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Do replacing distinct digits from distinct places of an algebraic irrational number necessarily make it a trancsendendal number? Since my question isn't worded well, therefore I would clarify it by means of a simplest example:

$x=\sqrt{2}=1.4142\color{blue}{1}3\color{green}{5}6237..\text{which is algebraic}$

Lets replace the $5^{th}$ digit by $2$ and $6^{th}$ digit by $7$

$y=1.4142\color{blue}23\color{green}76237$

$\large \text{Question}:$

Is $y$ necessarily a transcendental number?(Any examples to disprove this would be welcome)

PS: I am more interested in the general question.

Hashir Omer
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2 Answers2

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No, because $y - \sqrt 2 = r$ is a rational number: It has a terminating decimal expansion. Then $r + \sqrt 2$ is a sum of algebraic numbers, which is algebraic.

At a minimum, infinitely many digits must be changed.

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You're basically constructing the number $\sqrt 2 + 1 \cdot 10^{-5} + 2 \cdot 10^{-7}$ (the dots signify multiplication). An algebraic irrational number plus a rational number is, of course, still algebraic (albeit irrational). The same applies for changing any finite number of digits in the expansion.

Now if you're talking about changing an infinite number of digits, it really depends on the rule you're employing.

If you're (say) incrementing every fifth digit by one, then all you're doing is adding an infinite geometric series, which has a rational sum, so the result is still algebraic. But if you're adding something like the Champernowne Constant which is already transcendental, of course the result will be transcendental too.

Deepak
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