1. ## field axioms

I need to prove this: a,b,c ∈ and c ∉ 0 and if ac=bc then a=b
I am only allowed to use the axioms for

I have ac-bc=0
prove: ac-bc=c(a+(-b)) (I used the axiom associative)
because c ∉ 0 then (a+(-b))=0 so a=b

Can you tell if my pove is right and which axioms I need to use to prove this?
thank you

2. ## Re: field axioms

Odd. You titled this "field axioms" but Z is not a field!

Okay, from ac= bc, you get ac- bc= 0 and then you say
"ac-bc=c(a+(-b)) (I used the axiom associative)"
Well, first, that's the distributive axiom, not associative. But also you don't say that is equal to 0 which you should.
Now, which axiom, specifically, allows you to say that "if c(a- b)= 0 and c is not 0 then a- b= 0"?

3. ## Re: field axioms

ac - bc = 0, c ≠ 0 (given)

(a - b)c = 0 (the Distirbutive law!!!!)

(technically, this is actually:

ac - bc = ac + (-bc) = ac + (-b)c = (a + (-b))c = (a - b)(c), so we are using ANOTHER key fact:

a(-b) = -(ab). we should prove this, too (i know, it's a detour, but it never hurts to convince ourselves something really is true):

recall that -(ab) is the UNIQUE integer such that: ab + -(ab) = 0. so we need to show that:

ab + a(-b) = 0, which will then mean a(-b) has to be -(ab) (by uniqueness).

now ab + a(-b) = a(b + (-b)) (again, by the distributive law).

and so ab + a(-b) = a(b + (-b)) = a0.

NOW we need to show that a0 = 0 (no matter WHAT integer a is). yes, another detour.

a0 = a(0 + 0)....because 0 is an integer, and c + 0 = c for ANY integer c (including 0), so that 0 + 0 = 0.

again, by the distributive law, we have:

a0 = a(0 + 0) = a0 + a0.

thus a0 + -(a0) = (a0 + a0) + -(a0) (adding -(a0) to both sides, because if k = m, then k+n = m+n, for ANY integers k,m and n: sometimes this rule is given the awkward name of:

"equals to equals are equal")

but by DEFINTION of -(a0) we have a0 + -(a0) = 0, so we have:

0 = (a0 + a0) + -(a0)

0 = a0 + (a0 + -a0)) (NOW we have used the associative law)

0 = a0 + 0 (since a0 + -(a0) = 0, by the definition of additive inverse)

0 = a0 (since a0 + 0 = a0, by the identity law of addition for 0).

whew! this axiom stuff is harder than it looks!

now, having established that a0 = 0, we go back to our "first detour", and have:

ab + a(-b) = a0 = 0, and FINALLY, we conclude that a(-b) = -(ab).

THEN (and only now are we "fully justified"), we can say that:

ac - bc = (a - b)c (proving the distributive law for SUBTRACTION is more complicated than it looks!).

but ac - bc = 0. so

(a - b)c = 0.

now the next part is a little tricky: we want to prove that since c ≠ 0, a - b = 0.

to do this we need a SPECIAL property of Z, which is that it is an "integral domain" which means that:

if km = 0, then either k = 0, or m = 0. Z is a ring, and not ALL rings HAVE this property.

in this case our "k" is a - b, and our "m" is c.

if this property (also known as the "no zero divisors" property) is not one of your axioms, you have to use ORDER properties of Z:

for any integer k exactly ONE of the following is true:

k > 0
k < 0
k = 0 (the trichotomy law)

if k > 0, m > 0, then k+m > 0 (positivity of addition)
if k > 0, m > 0, then km > 0 (positivity of multiplication).

now, we need to argue by contradiction: suppose k,m ≠ 0. we need to PROVE km ≠ 0. there are 4 cases:

k > 0, m > 0. in this case km > 0, so km ≠ 0 (by trichotomy).
k > 0, m < 0. in this case, -m > 0, so k(-m) = -(km) > 0, so km < 0, and thus km ≠ 0.
k < 0, m < 0. this is the same as the previous case since km = mk (commutativity of multiplication).
k < 0, m < 0. this is sort of interesting: here -k > 0, -m > 0, so

0 < (-k)(-m) = -(k(-m)) = -(-(km)) = km.

why? well, we actually should prove that -(-a) = a. recall that -(-a) is the UNIQUE integer with -a + -(-a) = 0.

but -a + a = a + -a (by commutativity of addition), so

-a + a = a + -a = 0, so since a has that property, it has to be that -(-a) = a.

so since if k < 0, m < 0, km = (-k)(-m) > 0, we can safely conclude km ≠ 0.

NOW, at LONG LAST, we are in a position to say:

(a - b)c = 0 AND c ≠ 0 must mean a - b = 0 (for otherwise (a - b)c ≠ 0, as we showed above, no matter whether a-b and c are positive or negative).

and so, continuing our long and tiresome journey:

a - b = 0

a + (-b) = 0 (definition of subtraction is addition of an additive inverse)

(a + (-b)) + b = 0 + b

(a + (-b)) + b = b (since 0 + b = b + 0 = b, by the identity law for 0 in addition)

a + (b + (-b)) = (a + (-b)) + b = b (the first equality is by associativity of addition)

a + 0 = b (because b + (-b)) = 0, by definition of - b)

a = b (because a + 0 = a, why?)

THAT is the COMPLETE proof (with only a few minor steps glossed over). as you can see, it's quite involved.

4. ## Re: field axioms

a+0=a because 0 is nutural for the addition for every a in Z 0+a=a and a+0=a (sorry for my english I find it difficult to translate some math words into english)

to do this we need a SPECIAL property of Z, which is that it is an "integral domain" which means that:

if km = 0, then either k = 0, or m = 0. Z is a ring, and not ALL rings HAVE this property.

in this case our "k" is a - b, and our "m" is c.

if this property (also known as the "no zero divisors" property) is not one of your axioms, you have to use ORDER properties of Z:

for any integer k exactly ONE of the following is true:

k > 0
k < 0
k = 0 (the trichotomy law)

if k > 0, m > 0, then k+m > 0 (positivity of addition)
if k > 0, m > 0, then km > 0 (positivity of multiplication).

now, we need to argue by contradiction: suppose k,m ≠ 0. we need to PROVE km ≠ 0. there are 4 cases:

k > 0, m > 0. in this case km > 0, so km ≠ 0 (by trichotomy).
k > 0, m < 0. in this case, -m > 0, so k(-m) = -(km) > 0, so km < 0, and thus km ≠ 0.
k < 0, m < 0. this is the same as the previous case since km = mk (commutativity of multiplication).
k < 0, m < 0. this is sort of interesting: here -k > 0, -m > 0, so

0 < (-k)(-m) = -(k(-m)) = -(-(km)) = km.

This part I didn't learn at college, but I will learn it soon.
My teacher said I could use this axiom: voor every a,b in Z with ab=0 it follows that a=0 or b=0
So I see that (a-b)c=0 we say that a-b=0 or c=0 but we now c isn't equal to zero so we can say that a-b=0

Thank you very much for your help!

5. ## Re: field axioms

it turns out that there are "kinds of numbers" that go "in a circle", rather than "an infinite line" like the integers do.

because numbers like that exist, it is not a "universal property of rings" (which is the type of algebraic structure the integers ARE) that ab = 0 implies a = 0, or b = 0 (or maybe both).

it is, however, true for POLYNOMIALS with coefficients that are integers (for example). and polynomials with integer coefficients are very much LIKE integers (our arabic numeral system can be thought of as "polynomials in 10 (rather than x)" which explains why polynomial addition and multiplication are so "normal").

i point this out because for algebraic structure that ARE euclidean domains, we can use the very same proof we used above:

in a euclidean domain, if ac - bc = 0, then a = b. the advantage to this is that anything (no matter how "weird" the actual elements of our euclidean domain might be) that satisfies the same rules as the axioms we used in our proof above, also can use this proof. this is why things are often defined using axioms: you don't have to prove things over and over for each new "type of thing (structure)", prove it once as a consequence of the axioms and that takes care of all sorts of different cases at once.

it turns out that integers are fairly pervasive in mathematics. many (but not all) mathematical structures can be seen as "integers interacting with other things". but integers are simple, right? well, actually, no. if all we ever did was add, then yes, integers WOULD be simple. we could just "count everything". but multiplication throws a monkey wrench into the works.

for counting, everything is "1's", we just have different symbols because "5" is easier to write than 1+1+1+1+1. but for multiplying, everything is a combination of "prime numbers", and prime numbers are...unpredictable. there's doesn't seem to be a "pattern" to how often they occur (there is, sort of, but it's very complicated, and we can't prove it, yet).