Commutativity of Addition

**rualin** · July 8th 2007, 04:53 PM

Is commutativity of addition necessary as an axiom in the Field Axioms or can it be proved by the others?

**ThePerfectHacker** · July 8th 2007, 05:06 PM

Originally Posted by rualin

Is commutativity of addition necessary as an axiom in the Field Axioms or can it be proved by the others?

Yes it is necessay*.
It cannot be proved by others.

In fact, I can prove that it cannot be proved

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Define $\mathbb{H} = \mathbb{R}\times \mathbb{R}\times \mathbb{R} \times \mathbb{R} = \{(a,b,c,d)|a,b,c,d\in \mathbb{R}\}$ .

So any element in $\mathbb{H}$ is an ordered 4-tuple $(a,b,c,d)$ but it is more convient to write $a+bi+cj+dk$ . We define the following products:
$i^2 = j^2 = k^2 = -1$
And,
$ij=k \ \ jk = i \ \ ki = j \ \ ji = -k \ \ kj=-i \ \ ik = - j$

So for example if we need to multiply $(2+i)(1+j)$ we would write $2+i+j+ij$ and get $2+i+2j+k$ .

But if $(1+j)(2+i)=2+2j+i+ji=2+i+2j-k$ .
So $(1+j)(2+i)\not = (2+i)(1+j)$ .

I leave it to you to verify that $\mathbb{H}$ has the following properties:
$h_1+h_2=h_2+h_1$
$h_1+0=h_1$
$\forall h_1 \exists h_2 \ h_1+h_2=0$
$h_1+(h_2+h_3)=(h_1+h_2)+h_3$
$h_1(h_2h_3)=(h_1h_2)h_3$ --- > Boring and Long.
$h_1(h_2+h_3)=h_1h_2+h_1h_3$ ---> Boring and Long.
$(h_2+h_3)h_1=h_2h_1+h_3h_1$ --->Boring and Long.
$1h_1=h_11=h_1$
$\forall h_1\not = 0 \exists h_2 \ h_1h_2=h_2h_1=1$
This show it has everything EXCEPT $h_1h_2=h_2h_1$

*)If the field is finite it is not necesarry. This supprising result is not easy to prove.

**rualin** · July 8th 2007, 05:17 PM

According to a professor's notes I found online, Leonard Dickson discovered it can be proved from the other axioms. Here is how the proof in the notes goes:

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Commutativity of Addition Theorem
Let $F$ be any field. Then $+$ is a commutative operation on $F$ .

Proof: Let $x,y\in F$ . Then,

$(1+x)(1+y)=(1+y)(1+x)$ by commutativity of multiplication
$(1+x)\cdot 1 + (1+x)\cdot y = (1+y)\cdot 1 + (1+y)\cdot x$ by distributive law
$(1+x) + (1+x)\cdot y = (1+y) + (1+y)\cdot x$ by definition of multiplicative identity
$1 + (x + (1+x)\cdot y) = 1 + (y + (1+y)\cdot x)$ by associativity of addition
** $x + (1+x)\cdot y = y + (1+y)\cdot x$ by cancellation law for addition
$x+(1\cdot y) + (x\cdot y) = y + (1\cdot x) + (y\cdot x)$ by distributive law
$x+(y+(x\cdot y)) = y+(x+(y\cdot x)$ by definition of multiplicative identity
$(x+y)+(x\cdot y) = (y+x)+(y\cdot x)$ by associativity of addition
$(x+y)+(x\cdot y) = (y+x)+(x\cdot y)$ by commutativity of multiplication
** $x+y = y+x$ by cancellation law of addition

Hence, $+$ is commutative.

-----

** What doesn't completely convince me is the cancellation step but what do I know.

What do you think?

**ThePerfectHacker** · July 8th 2007, 05:25 PM

Originally Posted by rualin

What do you think?

I looked it over quickly and it looks right.

The only problem is the way the theorem is stated it should say let F be a field "expect for possibly commutativity of addition" then it must be commutative under addition.

But the problem is that is not the "right" definition.

In fact a more appropriate definition is that a field is a commutative division ring.

See Here for the definition of a field.

Note: I answered your question about commutativity of multiplication.

**rualin** · July 8th 2007, 05:39 PM

I understand. But in any case, if a Field (according to the actual definition) has all the properties a commutative division ring has and a "Field*" (the way it seems to be used in my notes) has those same properties, except that commutativity of addition is not taken as an axiom, but rather proved through the other axioms, do they serve the same exact purpose?

I can see why a Field being a commutative division ring is desirable over a "Field*" to keep the structure but I can also see how one would try to keep the axioms the fewest possible.

**ThePerfectHacker** · July 8th 2007, 06:36 PM

Originally Posted by rualin

do they serve the same exact purpose?

They are the exact same thing. In fact when I was in the 10th Grade we were learning about Groups and Fields just the most very basic defintions, and I still remember exactly how everything was defined. It was defined in the same way you defined it. But that was done to make it easier for the students.

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