Prove or disprove: If [$q\in\mathbb{Q}$ and $a+b = x+y+z = q$] then [$a,b,x,y,z ∈$ either $\mathbb{Q}⊻¬\mathbb{Q}$].

65 Views Asked by At

A small observation is that the unit integer $1$ can be split arbitrarily into two pieces: both pieces must exclusively be either rational or irrational (but not a heterogeneous mixture). This reasoning easily extends to all integers, and from there to all rationals. Furthermore, the two pieces of homogeneous type can be transformed into three pieces while necessarily retaining homogeneity, by dividing one of the original two parts into two (like in the beginning), but necessitation for sameness of class goes away if increasing to four total pieces (and possibly contradicts it, at this number of pieces if not higher) since you could then have three irrationals and one rational.

I wonder what the pattern is and if it extends to multiplication (I suspect it does) and other operations (such as exponentiation, tetration), and how to prove it maximally (i.e., in both directions if possible), and if it is equally true across all of complex numbers (i.e. dealing with $\{\mathbb{Q}\subseteq \mathbb{C}$ and $\mathbb{C}\text{\\}\mathbb{Q}\}$) or just the reals (i.e. $\{\mathbb{Q}\subsetneq\mathbb{R}$ and $\mathbb{R}\text{\\}\mathbb{Q}\}$).

1

There are 1 best solutions below

2
On

Homogeneity does not hold for three terms. For example, $0 + \sqrt{2} + (-\sqrt{2}) = 0$.

What does hold, for any number $n$ of terms $x_1, \dotsc, x_n$ such that $\sum_{i = 1}^n x_i$ is rational, is that the number of irrational terms $x_i$ cannot be exactly $1$; since if, say, $x_1, \dotsc, x_{n - 1}$ are rational, then $\sum_{i = 1}^n x_i - \sum_{i = 1}^{n - 1} x_i = x_n$ is rational.

Homogeneity does hold for multiplication, for two factors, but only for non-zero products. For a zero product, we have, for example, $0 \cdot \sqrt{2} = 0$.