Does the Set of Real Numbers Form a Group?

The phrase “the set of real numbers” does not specify a group by itself: a group consists of a set together with an operation. Under the usual addition operation, the real numbers form a group. Under usual multiplication, all real numbers do not, although the nonzero real numbers do.

If the question concerns a different operation on $\mathbb{R}$, its formula is needed to decide whether it gives a group. The checks below show how to handle the two standard operations.

What you must check

A group is a set $G$ with a binary operation that satisfies four conditions:

1. Closure: combining any two elements of $G$ gives an element of $G$.
2. Associativity: $(a*b)*c=a*(b*c)$ for all $a,b,c\in G$.
3. Identity: there is an element $e\in G$ such that $e*a=a*e=a$ for every $a\in G$.
4. Inverses: each $a\in G$ has an element $a^{-1}\in G$ with $a*a^{-1}=a^{-1}*a=e$.

A group operation must be specified: the same set can be a group under one operation and fail to be one under another.

The real numbers under addition

Consider $(\mathbb{R},+)$. It satisfies the group axioms:

• The sum of two real numbers is real, so addition is closed on $\mathbb{R}$.
• Addition is associative: $(a+b)+c=a+(b+c)$.
• The identity is $0$, since $a+0=0+a=a$.
• The additive inverse of $a$ is $-a$, since $a+(-a)=0$.

Therefore $(\mathbb{R},+)$ is a group. It is also abelian, meaning its operation is commutative: $a+b=b+a$.

For example, if $a=3.5$, its additive inverse is $-3.5$, and $3.5+(-3.5)=0$. This verifies the inverse condition for that element; the general inverse is $-a$ for every real $a$.

The real numbers under multiplication

For $(\mathbb{R},\cdot)$, multiplication is closed and associative, and the identity is $1$. But the inverse condition fails for $0$: there is no real number $x$ such that $0\cdot x=1$. Consequently, all real numbers under multiplication do not form a group.

Remove zero and consider $(\mathbb{R}\setminus\{0\},\cdot)$. This is a group:

• The product of two nonzero real numbers is nonzero.
• Multiplication is associative.
• The identity is $1$, which is nonzero.
• Every nonzero real $a$ has the multiplicative inverse $1/a$, also nonzero.

For instance, the multiplicative inverse of $-4$ is $-1/4$, because $(-4)(-1/4)=1$. This group is abelian as well, since real multiplication is commutative.

Set and operation	Group?	Identity	Inverse of an element
$(\mathbb{R},+)$	Yes	$0$	$-a$
$(\mathbb{R},\cdot)$	No	$1$	Fails for $a=0$
$(\mathbb{R}\setminus\{0\},\cdot)$	Yes	$1$	$1/a$

If the operation is unusual

Suppose an exercise defines a new operation, such as $a*b=a+b+ab$. Do not assume it behaves like addition or multiplication. Check the axioms for that exact formula. For this example, $-1*b=-1$ for every real $b$, so $-1$ cannot have an inverse that produces an identity distinct from $-1$; in fact, solving the identity equations gives identity $0$, and $-1$ has no inverse. Thus $(\mathbb{R},*)$ is not a group.

For a custom operation, it can be efficient to find the identity first by solving $a*e=a$ and $e*a=a$ for all $a$. Then solve the inverse equation $a*b=e$. A single element with no inverse is enough to show that the structure is not a group. Associativity still needs to be checked unless it follows from a known construction.

Common mistakes

• Naming a set without an operation: “$\mathbb{R}$ is a group” is incomplete. Write $(\mathbb{R},+)$ or name the intended operation.
• Assuming a familiar operation works on the whole set: Multiplication has identity $1$, but that does not supply an inverse for zero.
• Confusing a group with a field: $\mathbb{R}$ with addition and multiplication is a field, but group claims refer to one operation at a time. Its additive structure is a group, and its nonzero elements form a group under multiplication.
• Checking only identity and inverses: A proposed operation must also be closed and associative.

If an exercise gives a specific operation, substitute it into each axiom; the answer can differ from the standard addition and multiplication cases.