Let $r_1, r_2, \cdots, r_n$ be distinct rational numbers in the interval $(0,1)$. How to prove that in the space $\mathbb{R}$ over $\mathbb{Q}$ the numbers $2^{r_1}, \cdots, 2^{r_n}$ are independent?

  • $\begingroup$ Do you know how to prove that the square root of two is irrational? Notice that this corresponds to the case where $n=2, r_1 = 0, r_2 = \frac12$. $\endgroup$ Mar 27 '18 at 4:13
  • $\begingroup$ Yes, of course, I know. But how to prove it for $n=3$? $\endgroup$
    – SWalker
    Mar 27 '18 at 4:46

We may without loss of generality assume that

$i < j \Longrightarrow r_i < r_j; \tag 0$

since $0 < r_i \in \Bbb Q$, $1 \le i \le n$, we may write

$r_i = \dfrac{p_i}{q_i}, \; p_i, q_i \in \Bbb Z, 0 < p_i < q_i; \tag 1$

we choose $m = \text{lcm} \{q_i \mid 1 \le i \le n \}, \tag 2$

that is, $m$ is the least common multiple of the denominators $q_i$; as such, it is the least common denominator of the $r_i = p_i / q_i$, and so each $r_i$ may be written

$r_i = \dfrac{k_i}{m}, \; 0 < k_i \in \Bbb Z; \tag 3$

we note that (0) implies $k_i < k_j$ for $i < j$; if the numbers $2^{r_i}$ are linearly dependent over $\Bbb Q$, we have $\alpha_i \in \Bbb Q$, not all $0$, with

$\displaystyle \sum_1^n \alpha_i 2^{r_i} = 0; \tag 4$

we observe that at least two of the $\alpha_i \ne 0$, lest $2^{r_i} = 0$ for some $i$, impossible; using (3), (4) may be written

$\displaystyle \sum_1^n \alpha_i 2^{k_i / m} = 0; \tag 5$

since $r_i \in (0, 1)$ for all $i$, we have $k_1 > 0$ and $k_n < m$; thus we may factor $2^{k_1/m}$ out of (5) and, setting $t_i = k_i - k_1$, so that $t_1 = 0$ and $t_n < m$, write

$\displaystyle \sum_1^n \alpha_i 2^{t_i / m} = 0; \tag 6$


$f(x) = \displaystyle \sum_1^n \alpha_i x^{t_i} \in \Bbb Q[x]; \tag 7$

then (6) affirms that

$f(2^{1/m}) = 0; \tag 8$

we also have

$\deg f(x) \le t_n < m; \tag 9$

it is evident that $r=2^{1/m}$ satisfies

$g(x) = x^m - 2 \in \Bbb Q[x]; \tag{10}$

Eisenstein with $p = $ shows that $g(x)$ is irreducible over $\Bbb Q$; hence $2^{1/m}$ can satisfy no polynomial of degree less than $\deg g(x) = m$, but this contradicts (8), whence there can be no linear dependence such as (4). And we are done.

Note: in retrospect, having written up this proof, the division through by $2^{k_1/m}$ and the introduction of the $t_i$ don't really appear necessary; I must have been thinking along a different track when I introduced them; nevertheless, this doesn't affect the validity of the result, and it's less work to leave those steps in place than edit then out. End of Note.

  • $\begingroup$ In "Eisenstein with p= shows", what is p supposed to equal? $\endgroup$ Sep 30 '19 at 18:24

(1). Let $D$ be a positive common denominator of $r_1,...,r_n.$ Of course $1<D\in \Bbb N.$

Let $x=2^{1/D}.$

For $1\leq j\leq n$ let $r_j=N_j/D .$ We have $D>N_j\in \Bbb N$ and also $i\ne j\implies N_i\ne N_j.$

Suppose $q_1,..., q_n$ are rationals, not all $0,$ such that $\sum_{j=1}^nq_j2^{r_j}=0.$ We have $0=\sum_{j=1}^nq_jx^{N_j}.$

Now $f(y)=\sum_{j=1}^nq_jy^{N_j}$ is a polynomial in $y,$ of degree $d,$ where $1\leq d<D,$ with rational coefficients. We will return to $f(y)$ in (3), below.

(2). We quote an elementary theorem of Gauss: If $g(y)\in \Bbb Z[y]$ and $g(y)$ is irreducible over $\Bbb Z$ then $g(y)$ is irreducible over $\Bbb Q. $

The polynomial $y^D-2$ satisfies Eisenstein's Criterion so it is irreducible over $\Bbb Z,$ so by the above theorem it is irreducible over $\Bbb Q. $

(3). We have $x^D-2=0.$ Let $ h(y)$ be a non-$0$ member of $\Bbb Q[y]$ of smallest possible degree, such that $h(x)=0.$ Then for any $g(y)\in \Bbb Q[y],$ if $g(x)=0$ then $h(y)$ is a divisor of $g(y)$ in the ring $\Bbb Q[y].$

Referring back to $f(y)$ in (1): Since $f(x)=0,$ therefore $h(y)$ divides $f(y)$ so $\text {deg} (h)\leq \text { deg } (f)<D.$

But $x^D-2=0$ so $h(y)$ also divides $y^D-2$ in $\Bbb Q[y],$ with $\text {deg } (h)<D.$

This implies that $y^D-2$ is reducible in $\Bbb Q[y],$ a contradiction to (2).


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