= Solution
Let
$$
x=\sqrt[3]2+\sqrt[3]3.
$$
If $x$ were <rational number>[rational], then
$$
x^3=5+3x\sqrt[3]6
$$
would make $\sqrt[3]6=(x^3-5)/(3x)$ rational. But if $\sqrt[3]6=a/b$ in lowest terms, then $a^3=6b^3$, which is impossible by <unique prime factorization>: the exponent of $2$ on the two sides is respectively a multiple of three and one more than a multiple of three. Therefore
$$
\boxed{\sqrt[3]2+\sqrt[3]3\text{ is irrational}}.
$$
Use the given <convergent series>
$$
e-e^{-1}=2\sum_{n=0}^{\infty}\frac1{(2n+1)!}.
$$
Suppose this number were $a/b\in\mathbb Q$. Choose $N$ large enough that $b\mid(2N+1)!$. Multiplication by $(2N+1)!$ makes both the rational number and the partial sum through $n=N$ integers. Their difference
$$
R_N
=2(2N+1)!\sum_{n=N+1}^{\infty}\frac1{(2n+1)!}
$$
would therefore be an integer. It is positive, while
$$
0<R_N
\leq2\sum_{j=1}^{\infty}\frac1{(2N+2)^{2j}}
<1
$$
for sufficiently large $N$, a contradiction. Hence
$$
\boxed{e-e^{-1}\text{ is irrational}}.
$$
A <transcendental number> is a complex number that is not a root of any nonzero polynomial with rational, equivalently integer, coefficients. Let
$$
y=ae+be^{-1},
\qquad (a,b)\ne(0,0).
$$
If $a\ne0$ and $y$ were an <algebraic number>, then $e$ would satisfy
$$
aX^2-yX+b=0
$$
over the algebraic extension $\mathbb Q(y)$. By <transitivity of algebraic extensions>, $e$ would be algebraic over $\mathbb Q$, contradicting its transcendence. If $a=0$, then $b\ne0$ and $e=b/y$ would again be algebraic. Thus
$$
\boxed{ae+be^{-1}\text{ is transcendental}}.
$$
Back to article page
