Validity of Proof for 'Possibility of Subtraction' from Apostol 1

Question

I attempted a proof before reading the solution Apostol provides. I don't think it is valid but I am trying to determine why

Theorem 2, Possibility of Subtraction: Given $a \text{ and } b$, there is exactly one x such that $ a + x = b $ This is denoted by $b - a$. $\ 0-a$ is written as $-a$ and called the negative of a.

Proof from Apostol - Given $a$ and $b$, choose $y$ so that $a + y = 0$ and let $x = y + b$. Then $a + x = a + (y + b) = (a + y) + b = 0 + b = b$. Therefore there is at least one x such that $a + x = b$. But by Theorem 1.1 there is at most one such x. Hence there is exactly one.

I will also use theorem 1, previously proved

Theorem 1 $ \text{if } a + b = a + c \text{ then } b = c $

My attempt

$$ \begin{align} \text{Given } a, b \text{ let } &\quad a + x = b &\quad \text{ start }\tag{1}\label{eq1}\\ \text{there is exactly one } x \text{ such that } &\quad a + x = b &\quad\text{from theorem 1.1, show in 2.1}\tag{2}\label{eq2} \\ \text{ proof of (2) } \\ \text{ consider more than one x that satisfies (2) } &\quad a + x = b,\; a + x' = b &\quad\text{ start }\tag{2.1}\label{eq2.1}\\ \text{ then } &\quad a + x = a + x' &\quad\text{ }\tag{2.2}\label{eq2.2}\\ \text{ therefore x and x' are the same and (2) is proven } &\quad x = x' &\quad\text{ direct result of theorem 1.1 }\tag{2.3}\label{eq2.3}\\ \end{align}$$

Is this valid? If not, why?

My thoughts - one way that I could see it being invalid is that in (2.1) I break $a + x = b$ into two cases, x and x'. The form of the resulting cases is same as the original, so then, using my logic, you would have to break each case into two more cases on to infinity.

Appreciate the input!

Answer 1

You've got half of the answer: you've attempted to show that if such an $x$ exists, then there is at most 1. Presumably, Thereom 1.1 is something of the form "if $a + b = c + b$, then $a = c$.

What you haven't done is show that $x$ exists at all. For example, every part of your proof so far is entirely and completely valid if we restrict our variables to take values only in the natural numbers. However, the final result is clearly not true in the natural numbers: there is no natural number $x$ such that $2 + x = 1$.

Answer 2

No, this is not correct. First, you write “given $a,b$, let $a+x=b$”. This doesn't make sense. Besides, you don't explain why such a $x$ exists. And then you only prove that if such a $x$ exists, then it is unique. But, again, you don't prove that it exists.