(a) $\exists v_{2}v_{1}=v_{2}\cdot v_{2}$ ($r\ge0$ iff for some $x$ , $r=x^{2}$ ).

(b) $\exists v_{2}(v_{1}=v_{2}+v_{2}\wedge\forall v_{3}v_{2}\cdot v_{3}=v_{3})$ ($r=2$ iff for some $x$ , $r=x+x$ and for all $y$ , $x\cdot y=y$ ).

(c) Suppose we have a finite union $A_{1}\cup\ldots\cup A_{n}$ of intervals and $\alpha_{n}$ defines $A_{n}$ , then $\alpha_{1}\vee\alpha_{2}\vee\ldots\vee\alpha_{n}$ defines the union. So, we need only to show that one interval with algebraic endpoints is definable in $\mathfrak{R}$ .

Suppose an interval $A$ has endpoints $a$ and $b$ such that $\{a\}$ and $\{b\}$ are defined by $\alpha$ and $\beta$ . Then, let $\alpha'$ be $\alpha$ in which each variable $v_{i}$ is substituted by $v_{i+1}$ , and $\beta'$ be $\beta$ in which each variable $v_{i}$ is substituted by $v_{i+2}$ , and we have $$\exists v_{2}\exists v_{3}\exists v_{4}\exists v_{5}\exists v_{6}\exists v_{7}(\alpha'\wedge\beta'\wedge v_{4}=v_{6}\cdot v_{6}\wedge v_{5}=v_{7}\cdot v_{7}\wedge v_{1}=v_{2}+v_{4}\wedge v_{3}=v_{1}+v_{5})$$ defines the closed interval $[a,b]$ . We can further require in addition that $v_{1}\neq v_{2}$ and/or $v_{1}\neq v_{3}$ do define $(a,b]$ , $[a,b)$ and $(a,b)$ . For infinite intervals we can skip those part that restrict $v_{1}$ from that end. Therefore, we need only to show that for any algebraic number $a$ , the set $\{a\}$ is definable in $\mathfrak{R}$ .

First we show that for any rational number $a=\frac{m}{n}$ , $\{a\}$ is definable in $\mathfrak{R}$ . Indeed, $$\exists v_{2}(\underbrace{v_{1}+\ldots+v_{1}}_{n\mbox{ times}}=\underbrace{v_{2}+\ldots+v_{2}}_{m\mbox{ times}}\wedge\forall v_{3}v_{2}\cdot v_{3}=v_{3})$$ defines in $\mathfrak{R}$ the set $\{\frac{m}{n}\}$ .

Now, if an algebraic number $\{a\}$ is a root of equation $f(x)=c_{n}x^{n}+\ldots+c_{0}=0$ with integer coefficients, then let $A$ be an interval with rational endpoints such that $a$ is the only root of $f$ that lies within $A$ , and $\alpha$ be a wff that defines in $\mathfrak{R}$ the set $A$ (by this point we have already shown that such a set is definable in $\mathfrak{R}$ ). Then, $$\exists v_{2}((\underbrace{v_{2}+\ldots+v_{2}}_{c_{n}\mbox{ times}})\cdot(\underbrace{v_{1}\cdot\ldots\cdot v_{1}}_{n\mbox{ times}})+\ldots+(\underbrace{v_{2}+\ldots+v_{2}}_{c_{0}\mbox{ times}})=0\wedge\forall v_{3}v_{2}\cdot v_{3}=v_{3}\wedge\alpha)$$ defines in $\mathfrak{R}$ the set $\{a\}$ .