,?> if we know a=b, then for any c we know that a*c=b*c...

Can anyone tell me if i'm on the right track? I'm thinking field of rational numbers, but i'm not sure if that's the easiest... Thanks in advance!


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PostPosted: Sat, 5 May 2012 04:37:14 UTC 
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Should i use the field of rationals? If so, how do i write the notation? I know if i was working in a group i'd write:

"To solve, i will be in the group <Z,+> where + is the ordinary binary operation in Z."

So for this system in the field, would i write:

"To solve, i will be in the field <Q,+,x> where + and x are the ordinary binary operations in Q." ?


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PostPosted: Sat, 5 May 2012 18:40:49 UTC 
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Indeed you need a ring with 1 where 2 is invertible, that's about as minimal as I can see.

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PostPosted: Mon, 7 May 2012 21:04:22 UTC 
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so i'm just working in the ring <Z,x,+> of integers?


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PostPosted: Mon, 7 May 2012 21:05:49 UTC 
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ksmith630 wrote:
so i'm just working in the ring <Z,x,+> of integers?


NO! \mathbb{Z} is not a ring where 2 is invertible, however \mathbb{Z}[{1\over 2}] would be enough.

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PostPosted: Mon, 7 May 2012 21:11:23 UTC 
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Shadow wrote:
ksmith630 wrote:
so i'm just working in the ring <Z,x,+> of integers?


NO! \mathbb{Z} is not a ring where 2 is invertible, however \mathbb{Z}[{1\over 2}] would be enough.


I'm really sorry if i seem stupid but it's the notation that's getting me here.. :( What does Z(1/2) mean?


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PostPosted: Mon, 7 May 2012 21:15:16 UTC 
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ksmith630 wrote:
Shadow wrote:
ksmith630 wrote:
so i'm just working in the ring <Z,x,+> of integers?


NO! \mathbb{Z} is not a ring where 2 is invertible, however \mathbb{Z}[{1\over 2}] would be enough.


I'm really sorry if i seem stupid but it's the notation that's getting me here.. :( What does Z(1/2) mean?


It means polynomials in 1/2 with coefficients in \mathbb{Z}.

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PostPosted: Mon, 7 May 2012 21:19:05 UTC 
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Shadow wrote:
ksmith630 wrote:
Shadow wrote:
ksmith630 wrote:
so i'm just working in the ring <Z,x,+> of integers?


NO! \mathbb{Z} is not a ring where 2 is invertible, however \mathbb{Z}[{1\over 2}] would be enough.


I'm really sorry if i seem stupid but it's the notation that's getting me here.. :( What does Z(1/2) mean?


It means polynomials in 1/2 with coefficients in \mathbb{Z}.


Ahhh OK, can that be written any other way? It need to look like <R,x,+> etc... This "format" is completely new to me, not in our text... and i can't even find it online.


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PostPosted: Mon, 7 May 2012 21:22:37 UTC 
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ksmith630 wrote:
Shadow wrote:
ksmith630 wrote:
Shadow wrote:
ksmith630 wrote:
so i'm just working in the ring <Z,x,+> of integers?


NO! \mathbb{Z} is not a ring where 2 is invertible, however \mathbb{Z}[{1\over 2}] would be enough.


I'm really sorry if i seem stupid but it's the notation that's getting me here.. :( What does Z(1/2) mean?


It means polynomials in 1/2 with coefficients in \mathbb{Z}.


Ahhh OK, can that be written any other way? It need to look like <R,x,+> etc... This "format" is completely new to me, not in our text... and i can't even find it online.


It basically means you have everything with \mathbb{Z} and you have another element a which is the multiplicative inverse of 2. The format I wrote it in is the standard one, and is the simplest. You have the same + and * as before, just a new generator.

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PostPosted: Mon, 7 May 2012 21:28:52 UTC 
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Shadow wrote:
ksmith630 wrote:
Shadow wrote:
ksmith630 wrote:
Shadow wrote:
ksmith630 wrote:
so i'm just working in the ring <Z,x,+> of integers?


NO! \mathbb{Z} is not a ring where 2 is invertible, however \mathbb{Z}[{1\over 2}] would be enough.


I'm really sorry if i seem stupid but it's the notation that's getting me here.. :( What does Z(1/2) mean?


It means polynomials in 1/2 with coefficients in \mathbb{Z}.


Ahhh OK, can that be written any other way? It need to look like <R,x,+> etc... This "format" is completely new to me, not in our text... and i can't even find it online.


It basically means you have everything with \mathbb{Z} and you have another element a which is the multiplicative inverse of 2. The format I wrote it in is the standard one, and is the simplest. You have the same + and * as before, just a new generator.


K, so when i look for theorems and properties for each step of solving, should i look for those under "group, ring, field... etc"?


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PostPosted: Mon, 7 May 2012 21:35:22 UTC 
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ksmith630 wrote:
Shadow wrote:
ksmith630 wrote:
Shadow wrote:
ksmith630 wrote:
Shadow wrote:
ksmith630 wrote:
so i'm just working in the ring <Z,x,+> of integers?


NO! \mathbb{Z} is not a ring where 2 is invertible, however \mathbb{Z}[{1\over 2}] would be enough.


I'm really sorry if i seem stupid but it's the notation that's getting me here.. :( What does Z(1/2) mean?


It means polynomials in 1/2 with coefficients in \mathbb{Z}.


Ahhh OK, can that be written any other way? It need to look like <R,x,+> etc... This "format" is completely new to me, not in our text... and i can't even find it online.


It basically means you have everything with \mathbb{Z} and you have another element a which is the multiplicative inverse of 2. The format I wrote it in is the standard one, and is the simplest. You have the same + and * as before, just a new generator.


K, so when i look for theorems and properties for each step of solving, should i look for those under "group, ring, field... etc"?


The addition step is clearly groups, the addition is normal integers, so it's abelian. Then you need to be able to get rid of the 2, so to do this systematically you need a notion of multiplication and division, but clearly only 2 needs to be invertible.

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PostPosted: Mon, 7 May 2012 23:05:50 UTC 
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So when i start this, is it me OK to write:

"When solving this system i will be working in the group <Z,+,x> where + and x are normal binary ops for addition and mult in the integers"
?


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PostPosted: Tue, 8 May 2012 00:36:10 UTC 
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ksmith630 wrote:
So when i start this, is it me OK to write:

"When solving this system i will be working in the group <Z,+,x> where + and x are normal binary ops for addition and mult in the integers"
?


no, you need more than \mathbb{Z}, that's what I keep saying.

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PostPosted: Tue, 8 May 2012 04:40:48 UTC 
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Shadow wrote:
ksmith630 wrote:
So when i start this, is it me OK to write:

"When solving this system i will be working in the group <Z,+,x> where + and x are normal binary ops for addition and mult in the integers"
?


no, you need more than \mathbb{Z}, that's what I keep saying.


Hmm... if the OP has "cancel the 2" rather than "divide by 2" or "multiply by 1/2", then the integral domain \mathbb{Z} would do.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Tue, 8 May 2012 05:03:19 UTC 
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outermeasure wrote:
Shadow wrote:
ksmith630 wrote:
So when i start this, is it me OK to write:

"When solving this system i will be working in the group <Z,+,x> where + and x are normal binary ops for addition and mult in the integers"
?


no, you need more than \mathbb{Z}, that's what I keep saying.


Hmm... if the OP has "cancel by 2" rather than "divide by 2" or multiply by 1/2, then the integral domain \mathbb{Z} would do.


I agree, I opted for the more generalizable thing.

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