Posted by **Meherin** on Thursday, March 13, 2008 at 8:27pm.

I just need a description of the following:

Associative property of Multiplication

Associative Property of Addition

Commutative properties

Distributive Property

Identity property(I dont need to know much about this)

Zero property(i dont need to know much about this)

- Math -
**Damon**, Thursday, March 13, 2008 at 8:56pm
ADDITION:

associative a+(b+c) = (a+b) + c

commutative a+b = b+a

identity : there is a number 0 with

a+0=0+a = a

inverse a + -a = -a + a = 0

MULTIPLICATION:

associative a(bc) = (ab)c

commutative ab=ba

identity there is a number 1 with

a*1 = 1*a = a

inverse If a not zero then there is an 1/a such that a(1/a) = (1/a)a = 1

BOTH (the biggie!!)

distributive a(b+c) = ab + ac

- Math -
**Count Iblis**, Thursday, March 13, 2008 at 8:56pm
Associative property of Multiplication:

(x*y)*z = x*(y*z)

Associative property of Addition:

(x + y) + z = x + (y + z)

Commutative properties:

x*y = y*x

x + y = y + x

Distributive Property :

x*(y + z) = x*y + x*z

Identity property: There exists a number 1 such that:

1*x = x for all x.

Zero property: There exists a number 0 such that:

0 + x = x for all x

Inverse properties:

For every x there exists a number -x, such that:

x + (-x) = 0

For every x not equal to zero there exists a number x^(-1), such that:

x*x^(-1) = 1

Examples:

There can only be one 0. Proof suppose there were two numbers 0 and 0' that both satisfy the property that the zero element has to satsify, then:

0 + 0' = 0'

because 0 is a zero element.

But because 0' is also a zero element, you also have:

0 + 0' = 0

This means that 0 = 0'

Example:

(-1)*x = -x

Proof:

Let's check of (-1)*x satisfies the criterium of being the inverse (relative to addition) of x:

x + (-1)*x =

1*x + (-1)*x =

(1 + (-1))*x = (use that -1 is the inverse relative to addition of 1)

0*x = 0

So, we can conclude that (-1)*x is the inverse of x relative to addition, which means that (-1)*x = -x

- Math -
**Meherin**, Thursday, March 13, 2008 at 9:01pm
thank u soooooo much!

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