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Is it true that for every real number $x$ there exist transcendental numbers $\alpha$ and $\beta$ such that $x=\alpha-\beta$?

(it is true if $x$ is an algebraic number).

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    $\begingroup$ Sure, by counting. Look at the set $\{x-\alpha\}$ where $\alpha$ is transcendental. That is uncountable, hence they can't all be algebraic. $\endgroup$ – lulu Nov 6 '16 at 13:46
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    $\begingroup$ @lulu Your comment would qualify as an answer as is. $\endgroup$ – JiK Nov 6 '16 at 16:42
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    $\begingroup$ What do you mean by linearly independent? $\endgroup$ – Vim Nov 7 '16 at 10:26
  • $\begingroup$ @Vim I would assume this means that neither of α, β is a rational multiple of the other. But if this was in fact the case then either both α, β must have been rational multiples of x (in which case, just add an irrational multiple of x to both of them to get a new pair which is linearly independent) or x was 0 (in which case α, β have to be equal). $\endgroup$ – Robin Saunders Nov 7 '16 at 13:42
  • $\begingroup$ Note that if we apply the more stringent requirement that no rational multiples of α, β can differ by a rational number, then this rules out x itself being rational. On the other hand, we could have insisted instead that neither α, β was an algebraic multiple of the other, and then the argument above would still apply (with "irrational" changed to "transcendental"). $\endgroup$ – Robin Saunders Nov 7 '16 at 13:45
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Yes, since the set of algebraic numbers is countable.

Let $\mathbb{T}$ denote the set of transcendental numbers, and for $\alpha\in \mathbb{T}$ let $f(\alpha)=\alpha-x$. Then $f$ is an injection from $\mathbb{T}$ to $\mathbb{R}$, so its range $ran(f)$ is uncountable since $\mathbb{T}$ is. But since $\mathbb{R}\setminus\mathbb{T}$ (the set of algebraic numbers) is countable, this means $ran(f)$ contains some element of $\mathbb{T}$.

So let $\alpha\in\mathbb{T}$ be such that $f(\alpha)\in\mathbb{T}$, and set $\beta=f(\alpha)$; then $\alpha, \beta$ are transcendental and $\alpha-\beta=x$.

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If $x$ is algebraic, then take $\alpha=x\pi$ and $\beta=x(\pi-1)$.

If $x$ is transcedental, then take $\alpha=2x$ and $\beta=x$.

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    $\begingroup$ Need to treat x=0 separately (though the answer is clear in that case). $\endgroup$ – Ari Brodsky Nov 6 '16 at 16:45
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    $\begingroup$ I think it is better to write $x=(x+\pi)-\pi$ to avoid this case. $\endgroup$ – M.H.Hooshmand Nov 6 '16 at 16:52
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    $\begingroup$ @AriBrodsky: Yeah, it just occurred to me while watching Independence Day 2, thanks for pointing that out. For $x=0$, taking $\alpha=\pi$ and $\beta=\pi$ should suffice. $\endgroup$ – barak manos Nov 6 '16 at 17:27
  • $\begingroup$ @M.H.Hooshmand: Yes, I agree, though I'll leave the answer as is, with those comments there to remove any further doubt. Thanks. $\endgroup$ – barak manos Nov 6 '16 at 17:29
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    $\begingroup$ I am confused that this elegant, complete and easy to see (as no prior knowledge about countability or other number theory background is needed) answer is not accepted. Does anyone see a drawback of this constructive approach, as compared to the accepted answer? Am I just not getting it? $\endgroup$ – AnoE Nov 7 '16 at 15:21
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Once we know that $e$ is transcendental, we also know that $e^2$ and $e^2-e$ are transcendental because given a polynomial satisfied by either one we could find a polynomial for $e$. Then given algebraic $a \in \Bbb R$ we can write $a = (a+e)-e$. Given transcendental $a \in \Bbb R$ we can either write $a=(a+e)-e$ or $a=(a+e^2)-e^2$. We know that at least one of $a+e, a+e^2$ is transcendental because if they were both algebraic, so would $e^2-e$ be.

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Suppose there is some real number x that can't be written as the difference of transcendental numbers. Then for every transcendental number y there exists a unique algebraic number z = y+x. But this means there is an injective function f(a) = a+x from the transcendental numbers to the algebraic numbers, which implies the cardinality of the transcendental numbers is less than or equal to that of the algebraic numbers. But we know this is false because the cardinality of the algebraic numbers is strictly less than that of the transcendentals.

Q.E.D.

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If $x$ is algebraic and $a$ is transcendental, then $x+a$ is also transcendental and $(x+a)-a=x$. If $x$ is transcendental, then $2x$ is also transcendental and $2x-x=x$.

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