a) if $x+y = x+z$ , then $y=z$
b) if $xy=xz$ and $x$ not equal to $0$ , then $y=z$
c) $-(-x)=x$
d) if $x$ not equal to $0$ and $xy=x$ , then $y=1$
e) $(x^{-1})^{-1} =x$
I didn't understand what my lecturer teach about this topic ! So i really need some help ! Here are my answer for a , b ,c and d . Can anyone help me check ?
a) (-x) + x + y = (-x)+x+z
[(-x) + x ] + y = [ (-x) + x ] + z
0 + y = 0 + z
y=z
b) (1/x)(xy) = (1/x)(xz)
[(1/x).x] y = [ (1/x).z]
1.y = 1.z
y=z
c) -(-x) = x
-(-x) + (-x) = 0
add x to both side :
x + [ -(-x) + (-x) ]= x+ 0
x + 0 = [ x + (-x)] + [-(-x)]
x= 0+ -(-x)
=-(-x)
d) let xy=x , and x ≠ 0
∃ z ∈ R ∈ xz = zx =1
1=zx
1= z(xy)
1=(zx)y
1=1.y
1=y
You have to use the axioms to prove the statements. Sometimes it can be a bit tricky, but you should try it ourself, otherwise you don't learn how to use it.
For example the first can be done the following way: On axiom states that you can add $0$ to each number. So, $$ y=y+0 $$ Next, each element has an additive inverse, so you can use $x+(-x)=0$ and you get $$ y+0=y+(x+(-x)) $$ Use the associative property to get $$ y+(x+(-x))=(y+x)+(-x) $$ Now, you can use the equation $y+x=z+x$ and you get $$ (y+x)+(-x)=(z+x)+(-x) $$ And use the same steps as above to get $z$ at the end.