A function is additive if f(x+y) = f(x) + f(y). Intuitively, you might think that an additive function on R is necessarily linear, specifically of the form f(x) = kx. But assuming the axiom of choice, that is wrong, and the proof is rather simple: you just take a Hamel basis of R as a vector space over Q, and then you define your function f to be different in at least two distinct elements of the basis.(adsbygoogle = window.adsbygoogle || []).push({});

But my question is, does the same conclusion hold in the absence of the axiom of choice? Or does the negation of choice force the function to be linear?

Any help would be greatly appreciated.

Thank You in Advance.

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Can an additive function be nonlinear without AC?

Loading...

Similar Threads - additive function nonlinear | Date |
---|---|

I Derivative of a Real-Valued Function of Several Variables... | Feb 25, 2018 |

I Addition/Multiplication of Natural Numbers - Bloch Th. 1.2.7 | Jul 15, 2017 |

Numerical Approximation and addition of new data points | Feb 4, 2015 |

Additive but not Scalable f: R-->R | Jan 11, 2015 |

Measure theory question: Countable sub-additivity | Oct 3, 2013 |

**Physics Forums - The Fusion of Science and Community**