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Being interested in the very foundations of mathematics, I'm trying to build a rigorous proof on my own that $a + b = b + a$ for all $\left[a, b\in\mathbb{R}\right] $. Inspired by interesting properties of the complex plane and some researches, I realized that defining multiplication as repeated addition will lead me nowhere (at least, I could not work with it). So, my ideas:

  • Defining addition $a+b$ as a kind of "walking" to right $\left(b>0\right)$ or to the left $\left(b<0\right)$ a space $b$ from $a$. Adding a number $b$ to a number $a$ (denoted by $a+b$) involves doing the following operation:

    1. Consider the real line $\lambda$ and its origin at $0$. Mark a point $a$, draw another real line $\omega$ above $\lambda$ such what $\omega \parallel \lambda$ and mark a point $b$ on $\omega$. Now, draw a line $\sigma$ such that $\sigma \perp \omega$ and the only point in commom between $\sigma$ and $\omega$ is $b$. Consider the point that $\lambda$ and $\sigma$ have in commom; this point is nicely denoted as $a + b$.

(Note that all my work is based here. Any problems, and my proof goes to trash)

  • This definition can be used to see the properties of adding two numbers $a$ and $b$, for all $a, b \in\mathbb{R}$.

  • Using geometric properties may lead us to a rigorous proof (if not, I would like to know the problems of using it).

So, I started:

  • $a, b \in\mathbb{N}$:

$a+b = \overbrace{\left(1+1+1+\cdots+1\right)}^a + \overbrace{\left(1+1+1+\cdots+1\right)}^b = \overbrace{1+1+1+1+\cdots+1}^{a+b} = \overbrace{\left(1+1+1+\cdots+1\right)}^b + \overbrace{\left(1+1+1+\cdots+1\right)}^a = b + a$

(Implicity, I'm using the fact that $\left(1+1\right)+1 = 1+\left(1+1\right)$, which I do not know how to prove and interpret it as cutting a segment $c$ in two parts -- $a$ and $b$. However, this result can be extended to $\mathbb{Z}$ in the sense that $-a$ $(a > 0)$ is a change; from right to left).

  • $a, b \in\mathbb{R}$:

Here, we have basically two cases:

  • $a$ and $b$ are either positive or negative;
  • $a$ and $b$, where one of them is negative.

Since in my definition $-b, b>0$ means drawing a point $b$ to the left of the real line, there's no big deal interpretating it; subtracting can be interpreted now. So, it starts:

$a + b = c$. However, $c$ can be cut in two parts: $b$ and $a$. Naturally, if $a>c$, then $b<0$ -- many cases can be listed. So, $c = b + a$. But $c = a + b$; it follows that $a + b = b + a$. My questions:

Is there any problem in using my definition of adding two numbers $a$ and $b$, which uses many geometric properties? Is there any way to solve it from informality? Is there anything right here?

Thanks in advance.

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Axioms can't be proved. –  Gastón Burrull Jun 16 '12 at 0:34
Geometric definitions are fine, but if you really want to look into the foundations of mathematics you need to start with a set of axioms. I would recommend getting a book which guides you with this in the beginning. Your argument more or less implicitly assumes some axioms about the geometric properties of lines, which is similar to what Euclid did, but is not really in the same spirit as the axiomatic approach to mathematics. –  Michael Boratko Jun 16 '12 at 0:34
Generalization of PA to R dont have sense. Who is R? You must start with a definition of R. R is a field with order topology and complete. If you looking for another definition without word "field" you will need to prove that is an equivalent definition. –  Gastón Burrull Jun 16 '12 at 0:40
Many precalculus students don't have any idea what is the definition of real numbers, and trying to prove teorems without a formal previuos definition. Almost all precalculus students known the correct definition of the field Q, but never R (this fact requires some preparation in real analysis but I think that is the main problem in missunderstanding concepts in first undergraduate year) –  Gastón Burrull Jun 16 '12 at 0:47
@IanMateus: you can't prove anything about addition in real numbers until you define precisely what you mean by addition and real numbers, otherwise there's no basis to prove the theorem from. The simplest (rigorous) definition of real numbers is, I believe, by Dedekind cuts, which immediately supplies the definition of addition, and the proof that it is commutative is obvious (straight from the fact that addition of rationals is commutative, which in turn follows the fact for natural numbers, which is sometimes taken as an axiom, or else can be proved by induction). –  tomasz Jun 16 '12 at 0:50
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5 Answers

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First you need to define $\mathbb{R}$ in your construction!

To define $\mathbb{R}$, one way is to go about defining $\mathbb{N}$, then defining $\mathbb{Z}$, then defining $\mathbb{Q}$ and then finally defining $\mathbb{R}$. Once you have these things set up, proving associativity, commutativity of addition over reals essentially boils down to proving associativity, commutativity of addition over natural numbers.

As said earlier, one goes about first defining natural numbers. For instance, $2$ as a natural number is defined as $2_{\mathbb{N}} = \{\emptyset,\{\emptyset\} \}$. We will use the notation that $e$ is $1_{\mathbb{N}}$ and $S(a)$ be the successor function applied to $a \in \mathbb{N}$. Then we define addition on natural numbers using the successor function. Addition on natural numbers is defined inductively as $$a +_{\mathbb{N}} e = S(a)$$ $$a +_{\mathbb{N}} S(k) = S(a+k)$$ You can also define $\times_{\mathbb{N}},<_{\mathbb{N}}$ on natural numbers similarly.

Then one defines integers as an equivalence class (using $+_{\mathbb{N}}$) of ordered pairs of naturals i.e. for instance, $2_{\mathbb{Z}} = \{(n+_{\mathbb{N}}2_{\mathbb{N}},n):n \in \mathbb{N}\}$. You can similarly, extend the notion of addition and multiplication of two integers i.e. you can define $a+_{\mathbb{Z}} b$, $a \times_{\mathbb{Z}}b$, $a <_{\mathbb{Z}} b$. Addition, multiplication and ordering of integers are defined as appropriate operations on these set.

Then one moves on to defining rationals as an equivalence class (using $\times_{\mathbb{Z}}$) of ordered pairs of integers. So $2$ as a rational number, $2_{\mathbb{Q}}$ is an equivalence class of ordered pair $$2_{\mathbb{Q}} = \{(a \times_{\mathbb{Z}} 2_{\mathbb{Z}},a):a \in \mathbb{Z}\backslash\{0\}\}$$ Again define $+_{\mathbb{Q}}, \times_{\mathbb{Q}}$, $a <_{\mathbb{Q}} b$. Addition, multiplication and ordering of rationals are defined as appropriate operations on these set.

Finally, a real number is defined as the left Dedekind cut of rationals. i.e. for instance $2$ as a real number is defined as $$2_{\mathbb{R}} = \{q \in \mathbb{Q}: q <_{\mathbb{Q}} 2_{\mathbb{Q}}\}$$ Addition, multiplication and ordering of reals are defined as appropriate operations on these set.

Once you have these things set up, proving associativity, commutativity of addition over reals essentially boils down to proving associativity, commutativity of addition over natural numbers.

Here are proofs of associativity and commutativity in natural numbers using Peano's axiom.

Associativity of addition: $(a+b) + c = a + (b + c )$


Let $\mathbb{S}$ be the set of all numbers $c$, such that $ (a+b) + c = a + (b + c )$, $ \forall a,b \in \mathbb{N}$.

We will prove that $ e$ is in the set and whenever $k \in \mathbb{S}$, we have $S(k) \in \mathbb{S}$. Then by invoking Peano’s axiom (viz, the principle mathematical induction), we get that $\mathbb{S} = \mathbb{N}$ and hence $ (a+b) + c = a + (b + c )$, $ \forall a,b \in \mathbb{N}$.

First Step:

Clearly, $ e \in \mathbb{S}$. This is because of the definition of addition.

$ (a+b)+e = S(a+b)$ and $ a + S(b) = S(a+b)$

Hence $ (a+b)+e = a + S(b) = a+ (b+e)$

Second Step:

Assume that the statement is true for some $ k \in \mathbb{S}$.

Therefore, we have $ (a+b)+k = a+(b+k)$.

Now we need to prove, $ (a+b) + S(k) = a+(b+S(k))$.

By definition of addition, we have $ (a+b)+S(k) = S((a+b) + k)$

By induction hypothesis, we have $ (a+b)+k = a+ (b+k)$

By definition of addition, we have $ b + S(k) = S(b+k)$

By definition of addition, we have $ a+S(b+k) = S(a+(b+k))$

Hence, we get,

$ (a + b) + S(k) = S((a+b) + k) = S(a+ (b+k)) = a + S(b+k) = a+ (b + S(k))$

Hence, we get,

$ (a+b) + S(k) = a + (b+S(k))$

Final Step:

So, we have $ e \in \mathbb{S}$. And whenever, $k \in \mathbb{S}$, we have $S(k) \in \mathbb{S}$.

Hence, by principle of mathematical induction, we have that $\mathbb{S} = \mathbb{N}$ i.e. the associativity of addition, viz, $$(a+b) + c = a + (b+c)$$

Commutativity of addition: $ m + n = n + m$, $ \forall m,n \in \mathbb{N}$.


Let $ \mathbb{S}$ be the set of all numbers $ n$, such that $ m + n = n + m$, $ \forall m \in \mathbb{N}$.

We will prove that $ e$ is in the set $ \mathbb{S}$ and whenever $ k \in \mathbb{S}$, we have $ S(k) \in \mathbb{S}$. Then by invoking Peano's axiom (viz, the Principle Mathematical Induction), we state that $ \mathbb{S}=\mathbb{N}$ and hence $ m + n = n + m$, $ \forall m,n \in \mathbb{N}$.

First Step:

We will prove that $ m + e = e + m$ and hence $ e \in \mathbb{S}$.

The line of thought for the proof is as follows:

Let $ \mathbb{S}_1$ be the set of all numbers $ m$, such that $ m + e = e + m$.

We will prove that $ e$ is in the set $ \mathbb{S}_1$ and whenever $ k \in \mathbb{S}_1$, we have $ S(k) \in \mathbb{S}_1$. Then by invoking Peano's axiom (viz, the Principle Mathematical Induction), we state that $ \mathbb{S}_1=\mathbb{N}$ and hence $ m + e = e + m$, $ \forall m \in \mathbb{N}$.

To prove: $ e \in \mathbb{S}_1$

Clearly, $ e + e = e + e$ (We are adding the same elements on both sides)

Assume that $ k \in \mathbb{S}_1$. So we have $ k + e = e + k$.

Now to prove $ S(k)+ e = e + S(k)$.

By the definition of addition, we have $ e + S(k) = S(e + k)$

By our induction step, we have $ e + k = k + e$.

So we have $ S(e+k) = S(k+e)$

Again by definition of addition, we have $ k + e = S(k)$.

Hence, we get $ e + S(k) = S(S(k))$.

Again by definition of addition, $ p + e = S(p)$, which gives us $ S(k) + e = S(S(k))$.

Hence, we get that $ S(k+e) = S(k) + e$.

So we get,

$ e + S(k) = S(e+k) = S(k+e) = S(S(k)) = S(k) + e$.

Hence, assuming that $ k \in \mathbb{S}_1$, we have $ S(k) \in \mathbb{S}_1$.

Hence, by Principle of Mathematical Induction, we have $ m + e = e + m$, $ \forall m \in \mathbb{N}$.

Second Step:

Assume that $ k \in \mathbb{S}$. We need to prove now that $ S(k) \in \mathbb{S}$.

Since $ k \in \mathbb{S}$, we have $ m + k = k + m$.

To prove: $ m + S(k) = S(k) + m$.


By definition of addition, we have $ m + S(k) = S(m+k)$.

By induction hypothesis, we have $ m + k = k + m$. Hence, we get $ S(m+k) = S(k+m)$.

By definition of addition, we have $ k + S(m) = S(k+m)$.

Hence, we get $ m + S(k) = S(m+k) = S(k+m) = k + S(m)$.

We are not done yet, since we want to prove, $ m + S(k) = S(k) + m$.

So we are left to prove $ k + S(m) = S(k) + m$.

$S(k) +m = (k+e) + m = k + (e+m) = k + (m+e) = k + S(m)$.

Hence, we get $ m + S(k) = S(k) + m$.

Final Step:

So, we have $ e \in \mathbb{S}$. And whenever $ n \in \mathbb{S}$, we have $ S(n) \in \mathbb{S}$.

Hence, by Principle of Mathematical Induction, we have the commutativity of addition, viz,

$ m + n = n + m$, $ \forall m,n \in \mathbb{N}$.

We might think that associativity is harder/lengthier to prove than commutativity, since associativity is on three elements while commutativity is on two elements.

On the contrary, if you look at the proof, proving associativity turns out to be easier than commutativity.

Note that the definition of addition, viz $m + S(n) = S(m+n)$, incorporates the associativity $m+(n+e) = (m+n)+e$.

For commutativity however, we are changing the roles of $m$ and $n$, (we are changing the "order") and no wonder it is "harder/lengthier" to prove it.

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@PeterTamaroff: Yes. But I have written "Once you have these things set up, proving associativity, commutativity of addition over reals boils down to proving associativity, commutativity of addition over natural numbers." –  user17762 Jun 16 '12 at 1:26
The TL;DR effect tricked me! (+1) =) –  Pedro Tamaroff Jun 16 '12 at 1:27
@PeterTamaroff Feel free to comment. I have written this from whatever little I know by reading from online and other resources. So I would more than appreciate if you point out some mistakes or if you want to leave any other comment. –  user17762 Jun 16 '12 at 1:29
I really like the answer. It I guess you can say that $_\Bbb N e$ is the identity (I pressume). Maybe it is worth mentioning what an equivalence relation, and hence, what an equivalence class is, since $a = b$ and $ a \equiv b$ are equivalence relations known to most people in the math circle. –  Pedro Tamaroff Jun 16 '12 at 1:35
@PeterTamaroff I have now mentioned the identity. Also, the $_{\mathbb{N}}$ before $e$ goes with the $+$ sign and not with the $e$. I will soon add something on equivalence class. –  user17762 Jun 16 '12 at 1:39
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Unfortunately, your method is not rigorous. There are too many undefined notions running around. For example, you never define what $\mathbb{R}$ is in your construction, or what properties it has. To answer your question, in the usual construction of the real numbers, the fact that $a+b=b+a$ is taken as axiomatic (along with several other axioms, called field axioms).

You mention that you are a precalculus student. At this stage in your mathematical development, you should just take it on faith that you can add numbers. It's intuitively obvious, and there are much more interesting things you can be doing with your time than worrying about foundational issues. Learn some elementary number theory! Learn some combinatorics! Find a book of challenging elementary problems in mathematics and work through it! There is a time for rigor and proof, but it comes after developing fundamental intuitions about the subject. The whole point of the formalizations of the real numbers is to make our intuitions as precise and error-free as possible, so putting the formalization before the intuition is the wrong way to go about things here, in my opinion.

If you are still interested in foundational issues, there are many books available that cover the foundations of mathematics. You would probably want an introductory book on set theory. For an exposition of the construction of the reals and their properties, I recommend Foundations of Analysis by Edmund Landau. But without any experience with proofs, such books are quite difficult to understand.

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Please don't let the comments discourage you - it is good that you are trying to understand the fundamentals of Mathematics, and your effort is to be applauded.

In order to prove a theorem, one must have in place a set of axioms and definitions (as well as acceptable rules of logic) from which the theorem can be deduced. It is quite helpful to see some specific examples of how this is achieved before striking out on your own. In addition (pun intended), it is helpful to have a book (or better yet, a teacher) which can guide your proofs along.

The answer to your question is that your proof is missing some key parts, most importantly the axiomatic framework. Your proof seems to boil down to a proof by intuition, where you define addition in a geometric way and rely on the geometric intuition that no matter which way you perform your construction, the total length would be the same. This is not enough, and to understand why you need to see what is enough.

I would recommend starting a book on set theory, which often is the fist place one encounters proofs of this nature. This free online book looks accessible to someone with your background: http://math.boisestate.edu/~holmes/holmes/head.pdf

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You should first think how you define $\alpha$, a real number. One of the classical construction is defining it as a set. This is a construction based on Dedkind cuts, defined as follows:


A real number is a set of rational numbers $\mathrm {\mathbf A}$ such that

$(1)$ If $x \in \mathrm {\mathbf A}$ and $y < x$ then $y \in \mathrm {\mathbf A}$.

$(2)$ If $x \in \mathrm {\mathbf A}$, there exists another $y$ such that $y \in \mathrm {\mathbf A}$ and $y >x$ - viz, $\mathrm {\mathbf A}$ has no maximal element.

$(3)$ $\mathrm {\mathbf A}$ is not empty - viz $\mathrm {\mathbf A} \neq \emptyset$

$(4)$ $\mathrm {\mathbf A} \neq \mathbf Q$.

The set of all real numbers is noted $\mathbf R$

These sets are also called Dedekind cuts, honoring Dedekind, who first considered them. The classical example of a Dedekind cut is $\sqrt{ \mathbf {2}} = \{ x : x^2 < 2 \text{ or } x <0\}$. This set defines the real number $\sqrt 2 $.

Then we define the sum $ \large {\bf +}$ of two real numbers (different from $+$, the sum of rationals) as follows


If $\mathrm {\mathbf A} $ and $\mathrm {\mathbf B} $ are real numbers, then

$$\mathrm {\mathbf A} {\large {\bf +}}\mathrm {\mathbf B}= \{x:x=y+z \text{ ; for some } y \in \mathrm {\mathbf A} \text{ and some } z \in \mathrm {\mathbf B} \} $$

Note that with this definition, the proofs that

$${\bf{A}} {\large {\bf +}} {\bf{B}} = {\bf{B}} {\large {\bf +}} {\bf{A}}$$


$$({\bf{A}} {\large {\bf +}} {\bf{B}}){\large {\bf +}}{\bf{C}} = {\bf{B}} {\large {\bf +}} ({\bf{A}}{\large {\bf +}}{\bf{C}})$$

directly follow from the fact that for any $x,y,z$ rational numbers

$$\eqalign{ & x + y = y + x \cr & x + \left( {y + z} \right) = \left( {x + y} \right) + z \cr} $$

We can define then $\mathbf < $ for two real numbers, as:


If ${\bf{A}}$ and ${\bf{B}}$ are real numbers, then ${\bf{A}}\mathbf{<} {\bf{B}}$ means that ${\bf{A}}$ is contained in ${\bf{B}}$, but ${\bf{A}} \neq {\bf{B}}$.

Note that $\mathbf > $ $\mathbf \leq $ and $\mathbf \geq $ canall be defined analogously.

Then one can prove the following:

THEOREM (Spivak)

If $A$ is a set of real numbers, $A \neq \emptyset$ and $A$ is bounded avobe, then $A$ has a least upper bound, or supremum.

Note then that if we consider the set $\mathbf R$ along with $+$, $\leq$, $\cdot$ then

$(\mathbf 1)$ Then $(\mathbf R,+,\cdot)$ is a field, meaning the usual properties of addition and multiplication hold (along with the relations between them) , along with the existance of an identity for multiplication ($1$), and identity for the sum ($0$), and the existance of inverses for both operations (excluding $0$ in multiplication).

$(\mathbf 2)$ The field is ordered, in the sense the relation $\leq$ is a total order.

$(\mathbf 3)$ It is complete, in the sense that every set of real numers which is not the empty set and has an upper bound, has a least upper bound.

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Your example for $\sqrt{2}$ is slightly off, I believe. Don't you also need to include negative numbers less than $-2$? –  Potato Jun 16 '12 at 1:01
Your definition is little ambiguous. "A real number is a set of rational numbers" sounds like a subset of Q. –  Gastón Burrull Jun 16 '12 at 1:03
@GastónBurrull I don't know what amiguity you're talking about. –  Pedro Tamaroff Jun 16 '12 at 1:06
@PeterTamaroff Sorry, there's no typo. However, I fail to see how $\sqrt{ \mathbf {2}} = \{ x : x^2 < 2\}$ defines $\sqrt{ \mathbf {2}}$. –  Ian Mateus Jun 16 '12 at 1:07
@PeterTamaroff Perhaps I did not word my objection correctly. The Dedekind cut corresponding to to $\sqrt{2}$, as I learned it, should be $A = \{ a\in\mathbb{Q} : a^2 < 2 \lor a\le 0 \}$, which is larger than what you wrote. –  Potato Jun 16 '12 at 1:16
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Something of this kind was done by Hilbert a bit over a century ago, in his Foundations of Geometry. The axioms of the geometric substrate were laid out in great detail. Then an arithmetic was defined on the points of a particular line $\ell$, with addition defined in a way reminiscent of what you did, and then multiplication and division. For addition one picks an arbitrary point $O$ on $\ell$ to serve as $0$. For multiplication one needs another arbitrary point $P\ne O$, to serve as $1$.

Hilbert then showed that the points on the line, under these geometric operations, form what we now call a complete ordered field. This is of some interest, in that it shows that classical plane geometry, done correctly (that is, with the axioms missed by Euclid added), yields a structure isomorphic to the familiar coordinate plane $\mathbb{R}^2$.

However, in my opinion, this work of Hilbert had little long-term significance. Although it shows that the real numbers can be developed within a classical geometric framework, the standard approach continues to be the arithmetical one pioneered by Weierstrass, Dedekind, and Cantor. The natural numbers are either taken as fundamental or (later) defined set-theoretically. Then one uses set-theoretic tools to build in succession the integers, the rationals, and the reals.

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Cool! It makes me proud to see how everything is connected in Mathematics =) It is the kind of thing I was looking for. Thanks! –  Ian Mateus Jun 16 '12 at 1:34
Yes, everything is connected to everything else. –  André Nicolas Jun 16 '12 at 1:40
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