I was wondering if the sum of an algebraic and transcendental complex number is transcendental.

I was thinking if a is algebraic, and b is transcendental, then if a+b is algebraic, then a+b-a is also algebraic since algebraic numbers are closed under additive inverses, but b is transcendental.

Is this a correct approach?

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    $\begingroup$ Yes, that's the right idea. $\endgroup$ – William Stagner Mar 20 '15 at 5:07
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    $\begingroup$ Put another way, given $\Bbb C$, we see that algebraic numbers form a subgroup of $(\Bbb C,+)$, call it $A$. Now $a+b = b + a \in b + A \neq A$, since $b \not\in A$. Since cosets either coincide, or are disjoint, we see $a + b \not\in A$. $\endgroup$ – David Wheeler Mar 20 '15 at 8:35

Your proof is right. The sum of an algebraic and transcendental number must be transcendental, since if not, the sum minus the algebraic number would be algebraic, and can not be transcendental.


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