# Does $(x) = x$ follow from the axioms of the real numbers, or is it actually an axiom?

I'm working through a real analysis book. I'm still at the very beginning, and the current section is about the real number system. Specifically, the author lists 11 axioms (associativity, commutivity, identities, etc.).

However, I can't figure out how to show that $(x) = x$. Do we just assume this is true? By $(x)$, I just mean $x$ with parentheses.

The book I'm reading is Foundation of Mathematical Analysis by Richard Johnsonbaugh. I'm in chapter 3, The Real Number System.

This book does not go through the set-theoretic construction of the real number system, but rather it simply lists axioms. The axioms are as follows:

A1: There is a binary operation called addition and denoted + such that if $x$ and $y$ are real numbers, $x+y$ is a real number.

A2: Addition is associative. $$(x+y) + z = x + (y+z)$$ A3: Addition is commutative. $$x+y = y + x$$ A4: An additive identity exists. There exists a real number denoted 0 which satisfies $$x+0 = x = 0+x.$$

I could list the rest of the axioms, but I'm not sure it would help with this discussion.

• What is $(x)$ supposed to denote here?! – rschwieb Nov 16 '16 at 13:41
• @rschwieb (x) should stand for the cauchy-sequences converging to $x$. – Peter Nov 16 '16 at 13:42
• You could say $(x+0)+0=x+0$, so $(x)+0=x+0$, so $(x)=x$. But this is sort of meaningless. Parentheses are merely used to denote the order in which operations will be performed. Otherwise what does $x+y+z$ mean? Do we add $x$ and $y$ first or $y$ and $z$? (This order matters for some operations, but not addition of real numbers precisely because addition is associative.) Parentheses can be added and removed freely as long as they do not change the meaning of the expression. – kccu Nov 16 '16 at 14:08
• @rschwieb I think some of the confusion is that I didn't know (x) had other meanings than simple parentheses around a variable. – EternusVia Nov 16 '16 at 15:07
• No, $(x+0)+0=x+0$ follows from A4. This axiom says if you add $0$ To $x+0$ you get $x+0$ back. But we aren't required to write this as $(x+0)+0$ rather than $x+0+0$ to indicate the order in which the operations aren't performed. Again, the parentheses merely indicate order of operations, so you are free to remove them from $(x)$ since there are no operations in that expression. – kccu Nov 16 '16 at 16:21

I didn't know $(x)$ had other meanings than simple parentheses around a variable.

Parentheses are being used in these axioms purely to clarify the order in which operations are taking place. If you don't know an operation $\ast$ is associative, then something like $a\ast b\ast c \ast d$ makes no sense at all, since you only know what $a\ast b$ is for two things $a,b$. Parenthesis can be used to group pairs so that it's clear what to 'multiply' first: say $(a\ast b)\ast(c\ast d)$ or $a\ast(b\ast( c\ast d))$.

If you're running into something like $(a+0)$, then formally rewriting this as $(a)$ isn't a meaningful thing to do. The parenthesis are a statement about the operation that's taking place, and there's no operation within the string "$(a)$".

It would be just fine to say that $x+0=(x+0)$, in which case the parentheses are redundant and not saying anything.

Can't we only say that $(x+0)+0=(x+0)$ from the axioms?

Sure, you can: firstly the right hand side can be replaced by $x+0$ alone, so that you have

$$(x+0)+0=x+0$$

According to the axiom that says $y+0=y$ for all $y$, the above holds with $y=x+0=(x+0)$.

## Never mind this stuff below this line

original answer back when we thought the OP was stuck on something harder

When defining the real numbers using equivalence classes of Cauchy sequences of rational numbers (as it appears you are doing, denoting the equivalence class of a sequence $s$ as $(s)$) every element of your model of $\mathbb R$ is an equivalence class of the form $(s)$ for some rational Cauchy sequence $s$.

These equivalence classes are equal or unequal as sets, and there is no ambiguity like "$s=(s)$" at this point. However, there comes the time to look for where the rationals went in this process. They aren't literally in the set of equivalence classes, but there ought to be something that looks just like them.

Then one gets the idea that if we have $q\in \mathbb Q$, the sequence which is constantly $q$, call it $\vec q$, is a good candidate for representing $q$. But this sequence is not an element of the set of equivalence classes either, so it is not itself our new representative. Our representative ought to be $(\vec q)$, which is an equivalence class of rational Cauchy sequences. One can check that the map $\mathbb Q\to \mathbb R$ taking $q\mapsto (\vec q)$ is an injective ring homomorphism, so that we have found a copy of $\mathbb Q$ inside our model of $\mathbb R$.

At this point we say "hey you know what, among friends let's just say $q=(\vec q)$ and make the notation simpler." Everyone agrees, and you now have a new convention that lets you identify the rationals in your model of the reals. The equality isn't a mathematical one, it's just a conventional one that identifies elements of $\mathbb Q$ with their images in your model of $\mathbb R$.

Once this construction is over, you don't even really need to refer to the representatives $(s)$ at all. It's mainly important for proving that the real number axioms hold for the model. After that, it is "your old friend $\mathbb R$ and you can just say "let $r\in \mathbb R$ rather than referring to this underlying construction.

Something very similar is often done when defining $\mathbb C$ as addition and multiplication on $\mathbb R\times \mathbb R$. The element $(0,1)$ gets relabeled as $i$, and it turns out that the elements of the form $(r,0)$ for $r\in \mathbb R$ look exactly like $\mathbb R$, so it's OK to confuse $r$ with $(r,0)$ afterwards.

And when defining $\mathbb Q$ as equivalence classes of relation on $\mathbb Z\times \mathbb Z\setminus\{0\}$, we confuse an integer $z\in \mathbb Z$ with the equivalence class of $\frac{z}{1}$, etc.

$(x) = x$ is purely a point of mathematical grammar, and has nothing to do with real numbers or arithmetic. The parentheses don't actually mean anything: they just enforce grouping. e.g. because of the parentheses, one cannot read $(1+2)\times 3$ as the sum of $1$ and $2 \times 3$.

Actually, depending on the fine details of grammar, $(x)$ might even be grammatically incorrect; e.g. one form of the grammar for fully parenthesized arithmetic expressions only puts parentheses around the result of a binary operation, so $(x)$ doesn't make sense because there is no operation appearing in $x$.