Given two natural numbers a < b, we define b - a to be a natural number that satisfies a + (b - a) = b. Prove that this number is unique? Given two natural numbers $a < b$, we define $b - a\,$ to be a natural number that satisfies $a + (b - a) = b$. Prove that
this number is unique? That is, if some other number k claims that a + k = b, then k = b - a. (Recall the operation
subtraction or the additive inverse of a number is not defined for the natural numbers.)
How can i do this without the additive inverse. Otherwise I can easily show that if $x=b-a$ and $y=b-a$ that in fact $x=y$? 
 A: The standard way of proving that an object with a given property is unique is to show that if you have two objects with the property, then the two objects are in fact equal.
For example, to show that $0$ is the unique natural number with the property that $0+x=x$ for all natural numbers $x$, you can do the following:

Let $0$ and $0'$ be two natural numbers with the given property; that is, that $0+x=x$ for all natural numbers $x$, and that $0'+x = x$ for all natural numbers $x$. We wish to prove that $0=0'$.

(In the above case, you could do so by then noting that $0+0'=0'$, because $0$ has the desired property, and that $0+0'=0'+0 = 0$, because $0'$ has the property, and therefore that $0=0'+0 = 0+0' = 0'$, obtaining the conclusion we wish to establish).
Similarly here. I would state it as follows:

Let "$b-a$" represent a natural number $k$ such that $a+k = b$, if such a natural number exists. Prove that if $b-a$ exists, then it is unique. 

So, assume that $k$ and $\ell$ are natural numbers that both have the property: that means that $a+k=b$ and that $a+\ell = b$. We wish to show that $k=\ell$.
Since $a+k=b=a+\ell$, then we have $a+k=a+\ell$. By the cancellation property of addition, we conclude that $k=\ell$< as desired. $\Box$
Thus, when "$b-a$" exists, it is unique. (And of course, such a $k$ exists if and only if $a\lt b$, assuming your natural numbers do not include $0$ [mine do]). 
A: You need to know that if $a+b=a+c$, then $b=c$. You can prove this by induction on $a$.
Since $a+(b-a) =a+k$, it follows that $k=b-a$. 
