proof involving rational number

Question

Show that a real number $q$ is rational if and only if there are three distinct integers, $n_1, n_2, n_3$, such that $q + n_1, q + n_2, q + n_3$ forms a geometric progression.

Answer 1

First, suppose $q$ is a rational number, so that $q=m/n$ where $m,n\in \mathbb{N}$. Then let $n_1=0,n_2 = mn, n_3 = 2mn+mn^3$.

Conversely, suppose we have such $n_1,n_2,n_3$. Then $$\frac{q+n_3}{q+n_2}=\frac{q+n_2}{q+n_1}$$ $$\implies (n_1+n_3)q+n_1n_3 = 2n_2q+n_2^2$$ $$\implies q = \frac{n_2^2-n_1n_3}{n_1+n_3-2n_2}$$ (noting that $n_1+n_3\neq 2n_2$)

Answer 2

IF

Assume that $q+n_1$, $q+n_2$, and $q+n_3$ form a geometric progression.

Then, $(q+n_2)^2 = (q+n_1)(q+n_3)$.

Expanding and cancelling: $2n_2q+n_2^2 = (n_1+n_3)q+n_1n_3$.

Rearranging: $q=\dfrac{n_2^2-n_1n_3}{n_1+n_3-2n_2}$, hence $q$ is rational.

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If $n_1+n_3-2n_2=0$, i.e. $2n_2=n_1+n_3$, then we have $n_2^2 = n_1n_3$, then $(2n_2)^2=4n_1n_3$, then $(n_1+n_3)^2-4n_1n_3=0$, then $(n_1-n_3)^2=0$, meaning that $n_1=n_3$, a contradiction.

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ONLY IF

See the other answer.

Answer 3

let $$a_1=q+n_1$$,$$a_2=(q+n_1)x$$ and $$a_3=(q+n_1)x^2$$ eliminating the $x$ we obtain $$q+n_3=(q+n_1)\left(\frac{q+n_2}{q+n_1}\right)^2$$ and from hier we get:$$q=\left(\frac{n_2^2-n_1n_2}{n_1+n_3-2n_2}\right)$$ and this is surely rational.