Give an L-formula ϕ(x, y) that defines the less-than relation < over R in M.? Let L = {+, ·, P} where + and · are binary function symbols and P is a unary predicate symbol, and let M be an L-structure where its domain |M| is the 
set R of real numbers, + and · are the usual addition and multiplication over R 
respectively, and P = {r ∈ R | r > 0}. 
(i) Give an L-formula ϕ(x) that defines the set {1} ⊂ R in M. 
(ii) Give an L-formula ϕ(x, y) that defines the less-than relation < over R in M.
(iii) Give an L-formula ϕ(x) that defines the set {√2} ⊂ R in M. 
(iv) Give an L-formula ϕ4(x) that defines the interval [1,√2) ⊂ R in M. 
My formula for part (i) is: ∃x∀y(x·y ≡ y), and (iii): ∃x(Px ∧ x·x ≡ 2) 
Is this what the question is asking for? If so, I'm unable to answer part (iii) and therefore (iv). Thank you!
 A: Answers in the form of guidance as it seems like what you're asking for:
A formula $\varphi(x)$ which defines a set needs to be open. We say that $\varphi(x)$ defines the set ${1}$ if $M\models \varphi(c)$ if and only if $1=c \in R$. Thus your answer to (i) is not really correct but it is if you remove the quantifier for $x$.
i.e.  $\forall y (x \cdot y = y)$ is a good answer.
Regarding your answer to (iii) you have a problem even if you remove the quantifier, since $2$ is not a constant in the language, this $\varphi$ you have created is not an $L-$formula. Remember that L formulas use only symbols from the language, together with quantifiers, variables and connectives.
Instead use the following strategy for (iii): Let $\varphi(x)$ be the formula you used to define ${1}$ in exercise (i). The formula $\forall y(\varphi(y) \rightarrow y+y = ...)$ then say that "for each element satisfying $\varphi$ (i.e. only $1$) 1+1 equals something. Thus we may using this construction say the number 2 without having it in the language. Combine this way with your thoughts on the formula for (iii) and you get the answer.
For (ii): Expressing that $x< y$ is the same as saying that there exists a positive number $z$ such that $x+z = y$. Now use the relation P to express this.
For (iv): combine the other exercises together to state that $x$ is less than $\sqrt(2)$ and greater than or equal to $1$.
A: Thanks Ove!
I think these make sense.
φ1(x): ∀y(x⋅y = y)
φ2(x,y): ∃z(P(z) ∧ (x+z = y))
φ3(x): ∀y(φ1(y) → ((y+y) = (x⋅x))) ∧ P(x)
φ4(x): (∀y∀z(φ3(y) ∧ φ2(x,y) ∧ φ1(z) ∧ φ2(z,y))) ∨ φ1(x)
Please correct me if I'm wrong! :)
