Context: Rings.

Are $x \cdot 0 = 0$ and $x \cdot 1 = x$ and $-(-x) = x$ axioms?

Arguably three questions in one, but since they all are properties of the multiplication, I'll try my luck...

  • 7
    $\begingroup$ your second one isn't usually true... $\endgroup$ Jul 21, 2010 at 19:47
  • 3
    $\begingroup$ Axioms for what? $\endgroup$ Jul 21, 2010 at 20:48

5 Answers 5


I will assume this is in the context of rings (e.g., real numbers, integers, etc). In this case, the axiom defining $0$ is that $x + 0 = x$ for all $x$. $x*0 = 0$ is a result of this since we have $x*0 = x*(0+0) = x*0 + x*0$ which implies $x*0 = 0$ (canceling one of the $x*0$'s).

I am guessing that for the second one you mean $x*1 = x$. This is a definition (axiom).

The third one is a consequence of the definition of $-x$ being the element such that $x + (-x) = 0$. For then we have $(-x) + x$ is also zero so that $x$ is the negative of $-x$.

  • $\begingroup$ Your assumption is correct! $\endgroup$ Jul 21, 2010 at 19:58
  • 2
    $\begingroup$ in other words, only the second one is an axiom. $\endgroup$
    – mau
    Jul 21, 2010 at 20:04
  • 3
    $\begingroup$ @guillermooo - Better to edit your question to state the assumption explicitly. P.S. That's you from Sublime right? :) $\endgroup$
    – Edan Maor
    Jul 22, 2010 at 10:28
  • $\begingroup$ It's me, yeah! :) $\endgroup$ Jul 22, 2010 at 22:36

The question is more profound than is initially seems, and is really about algebraic structures. The first question you have to ask yourself is where you're working:

In general, addition and multiplication are defined on a structure, which in this case is a set (basically a collection of "things") with two operators we call addition (marked $+$) and multiplication (marked $\cdot$ or $\times$ or $\ast$ or whatever). If this structure holds some properties, which are sometimes called axioms, then it is called a unit ring. The properties are:

  1. The set is closed under the operator $+$. That is, if $a$ and $b$ are in $R$, then $a+b$ is also in $R$.
  2. The set has a member which we mark as $0$. It has the properties that for every $a$ in $R$, $a+0 = 0$ and $0+a = 0$.
  3. The operation $+$ is commutative: $a+b = b+a$.
  4. The operation $+$ is associative: $(a+b)+c = a+(b+c)$.
  5. Every member has an additive inverse: for every $a$ in $R$ there is some $b$ in $R$ such that $a+b = 0$ (we mark $b$ as $-a$).
  6. The set is closed under the operator $*$. That is, if $a$ and $b$ are in $R$, then $a*b$ is also in $R$.
  7. The set has a member which we mark as $1$. It has the properties that for every $a$ in $R$, $a*1 = a$ and $1*a = a$.
  8. The operation $*$ is associative: $(a*b) * c = a * (b*c)$.
  9. Multiplication is distributive over addition: $a * (b+c) = a*b + a*c$ and $(a+b) * c = a*b + a*c$.

While this is a long list, and introduces the operator $+$ which is not even explicitly mentioned in the question, these properties are quite natural. For example, the integers $\{ \ldots, -2, -1, 0, 1, 2, \ldots\}$ we all know and love indeed form a ring. The real numbers also form a ring (in fact they form a field, which means they hold even more properties).

In regard to your question, the identity $x * 1 = x$ (I assume that's what you meant) is in fact an axiom - it is axiom 7. However, the other two identities are results of the other axioms.

First identity: We use axioms 2 and 9 to get $0 * x = (0+0) * x = 0*x + 0*x$ and then by adding $-(0*x)$ (the additive inverse of $0*x$, from axiom 5) to both sides, $0 = 0*x$.

Second identity: As stated in axiom 5, $-(-x)$ is just a notation used which means "the additive inverse of $-x$". To show that $-(-x) = x$ we need to show that $x$ is in fact the additive inverse of $-x$, or in other words that $x + -x = 0$ and $-x + x = 0$. But that's just what axiom 5 says, so we're done.

Last point: You might be wondering why did we have to go and introduce addition to answer a question about multiplication? Well, it so happens that without addition the other two identities are simply not true. For example, if we look at the positive integers $\{1, 2, 3, \ldots\}$ with only multiplication, then there is no $0$ there! Simply put, this is because the positive integers do not form a ring.


Let $A$ be a ring (which I will assume commutative).

Only the second sentence is an axiom: multiplication has a (provably unique) neutral element, id est, $a \in A$ such that $ax = xa = x$ for every $x \in A$; it's called one and its symbol is $1$.

The others are consequences of axioms:

  • The 3rd sentence comes from the existence of a symmetric to every element in the ring: for every $x \in A$, there is a (unique) element $y \in A$ such that $x + y = y + x = 0$. It is denoted $–x$. From this equality, also comes that the symmetrical of $y$ can only be $x$.

  • Analogously to the multiplication axiom above, addition has a neutral element, a (unique) number $b$ such that $x + b = b + x = x$ for each $x \in A$; it's called zero and its symbol is $0$. The first sentence is deduced by using the inverse additive (for some $a$), distributivity and multiplicative associativity axioms: $x · 0 = x · (a + (–a)) = xa + x(–a) = xa + (–xa) = 0$.


x*1 =x is axiom.

The rest is not.

Let me pencil push the rest :D. It's been a while.

=x*0+0 ' adding 0 does no harm
=x*0+(x*0+(-(x*0))) ' 0 is the sum of x* 0 and it's additive inverse
=(x*0+x*0)+(-(x*0)) ' associativity.
=x*(0+0)+(-(x*0)) 'distributive
=x*(0)+(-(x*0)) '0+0 is 0
=x*0+(-(x*0)) '(0) is just 0
=0 ' the sum of x*0 and it's own additive inverse is 0

Notice that I add +(-(x*0)) to mean add by the additive inverse of (x*0).

From now one I would write +(-x) by simply -. So we define a subtraction as adding by the additive inverse. (is this the standard definition of substraction?)

Now, let's do -(-x)

=-(-x)+0 'adding 0 does no harm
=-(-x)+(-x+x) '0 is (-x+x) by definition of -x.

I wonder if the axiom only specify that only (x+ -x) is zero, then things can get more tricky. We would need the commutative axiom as reinforcement

=(-(-x)+(-x))+x ' by associative. 
=0+x ' the additive inverse of (-x) added by (-x) is 0
=x 'by definition of 0

I sort of like this kind of derivation because it follows the pattern of

A=...=...=...=...=...C which is better than to go half way and then use logic to imply stuffs.


Answering what you asked, the question is poorly formed because you didn't specify the theory you are talking about. To do that, well, you must define symbols, deduction rules... and axioms.

Answering what you probably meant to ask, no. Those are properties that $-$ and $\times$ have in the $\mathbb Z$ set, not axioms. They can be deduced from the operations themselves.


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