17 июн. 2012 г. · Thus (x−π)(x−e)=x2−(e+π)x+eπ cannot have rational coefficients. So at least one of e+π and eπ is irrational.

7 янв. 2015 г. · e and π are rather peculiar numbers. It turns out that, in addition to being irrational numbers, they are also transcendental numbers. Basically ...

21 мар. 2011 г. · It is not known whether π+e is irrational, nor whether π×e is irrational. See #22 here.

24 июл. 2016 г. · Thus at least one of e+π and eπ is irrational. As @Henning Makholm pointed out, actually at least one of them is not just irrational but ...

4 дек. 2014 г. · It is a conjecture that the statement is false, ie that π+e and π−e are irrational. According to Wikipedia this remains unproven.

4 окт. 2016 г. · We know that e is irrational and that π is also irrational. Then is π⋅e rational or irrational? algebra-precalculus · irrational-numbers · pi.

27 сент. 2014 г. · It is not currently known whether or not π+e is rational. It is also not known whether or not πe is rational. However, we can say that π+e ...

9 июл. 2016 г. · Whether πe is rational or not is a well-known open problem. It turns out that eπ is irrational, but the proof uses a rather difficult theorem.

8 февр. 2017 г. · Error in proving e+π is irrational. · The only thing I've found so far is a slight working error. 5∗21n is only continuous besides at n=0. – ...

23 дек. 2017 г. · No. The link shows that en is irrational for any natural number n. But π2 isn't a natural number, so that result doesn't apply.