Existence of a real number such that when added to seq of real gives all irrationals

Let $a_1,a_2,...,a_n$ be arbitrary real numbers. Prove that there exists a real number $x$ such that each of the numbers $x+a_1,x+a_2,...,x+a_n$ is irrational.

It’s easy to see that if all $a_1,a_2,a_3,...$ are rational numbers then any irrational number will do the job because the sum of rational+irrational is always an irrational number.

To prove this let’s assume that there exists an irrational number $\alpha$ and we generate the following numbers $\alpha,2\alpha,3\alpha,...,n\alpha,,(n+1)\alpha$ and we see that all these numbers are irrationals since rational $\times$ irrational = irrational. Our claim is that some $x \in \{\alpha,2\alpha,3\alpha,...,n\alpha,,(n+1)\alpha \}$ will satisfy $x+a_1,x+a_2,...,x+a_n$ is irrational.

Suppose this is not true then it means that there is at least one rational number, we create the following table

$\alpha+a_1, \alpha+a_2, \alpha+a_3,..., \alpha+a_n$

$2\alpha+a_1, 2\alpha+a_2, 2\alpha+a_3,..., 2\alpha+a_n$

$:,:,...:$

$n\alpha+a_1, n\alpha+a_2, n\alpha+a_3,..., n\alpha+a_n$

$(n+1)\alpha+a_1, (n+1)\alpha+a_2, (n+1)\alpha+a_3,..., (n+1)\alpha+a_n$

Thus the table consists of at least $n+1$ rational numbers and since it has $n$ columns, at least two of these rationals are in the same column. In other words there exists positive integers $i,j$ and $k, i \ne j$ such that $i\alpha+a_k$ and $j\alpha+a_k$ are rationals. But then the number $(i-j)\alpha=(i\alpha+a_k)-(j\alpha+a_k)$ is also rational, and this contradicts the irrationality of $\alpha$

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