Throughout school we are taught that when something is multiplied by 1, it equals itself.

But as I am learning about higher level mathematics, I am understanding that not everything is as black and white as that (like how infinity multiplied by zero isn't as simple as it seems).

Is there anything in higher-level, more advanced, mathematics that when multiplied by one does not equal itself?

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    $\begingroup$ If the Multiplicative Identity is something else then 1 then there you have it. $\endgroup$
    – SAJW
    Commented Oct 2, 2016 at 3:05
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    $\begingroup$ One thing that is sort of an example is when you manipulate some complicated expression by multiplying by $\frac{f(x)}{f(x)}$. This does equal $1$ everywhere where $f(x)\ne 0$, but it is of course undefined where $f(x)=0$. So sometimes we change the domain of our function by multiplying by this version of "$1$". $\endgroup$
    – user137731
    Commented Oct 2, 2016 at 3:20
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    $\begingroup$ Somewhat pedantically, something is always equal to itself, even if it were say multiplied by $7$. What is different in that case is the result of the multiplication, not the something being multiplied. Which is why I would say "when something is multiplied by $1$, the result equals the original thing". But of course the looser formulation is common and poses no difficulty. $\endgroup$ Commented Oct 2, 2016 at 7:59
  • $\begingroup$ As an aside, I see no reason why "infinity multiplied by zero" would seem simple at any stage of one's development. $\endgroup$
    – user360874
    Commented Oct 3, 2016 at 5:39
  • $\begingroup$ "like how infinity multiplied by zero isn't as simple as it seems" Infinity is not a number (at least in traditional number systems). It can't be multiplied by anything. You could potentially take a limit as one factor increases without bound, but that isn't the same thing. $\endgroup$
    – jpmc26
    Commented Oct 3, 2016 at 6:50

6 Answers 6


Usually, if there is an operation called multiplication, it is defined as having an identity element. We call that element $1$. When we do that, we define $1\times x = x \times 1 = x$. Sometimes we only define one of the equalities because we have the power to derive the other one. If we don't have a $1$, we don't have a multiplicative identity.

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    $\begingroup$ So it is intrinsic to the operation of multiplication? $\endgroup$
    – esote
    Commented Oct 2, 2016 at 3:09
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    $\begingroup$ @Anonymous Pretty much. The $1$ is the identity, which might not be $1$. Look up what a mathematical group is. $\endgroup$ Commented Oct 2, 2016 at 3:16
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    $\begingroup$ You have to look at the definition of the operation. As a kid, the only multiplication is on the naturals and $1$ is the identity. Later we extend that to the rationals, the reals, and the complex numbers. In all of those cases, $1$ is the identity. We can also define arbitrary operations on arbitrary sets. When we get to that stage, it would be perverse to call an operation multiplication that did not have an identity and you would call that identity $1$. $\endgroup$ Commented Oct 2, 2016 at 3:16
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    $\begingroup$ "So it is intrinsic to the operation of multiplication?" - I'd rather say it's intrinsic to the 1: The 1 is usually defined exactly as the element that has the property of not changing other elements when being multiplied with them (and I don't know this as "identity element", but rather as the neutral element - similar to 0 being the neutral element for addition) $\endgroup$
    – Marco13
    Commented Oct 2, 2016 at 4:59
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    $\begingroup$ @OFRBG: And even with matrices, the identity in a matrix ring of interest is often called $1$. e.g. on the bottom of page 2 of Matrix Rings in my copy of Jacobson's algebra 1, quote, We define the unit matrix $1$ by $$1 = \left( \begin{matrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & \cdots & 1 \end{matrix} \right)$$ endquote. It may look odd at first, but it really doesn't lead to confusion, since nearly any situation where you might interpret $r$ either as a scalar or as a scalar matrix, expressions involving $r$ mean the same thing either way. $\endgroup$
    – user14972
    Commented Oct 2, 2016 at 16:45

As Ross Millikan has pointed out, mathematicians like to call multiplicative identities $1$ because we like things to behave familiarly.

There are exceptions to this though, especially when we're working with sets of numbers that are endowed with structures other than the ones we usually use. The easiest example that comes to mind is the set $S=\{0,2,4,6,8\}$ with addition and multiplication defined modulo $10$ (in other words, if I multiply or add numbers and they exceed $10$, then I take the remainder dividing by $10$. So $6+8\equiv 4$ modulo $10$ since $14=10+4$, while $4\cdot 8\equiv 2$ modulo $10$ since $4\cdot 8=32=3\cdot 10+2.$) If you sit down and multiply each element of $S$ by $6$, you will find that $6$ is the multiplicative identity, not $1$ (which isn't even in the set).

This is kind of cheating though because in some sense $S$ with this structure is "the same" as $\{0,1,2,3,4\}$ with addition and multiplication defined modulo $5$, and in this case $1$ is is the multiplicative identity.

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    $\begingroup$ This is a very good point. The field axioms say there must be a distinguished element $1$ which is the multiplicative identity. You have a field and the distinguished element is named $6$. I could also define a five element field with elements $\{0, george, bill, jim, steve\}$ As long as I get the addition and multiplication tables right it will work. As you indicate in the last paragraph, there will be an isomorphism onto the field of integers modulo $5$. Depending on how I match them up, any of the boys could be $1$, the multiplicative identity. $\endgroup$ Commented Oct 2, 2016 at 3:36
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    $\begingroup$ While interesting, this does not answer the original question. That question was whether multiplication by $1$ ever does not give back the original value, not whether there are other things than $1$ that have the property that multiplication by them is the identity. $\endgroup$ Commented Oct 2, 2016 at 8:03
  • $\begingroup$ @MarcvanLeeuwen: Well the question is based on a false assumption that the symbol "1" refers to a particular object (presumably the real number $1$), whereas it actually is merely a symbol and is usually defined axiomatically to satisfy $1 \cdot x = x \cdot 1 = x$ for every $x$ in the relevant domain. Of course, it is clearer to write something like "$1_S$" to indicate the domain $S$. This answer gives a clear example of a ring which is built from some integers but has a different identity from that of the integers (which is $1_\mathbb{Z}$). $\endgroup$
    – user21820
    Commented Oct 3, 2016 at 7:40

I don't think there's really any context where we want to relax the axiom $1a = a$. But sometimes, we want to relax the axiom $0a=0$. I've seen the terminology almost-semiring used to describe algebraic structures where addition and multiplication behaves as expected, except that $0a = 0$ might not be true. Almost-semirings arise naturally: for example, suppose $X$ is a set, $S$ is a semiring, and consider the collection of all partial functions $X \rightarrow S$. This is an almost-semiring, but if $X$ is non-empty, it won't satisfy $0a=0$. To see this, let $a$ denote any non-total function $X \rightarrow S$. The big difference between $1a=a$ and $0a=0$ is that $a$ occurs exactly once on each side of $1a=a$ (it's a "balanced" identity), whereas this isn't true of $0a=0$. This means that the identity $1a=a$ can be expressed using operads, whereas $0a=0$ cannot. The end result is that while most ways of building constructions will preserve the identity $1a=a$, many will not preserve $0a=0$. For another, similar example, try doing algebra in the powerset of a semiring. You'll quickly notice that $1A=A$ holds, but that $0A=0$ doesn't.


A key aspect to this question is realizing that 1 is nothing more than a symbol with a vertical line in it. It has no intrinsic meaning. It is mathematicians who choose to give it meaning.

As many have pointed out, it is very common to assign the multiplicative identity element the symbol 1. This is the symbol for the multiplicative identity in our usual arithmetic, and it turns out that this is very convenient for people to remember. However, it's just a symbol. Your multiplicative identity could be if you wanted. There might be some grumbling about your symbol choices, but it's legal.

Now 1 is also the symbol given to $Su(0)$, that is "the successor to 0". This meaning for the symbol 1 comes from addition, rather than multiplication. It happens to be that, in normal arithmetic, the number that comes after 0 ($Su(0)$) and the multiplicative identity are the same number. If I may borrow Glare's excellent example of modulo 10 addition and multiplication over the set $\{0, 2, 4, 6, 8\}$, the successor of 0 is 2, but the multiplicative identity on this ring is 6.

One valid reason you may see a lack of the symbol l is because of this situation. Because people often think about the number after 0 and the multiplicative inverse as being the same thing, one may choose not to use that symbol if it could cause confusion. In Glare's example, the successor to 0 and the multiplicative inverse are different. Maybe this is a good time to not use 1. (That being said, 1 as a multiplicative inverse is very common, so even though I say you could choose not to use it that way... people will).

Now I used numbers in that example. I used them for two reasons. One is because that's how Glare presented them in his answer. The other is because you and I are both very comfortable with how those symbols operate. I could have had addition and multiplication over $\{☀,☁,☂,☃,☄\}$ and provided you the following definitions for the addition and multiplication operators:

add  ☀ ☁ ☂ ☃ ☄       mul  ☀ ☁ ☂ ☃ ☄     
  ☀  ☀ ☁ ☂ ☃ ☄         ☀  ☀ ☀ ☀ ☀ ☀
  ☁  ☁ ☂ ☃ ☄ ☀         ☁  ☀ ☂ ☄ ☃ ☃
  ☂  ☂ ☃ ☄ ☀ ☁         ☂  ☀ ☄ ☃ ☃ ☁
  ☃  ☃ ☄ ☀ ☁ ☂         ☃  ☀ ☃ ☃ ☃ ☃
  ☄  ☄ ☀ ☁ ☂ ☃         ☄  ☀ ☃ ☁ ☃ ☂

The resulting math would be the same, but your anger at me for using nonstandard symbols might be justified. By using the common symbols 0, 2, 4, 6, and 8, in an environment where their behavior is very similar to how they are used in normal arithmetic, the whole process goes a lot smoother!

  • $\begingroup$ This is the only answer so far that makes very clear the distinction between the symbol "1" and the (natural) number $1$! $\endgroup$
    – user21820
    Commented Oct 3, 2016 at 7:47

Outside of the usual integers / rational numbers / real numbers that we learn about in grade school, the symbol "$1$" is often used as a placeholder for the multiplicative identity in a given algebraic structure, such as a ring or field.

To build such a structure, we start with some set, which we'll call $S$. Next, we define "operations" on this set, which are functions that input any two elements of $S$ and output a single element in $S$. Typically, we call these operations "addition" and "multiplication" and use the standard $\times$ and $+$ symbols (though these can be defined quite differently from grade school addition/multiplication). Finally, the mathematician will specify a list of axioms that this set and the operations should satisfy, such as associativity, commutativity, and so forth.

Among these axioms will often be a statement along the lines of "There exists a special element in $S$, which we will call the multiplicative identity and denote by the symbol "$1$" which has the property that $y \times 1 = 1 \times y = y$ for any element $y \in S$.


It may not be what you're looking for, but I believe this is related. In the case of limits, repeated multiplication by a limiting value of $1$ can have surprising effects.

Consider this: $$\left(1+\frac{1}{n}\right)^{n}=\left(1+\frac{1}{n}\right)\left(1+\frac{1}{n}\right)\left(1+\frac{1}{n}\right)\ldots$$

For large values of $n$, this appears to approach: $$\left(1+0\right)\left(1+0\right)\left(1+0\right)\ldots=1\cdot1\cdot1\ldots=1$$ $$1^{\infty}\stackrel{?}{=}1$$

However, this is not the true value of the limit. In reality: $$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}=e\gt1$$

It appears that infinite repeated multiplication by $1$ somehow becomes another value entirely. This is why $1^\infty$ is considered an indeterminate form.

  • $\begingroup$ It's repeated multiplication by a value approaching 1. Not by 1. $\endgroup$ Commented Oct 2, 2016 at 19:46
  • $\begingroup$ @immibis Yes, that's why I said it appears that way. $\endgroup$
    – crb233
    Commented Oct 2, 2016 at 19:50
  • $\begingroup$ This fits quite well with the OP's example of infinity times zero. $\endgroup$
    – Owen
    Commented Oct 3, 2016 at 0:11

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