# Homework Help: Field Proofs (just needs revision)

1. Oct 2, 2011

### JPanthon

1. The problem statement, all variables and given/known data

Thanks to everyone who has helped me so far - I'm very grateful.

(1) Prove that the multiplicative inverse in any field is unique
(2) Prove the cancellation law | ab = ac => b=c
(3) Prove (-1)a = -a

2. Relevant equations

The field axioms: http://mathworld.wolfram.com/FieldAxioms.html

3. The attempt at a solution

(#1)
Let: a is an element of a field
Let: b and c are multiplicative inverses of a, such that, (1) ab = 1, and (2) ac = 1

Proof.

ab = 1
(ab)/b = (1/b)
a (b/b)= (1/b)
a 1 = (1/b)
a = (1/b)

(2)

ac = 1
(1/b)c = 1
b((1/b)c)) = b(1)
(b/b)c = b
1c = b
c = b

QED

(#2)

Let: a, b are elements of F

Let: z is the mutliplicative inverse of x

Proof.

ax = bx
z(ax) = z(bx)
z(xa) = z (xb)
(zx)a = (zx)b
1a = 1b
a = b

QED

(#3)
Let: a is in a field
Let: -1 is the additive inverse of 1

Proof.

(-1)a = (-1 + 0) a
(-1)a = (-1 + (-1 + 1)) a
(-1)a = -1a + (-1a + 1a)
(-1)a = -1a + (1(-a + a))
(-1)a = -1a + (-a + a)
(-1)a = 1(-a) + (-a + a)
(-1)a = -a + (-a + a)
(-1)a = -a + (a + (-a))
(-1)a = (-a + a) + (-a)
(-1)a = 0 + (-a)
(-1)a = -a

QED

Thank you very much!
Please be as critical as possible, I really want to learn.
This is for a first year algebra class.

J.Anthony

2. Oct 2, 2011

### ArcanaNoir

for your part 2, I thought you were trying to show b=c. so why do you show a=b?

For proof 1 (and any other) it's slicker to show the proof using this=this=this=this=conclusion than to have it set up as individual lines as you did. try starting with:
let: b and c are multiplicative inverses of a with e the identity.
b=b*e=b*(a*c)=......=c

Try doing two similarly starting with b= , the crucial step will be when you get a*b as part of a line and you replace it with a*c since it's part of your assumption, then decompose back to c.

3. Oct 3, 2011

### Deveno

suppose ab = 1 and ac = 1.

since fields are commutative, ba = 1.

thus b = b1 = b(ac) = (ba)c = 1c = c.

see how much shorter this is?

ab = ac
z(ab) = z(ac)
(za)b = (za)c
1b = 1c
b = c (this also saves one line, since you don't need to "switch factors").

3) isn't very good. in these lines:

(-1)a = -1a + (-1a + 1a)
(-1)a = -1a + (1(-a + a))

you assert that -1a = 1(-a), which is what you are trying to prove.

first we show that if x + x = x, x = 0.

x+x = x
(x+x) + -x = x + -x = 0
x + (x + -x) = 0
x + 0 = 0
x = 0.

next we show that 0a = 0:

0a = (0 + 0)a = 0a + 0a, so by the above, 0a = 0.

finally, the proof that (-1)a = -a:

(-1)a + a = (-1)a + 1a = (-1 + 1)a = 0a = 0. so
((-1)a + a) + -a = 0 + -a = -a
(-1)a + (a + -a) = -a
(-1)a + 0 = -a
(-1)a = -a.