[faithHunter]

I think this image is relevant:

[mGGMueX]

I have one.

Derive the commutativity of addition from the 4 addition axioms closure, associativity, zero, inverses.

[SayfT]

zero would be a + 0 = a
associativity should be (a+b) + c = a + (b+c)
inverses would be a + -a = -a + a

and closure is escaping me for the moment ><

[AxS.pinealan]

what is commutativity -__-''
looks like so really basic stuff when i searched on wiki

[SayfT]

how stuff is added, at least thats how I understand it, multiplicativity (is that even a word?!) would be how stuff is multiplied, or the axioms that multiplication is based on.


fyi: I could be wrong :P

[Pox]

Commutativity is the fact that a+b=b+a for all a,b.

Quick Comments

	RVLiveR:	not if 1+1=1 (see proof above)
	FaDeBadger:	That "proof" is false

[Mukade]

An operation is commutative if a x b = b x a where x is some operation. Like, you can say addition is commutative for a,b e C.

[mGGPrometheus]

I don't think you can derive commutativity from only closure, associativity, identity and inverses.

For example consider the general linear group consisting of nxn matrices with nonzero determinant.
- It is closed (multiplying two matrices with nonzero determinant results in a matrix with nonzero determinant)
- It is associative (matrix multiplication is associative)
- It has an identity
- It has inverses
And yet any two matrices in the general linear group need not commute.

[mGGMueX]

Originally Posted by mGGTitan

I don't think you can derive commutativity from only closure, associativity, identity and inverses.

For example consider the general linear group consisting of nxn matrices with nonzero determinant.
- It is closed (multiplying two matrices with nonzero determinant results in a matrix with nonzero determinant)
- It is associative (matrix multiplication is associative)
- It has an identity
- It has inverses
And yet any two matrices in the general linear group need not commute.

Matrix multiplication is not commutative but addition is.

It is definitely possible to derive from the 4 stated axioms.

[Pox]

Originally Posted by mGGMueX

Matrix multiplication is not commutative but addition is.

This is exactly the point - both (GL,*) and (R,+) are groups (i.e. satisfy the four given axioms) but (GL,*) is not commutative - so the group axioms are not enough to prove that (R,+) is commutative, you need some extra fact.

[Mukade]

Commutativity is just a property. Of course everything isn't commutative, one being matrix multiplication, normal division, etc.

[ToRSpookToR]

shama lama ding dong

[Mukade]

As for deriving commutativity from the peano axioms, that's actually fairly tricky, but it can be done. The trick is to prove associativity first with induction, and then a cute little proof for commutativity follows (that I don't remember).

[mGGMueX]

You guys failed the test... I was hoping to get some help revising this

I'll post the proof later if noone can figure it out.

[Mukade]

Originally Posted by mGGMueX

You guys failed the test... I was hoping to get some help revising this

I'll post the proof later if noone can figure it out.

I have it in a textbook, lol

[mGGPrometheus]

Originally Posted by mGGMueX

You guys failed the test... I was hoping to get some help revising this

I'll post the proof later if noone can figure it out.

I'd like to see that... from what I've learnt in group theory, any group with those 4 axioms can either be commutative or noncommutative (in fact those 4 axioms are the defining properties of a group)

Btw for anyone who likes calculus, try this simple integration problem =P

Quick Comments

Zepph:

Oh man...memories. 1/cosx is secx? so its something like log secx - tanx + C ? gosh i hardly remember

[Mukade]

For addition anyway, but the other operations are similarly easy.

[Pox]

I'll give anyone $100 for a singular solution to the 3+1-dimensional navier-stokes equations with a smooth initial condition. It's, uh, my homework.

Quick Comments

AxS.pinealan:

I dont even know what is navier-stokes equation

[Mukade]

Originally Posted by Av.ToRPox

I'll give anyone $100 for a singular solution to the 3+1-dimensional navier-stokes equations with a smooth initial condition. It's, uh, my homework.

Sorry dude, beyond me.

[AxS.pinealan]

Originally Posted by mGGTitan

Btw for anyone who likes calculus, try this simple integration problem =P

lnlsecx+tanxl + C
thats an identity in my textbook which I am supposed to memorize =P