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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!

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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 $2 + x = 1$.

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  • $\begingroup$ Ok I think I understand, Apostol proves it by first stating "choose y so that a + y = 0" which I think can be read as "there exists a y s.t. a + y = 0 (existence of negatives)". We don't have to do this for a or b because it's given. Then he says "let x = y + b" which i think can be read as "there exists an x s.t. x = y + b (valid because of the addition operation that we're assuming real numbers have)". So the key point here is that he states there must be an x that exists if the givens and the axioms hold true, and from that, there must also be a y that exists? $\endgroup$ – Jake Kirsch Jan 11 at 20:52
  • $\begingroup$ Apostol's $y$ exists by some presumably-previous result (or axiom maybe?), which is what you're calling "existence of negatives". He then picks his $x$ as something that you know exists ($y + b$), then checks that it satisfies the required equation, and that it's the only one that does, though he abreviates the latter down to essentially nothing: your attempt gives that latter step in full. $\endgroup$ – user3482749 Jan 11 at 20:55
  • $\begingroup$ Yes, he states several axioms of the real number system, as that is the starting point for the way he develops calculus for the reader. The existence of negatives is one of those, I believe it's also known as the additive inverse. I think it's making sens for me now, thank you for the help! $\endgroup$ – Jake Kirsch Jan 11 at 20:59
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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.

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