- #1
em2008
- 1
- 0
I realize this is really simple but I'm stuck, I've worked on it for awhile All that needs to be done is proving that -(-x)=x. We are allowed to use:
1) (x+y)+z=x+(y+z)
2) (xy)z=x(yz)
3) x(y+z)=xy+xz
4)There is a unique element 0 element of real numbers s.t. 0+x=x for all x
5) For each x an element of real numbers there is a unique y element of real numbers s.t. x+y=0 and we write y=-x
6) There is a unique element 1 an element of the real numbers s.t. x(1)=x for all x element of r and 0 does not equal 1
7) For each x an element of the real numbers with x not equal to 0 there is a unique element y an element of the real numbers s.t. x(y)=1 and or x(1/x)=1
8) x is greater than y implies x+z greater than y+z
9) x less than y and y less than z implies x less than z
10) For x,y an element of the real numbers exactly one of the following is true: x greater than y, y greater than x, or x=y
11) x greater than y and z greater than 0 implies xz less than yz
1) (x+y)+z=x+(y+z)
2) (xy)z=x(yz)
3) x(y+z)=xy+xz
4)There is a unique element 0 element of real numbers s.t. 0+x=x for all x
5) For each x an element of real numbers there is a unique y element of real numbers s.t. x+y=0 and we write y=-x
6) There is a unique element 1 an element of the real numbers s.t. x(1)=x for all x element of r and 0 does not equal 1
7) For each x an element of the real numbers with x not equal to 0 there is a unique element y an element of the real numbers s.t. x(y)=1 and or x(1/x)=1
8) x is greater than y implies x+z greater than y+z
9) x less than y and y less than z implies x less than z
10) For x,y an element of the real numbers exactly one of the following is true: x greater than y, y greater than x, or x=y
11) x greater than y and z greater than 0 implies xz less than yz