Assume x,y e N. If y != 0 (not equal) , then x + y > x.
Again, this is obvious, but Im having trouble giving a solid mathematical proof.
is this for a set theory class? i just want to know how rigorous your proofs have to be. here are some simple approaches, they probably wouldn't hold up in a set theory class, but they'd be ok in most other classes:
(1) let x be fixed and do induction on y
(2) a proof by contradiction. show that is false if y is not zero.
(3) since y is a natural number that is not zero, it must be greater than 0 (since zero is the least element). so we have y > 0. now add x to both sides.
For that you need the following 2 axioms and a definition:
1) Axiom: for all x,yεN ...x+y'=(x+y)'...........where y' is the successor of y,hence ,if y=5 then y' =5'=6....and in symbols:
........ .................................................. ..........................................1
2) Axiom: for all xεN,....0+x=x............and in symbols.....
.......... .................................................. ........................................2
3) Definition: for all r,s εN ......r<s <===> there exists uεN and (r+u)'=s ,and in symbols:
....... .................................................. .................................................. ..3
Now for the proof:
NOTE we want to prove x<x+y.According to the definition (2) we must prove that there exists z such (x+z)'=x+y.
So,since ,0<y and by definition (2) there exists z such that (0+z)'=y ,or z'=y by using axiom (2)............................................... .................................................. 4
But ..........x+y=x+y,or x+z'=x+y by using (4),or (x+z)'=x+y by using axiom (1)
Hence there exists z such that (x+z)'=x+y which implies :
..................................x<x+y........... ..................................................