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Suppose $\ \pi+e\ $ is a transcendental number.

Can we conclude whether $\ \pi-e\ $ is rational, algebraic irrational or transcendental ?

If I understood the consequences of Schanuel's conjecture correctly, it implies that $\ \pi-e\ $ is trancendental. But can we also say something without any further unproven conjecture ?

Bill Dubuque
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Peter
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1 Answers1

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I claim that that statement "if $\pi+e$ is transcendental, then $\pi-e$ is transcendental" implies that $\pi-e$ is transcendental. This shows that the knowledge that $\pi+e$ is transcendental is of no help in concluding that $\pi-e$ is transcendental.

Assume that if $\pi+e$ is transcendental, then $\pi-e$ is transcendental. Separately, we can see directly that if $\pi+e$ is algebraic, then $\pi-e = (\pi+e)-2e$ is transcendental (since we know $e$ is transcendental). Since $\pi+e$ is either transcendental or algebraic, we conclude that $\pi-e$ is transcendental.

Julian Rosen
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  • I do not see how this answer shows that there is no way to determine the state of $\ \pi-e\ $ , assuming $\ \pi+e\ $ is transcendental. – Peter Apr 14 '19 at 14:33
  • I am proving that if we had a way to prove $\pi-e$ is transcendental on the assumption that $\pi+e$ is transcendental, then we would also have a proof that $\pi-e$ is transcendental without the assumption that $\pi+e$ is transcendental. – Julian Rosen Apr 14 '19 at 14:34
  • OK, I think, I understand now. – Peter Apr 14 '19 at 14:35