Integers do not form Field

Corollary of Invertible Integers under Multiplication

The integers $\struct {\Z, +, \times}$ do not form a field.

Proof

For $\struct {\Z, +, \times}$ to be a field, it would require that all elements of $\Z$ have an inverse.

However, from Invertible Integers under Multiplication, only $1$ and $-1$ have inverses (each other).

$\blacksquare$

Examples

Example: $2$

Take $2$, for example.

In the field of rational numbers, we have that $2 \times \dfrac 1 2 = 1$ and so $2$ has an inverse in $\Q$.

But that inverse is not in $\Z$.

Sources

• 1964: Iain T. Adamson: Introduction to Field Theory ... (previous) ... (next): Chapter $\text {I}$: Elementary Definitions: $\S 1$. Rings and Fields: Example $1$
• 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $4$: Fields: $\S 15$. Examples of Fields: Example $17$
• 1973: C.R.J. Clapham: Introduction to Mathematical Analysis ... (previous) ... (next): Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Example $3$
• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): field: 1.
• 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): field: 1.