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I am beginning an introductory college math course to catch up from my bad high school education. This is one my first proofs.

Prove that $x - a = b$ can be rewritten as $x = b + a$.

We have been given the properties of the operations of the set of the real numbers (not sure how to latex that).

My proof is this:

$x - a - (-a) = b - (-a)$

$x - 0 = b + a$

$x = b + a$

I'm not completely sure this is correct.

I have another, more important and general doubt. In the proof I use the fact that adding something to both sides of an equation does not change the equation. Do I need to prove this, since no proof has been given in this course, if I want to use it? We are proving very intuitively obvious theorems, so I'm not sure what other intuitively obvious theorems I can use without proving first!

Here's an attempt:

Theorem: adding $x \in R$ to both sizes of an equation does not change the equation.

If $a, b$ are real numbers and $a = b$, then $a$ and $b$ are the same. $a + x = b + x$ can then be rewritten as $a + x = a + x$, since a = b. Both sides are the same, so $a + x = b + x$.

Here I'm not sure how to say that a bunch of operations in the real numbers is a real number. This also should work for all operations we haven't mentioned yet: if $a^{2/3} = b^{4/7}$, then $a^{2/3} + c = b^{4/7} + c$. I'm not sure if this complicates things, but I have no idea how to say this either way.

Let's also ignore for a second that this is an introductory course. Would I need to prove this if I was asked to prove the theorem in an exam?

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    $\begingroup$ When proving something as elementary as "$x - a = b$ can be rewritten as $x = b + a$" (and here I don't mean "elementary" as "easy"), it is very important to be acutely aware of what exactly you have to work with, what rules you have at your disposal. I assume one rule is that you can add the same number to both sides of an equation or equality, but do you really have a rule that allows you to subtract the same thing from both sides? Do you have a rule which lets you go directly from $x-a-(-a)$ to $x-0$, or are there actually intermediate steps via, for instance, the expression $x+(-a)-(-a)$? $\endgroup$ – Arthur Sep 14 '17 at 13:16
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    $\begingroup$ +1 for Arthur's comment: "it is very important to be acutely aware of what exactly you have to work with, what rules you have at your disposal." $\endgroup$ – user9464 Sep 14 '17 at 13:18
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    $\begingroup$ What I have written in my comment above might seem pedantic, and in a sense it is, but on the other hand this is in principle the level that any proof is done. You can only use a shortcut once you've proven that the shortcut always works. And even then, each time you use the shortcut, you are really doing it the long way, you're just not writing it down. Of course, if you were to prove a theorem like Fermat's last theorem without shortcuts, the proof would probably fill a whole library and be completely unreadable. So we do need the shortcuts in practice. But be sure that it's justified. $\endgroup$ – Arthur Sep 14 '17 at 13:18
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    $\begingroup$ "Let's also ignore for a second that this is an introductory course. Would I need to prove this if I was asked to prove the theorem in an exam?" You would definitely like to ask your professor. $\endgroup$ – user9464 Sep 14 '17 at 13:22
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    $\begingroup$ Above, you seem to have written that you are unable to Latex the properties of the operations on the set of real numbers. If you could not do this, then either take a photo and upload it here, or look for a webpage with the rules, take a screenshot, and upload that. Because without the rules, we don't know what's right and what's wrong in your proof. $\endgroup$ – Teresa Lisbon Sep 14 '17 at 13:23
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$x - a - (-a) = b - (-a)$

What is "$-$"? Well, we do learn it to be an operation, and indeed you can define subtraction of reals because $\mathbb R$ is additive group, and then by definition it means that for all real $x$ there exists a real denoted by $-x$ with property $x+(-x) = -x+x = 0$. This allows us to define subtraction:

$$x-y:=x+(-y).$$

However, there are subtleties here, the first one being that additive inverse is unique (which can be easily proved from group axioms) and therefore the above will really be well defined operation.

But, what properties of subtraction do you know? For example, it is neither associative nor commutative and, because of that, ill-suited for proofs like these where you need to carefully pay attention to axioms.

More importantly, it is a mystery why you would choose to subtract $-a$ instead of just adding $a$. Note that $x-a-(-a) = x-a+a$.

Furthermore, what does $x-a-(-a)$ really mean? It should actually be $(x-a)-(-a)$ and after that one should invoke associativity to simplify.

With the above comments in mind, what I would do is the following:

\begin{align} x-a = b &\implies x+(-a) = b\\ &\implies (x+(-a))+a = b+a\\ &\implies x+(-a+a) = b+a\\ &\implies x+0 = b+a\\ &\implies x = b+a\\ \end{align}

You should be able to recognize what axiom was used in each line, except maybe the second line. So, what is going on there?

Well, reals come equipped with binary operation $+\colon\mathbb R\times\mathbb R\to \mathbb R$, either by construction or from axiomatic definition. Binary operation is by definition a function, so, for any real number $a$, we can define a new function $f_a\colon\mathbb R\to\mathbb R$ with formula $f_a(x) = x + a$. It is important that $f_a$ is function, since all functions must satisfy $$x = y\implies f(x) = f(y)$$ and letting $f = f_a$, we get

$$x = y \implies x + a = y + a,$$

which is what we did in the above second line.

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  • $\begingroup$ "More importantly, it is a mystery..." I used the opposite of $-a$ incorrectly. I was probably confused by thinking of something along the lines of $x + (-a) + (-(-a))$. One of the properties of + we are given is that for every $a \in R$ there exists $-a$ such that $a + (-a) = 0$. $\endgroup$ – user480624 Sep 14 '17 at 14:35
  • $\begingroup$ As for the fact that $+$ is a function, that is actually what I was missing. We actually did define binary operations just like you did here, but I didn't think about the fact that this means that I can use the properties of functions. By the way, am I using "property" correctly in this comment? $\endgroup$ – user480624 Sep 14 '17 at 14:36
  • $\begingroup$ @user8063888, yes, $x=y\implies f(x) = f(y)$ is one of the defining properties of a function. The other one is that it is left-total. $\endgroup$ – Ennar Sep 14 '17 at 14:46
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I encourage you to ask yourself the following questions:

  1. What do you mean for "="?

  2. What is $+$? It's first of all a $\textbf{function}$ $$+:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$$ So for any couple $(a,b)$ there is one and only one value $a+b$.

The fact that $+$ send a couple of real numbers to a real number (as it happens for example for $\mathbb{Z}$ or $\mathbb{Q}$) is related to a not so trivial question: what is a real number?

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Here is a proof with all steps spelled out:

$$ \begin{array}{rcll} x-a &=& b & \qquad\text{hypothesis} \\ (x-a)+a &=& b+a & \qquad\text{add $a$ to both sides} \\ (x+(-a))+a &=& b+a & \qquad\text{definiton of subtraction} \\ x+((-a)+a) &=& b+a& \qquad\text{associativity of addition} \\ x+0 &=& b+a & \qquad\text{definition of additive inverse} \\ x &=& b+a & \qquad\text{definition of additive neutral element} \\ \end{array} $$

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  • $\begingroup$ How do you define your variables? What stops you from trying to apply + function to a mixture of reals and integers? $\endgroup$ – James Arathoon Sep 14 '17 at 13:45
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To your second question. When you have two terms $t_1$ and $t_2$ which are equal $$ t_1 = t_2 $$ then since a function $f$ is well-defined we also have $$f(t_1) = f(t_2)$$ or in other words $$t_1 = t_2 \quad\Rightarrow \quad f(t_1) = f(t_2)$$ Your particular function is $f:\mathbb{R} \to \mathbb{R},\; x \mapsto x+a$. Since this function has an inverse function which is $f^{-1}: \mathbb{R}\to \mathbb{R},\; x \mapsto x-a$. We conclude: $$t_1 = t_2 \quad\Rightarrow \quad f(t_1) = f(t_2) \quad\Rightarrow \quad f^{-1}(f(t_1)) = f^{-1}(f(t_2)) \quad\Rightarrow \quad t_1 = t_2. $$ In a more compact notation this means $$t_1 = t_2 \quad\Leftrightarrow \quad f(t_1) = f(t_2)$$ which should vanish your concerns.

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As you say, $$x-a=b\implies (x-a)+a=b+a$$ because both members represent the same values before and after addition (actually, $\iff$ holds.)

Then by properties of the addition,

$$(x-a)+a=x+(-a+a)=x+0=x.$$

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