Showing the predicate $n \leq \sqrt2 < n+1$ is primitive recursive
This relation can be written as 
$$n \leq \sqrt{2} \text{ and } n+1  &gt; \sqrt{2}$$
or equivalently
$$ ((n^2 &lt; 2) \text{ or } (n^2 = 2)) \text{ and } ((n+1)^2  &gt; 2)$$
In terms of functions
$$h(n) = and[or[ls(sqr(n),2), eq(sqr(n),2)], gr(sqr(n+1),2)]$$
So, we have exponentiation (squaring), increment by $1$ (i.e. $n+1$), less than, equal to, greater than, conjunction, and disjunction. They are all primitive recursive whose composition result in the primitive recursive predicate $h(n)$ equal to $1$ if $n \leq \sqrt{2} \text{ and } n+1  &gt; \sqrt{2}$, and $0$ otherwise.

On @chi's hint:

In fact this predicate is true for only $n=1$. So this predicate is drastically simplified and takes the following form $$h(n)=eq(1,n)$$