Euler Phi Function of 1

Theorem

$\map \phi 1 = 1$

where $\phi$ denotes the Euler $\phi$ function.


Proof

The only (strictly) positive integer less than or equal to $1$ is $1$ itself.

By Integer is Coprime to 1, $1$ is coprime to itself.

Hence, by definition, there is exactly $1$ integer less than or equal to $1$ which is coprime with $1$.

Hence the result.

$\blacksquare$


Sources

  • 1964: Walter Ledermann: Introduction to the Theory of Finite Groups (5th ed.) ... (previous) ... (next): Chapter $\text {I}$: The Group Concept: $\S 6$: Examples of Finite Groups: $\text{(iii)}$: $(1.29)$
  • 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $27$
  • 2008: David Joyner: Adventures in Group Theory (2nd ed.) ... (previous) ... (next): Chapter $2$: 'And you do addition?': $\S 2.1$: Functions: Example $2.1.1$
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