You met the rule a \times 1 = a and probably thought: obviously. One is one. How could it ever fail to be the thing that leaves numbers alone? But the word "identity" is doing more work than you think. It is not the number 1 that is special — it is the role 1 plays inside the real numbers. In other systems, the role is played by something else entirely. And in some systems, nothing plays the role at all.

The real claim is about a role, not a number

In the real numbers, the multiplicative identity is the unique element e such that

a \times e = e \times a = a

for every a \in \mathbb{R}. That element turns out to be 1. But notice the definition does not name the number 1 anywhere. It names a role — "the thing that leaves every element unchanged under multiplication." The real-number system happens to have such an element, and we call it 1. Inside \mathbb{R}, under ordinary \times, the rule never fails.

So if the question is "within real-number multiplication, can 1 fail as the identity?" — the answer is flatly no. The field axioms of \mathbb{R} guarantee it.

The interesting question is what happens when you leave \mathbb{R}.

The identity changes when the system changes

Consider 2 \times 2 matrix multiplication. The "numbers" are now matrices, and \times is matrix multiplication. What matrix I satisfies A \times I = I \times A = A for every A? Not the matrix whose entries are all 1. It is the matrix

I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

called the identity matrix. Multiply any 2 \times 2 matrix by this and you get the same matrix back. The number 1 has been replaced by a matrix that plays the same role. Same rule, different identity.

Why: the definition of identity is "leaves every element alone under the operation." In matrix world, the element that does that isn't the number 1 — it is a matrix with 1s on the diagonal and 0s elsewhere. The role stays, the occupant changes.

The pattern keeps going:

In each case the role is the same — the element that leaves everything alone — but the thing filling the role is whatever the system demands.

Different systems have different multiplicative-role identitiesFour columns, each labelled with a mathematical system and the element that plays the identity role inside it. Column one: real numbers, identity equals the number one. Column two: two by two matrices, identity equals the two by two identity matrix. Column three: functions under composition, identity equals the function f of x equals x. Column four: sets under union, identity equals the empty set. A footer line says: same role, different occupant. Real numbers under × 1 a × 1 = a 2×2 matrices under matrix × [1 0] [0 1] A × I = A Functions under ∘ f(x) = x g ∘ f = g Sets under ∪ ∅ A ∪ ∅ = A same role — different occupant
Each system has its own identity for its own "multiplication-like" operation. The number $1$ is the identity only when the operation is ordinary multiplication of real numbers. Swap the operation and the identity morphs to fit.

Try it: spot the identity

Pick an operation and an identity candidate. Does the candidate leave every input alone?

Interactive: slide a to see that a × 1 = a alwaysA number line from negative five to five with a single draggable red point labelled a. Two readouts above the line show the current value of a and the product a times one. No matter where you drag the point, the two values match — confirming that one is the multiplicative identity for real numbers. −5 0 5 ↔ drag the red point
Drag $a$ anywhere along the line. The product $a \times 1$ tracks $a$ exactly. This is what "identity" means — perfect tracking, no distortion, zero effect.

Can an identity fail to exist at all?

Yes. Consider the set of positive even integers \{2, 4, 6, 8, \dots\} under ordinary multiplication. Is there an element e in this set with a \times e = a for every a? The only real number that does the job is 1, and 1 is not in the set. So this system has no multiplicative identity — the role is vacant.

Consider the set of 2 \times 2 matrices with determinant zero under matrix multiplication. There is no singular matrix that leaves every singular matrix alone — again the role is vacant. Systems like this are studied under names like semigroup (associative operation, no identity required) or magma (just an operation, no guarantees) in abstract algebra.

Why: having an identity is a property a system might or might not have. The field axioms of \mathbb{R} bake this property in — that is what makes \mathbb{R} a field. Weaker systems can skip it.

What about zero — does it ever play identity?

Zero is the identity for addition, not multiplication. a + 0 = a is the definition of additive identity. This is a separate role entirely. One system can have two identity elements, one per operation — that is exactly how \mathbb{R} works, with 0 for + and 1 for \times. In multiplication, zero behaves in the opposite way: a \times 0 = 0 absorbs every input into itself. That makes zero an absorbing element for multiplication, not an identity. Same symbol, very different role.

The takeaway

"Is 1 always the multiplicative identity?" is really two questions in disguise.

  1. Inside real-number multiplication, is 1 the identity? Yes, always. That is a field axiom and never fails.
  2. Is the number 1 always what plays the multiplicative-identity role? No. The role belongs to whatever element of the system leaves all others alone under the operation — matrices, functions, sets, and strings each have their own occupant for the role, and some systems have no occupant at all.

When your textbook writes "the multiplicative identity is 1," it is making a statement about the real numbers, not about the number 1 itself. Once you start working with matrices or functions or sets, you look for whatever is playing the role in that system, and call that thing the identity.

Related: Operations and Properties · Identity and Inverse: The Two 'Undo' Buttons of Arithmetic · Number Systems · Real Numbers and Their Properties