I'm reading Matiyasevich book Hilbert's Tenth Problem, and I've got a doubt on Chapter 6, Section 6.4:
When he defines the relation: For $a,b \in \mathbb{N}$
$explog(a,b)\Leftrightarrow \exists x \in \mathbb{N} \left[x > b+1 \ \wedge \ \left( 1+ \frac{1}{x} \right)^{xb} \leq a+1 < 4\left( 1+ \frac{1}{x} \right)^{xb} \right]. $
He claims that it is Diophantine. But how it can be Diophantine since $\left( 1+ \frac{1}{x} \right)^{xb} $ is not an integer? Any hint would be really appreciate.
For example: $$ \left(1+\frac{1}{x}\right)^{xb} \leq a+1 \iff \frac{(x+1)^{xb}}{x^{xb}} \leq a+1 \iff (x+1)^{xb}\leq x^{xb}(a+1) $$ So once you know that exponentiation and inequalities are Diophantine, this condition is also Diophantine.