1. ## Another proof

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.

2. Originally Posted by jzellt
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 $x + y \le x$ 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.

3. Originally Posted by jzellt
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.
You prove it by induction. The base case is $y=1$, it is true because $x < x+1$ since $x\in x \cup \{ x\}$.
If $x then $x<(x+y)+1 = x+(y+1)$.

4. Originally Posted by jzellt
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.

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:

........ $\forall x\forall y[ x+y'=(x+y)']$.................................................. ..........................................1

2) Axiom: for all xεN,....0+x=x............and in symbols.....

.......... $\forall x( 0+x=x)$.................................................. ........................................2

3) Definition: for all r,s εN ......r<s <===> there exists uεN and (r+u)'=s ,and in symbols:

....... $\forall r\forall s[r.................................................. .................................................. ..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 $y\neq 0$ ,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........... ..................................................