Questions 5-6 Pre-Write

5. Ambiguity of "formulas"
a. Without the proper parenthesis, it is unclear in which order the formula should be conducted.
((ϕ∧ψ)∨θ) is read as θ and either ϕ or ψ, so either θ ϕ or θ and ψ
(ϕ∧(ψ∨θ)) is read as ϕ or ψ and θ
We know that $\sqrt{2}$ is irrational. So, if A=$\sqrt{2}$ and B =$\sqrt{2}$ satisfy the theorem, then we are done. If they do not, then $\sqrt{2}$$\sqrt{2} is irrational, so let a be this number. Then, letting B=\sqrt{2}, it is easy to verify that aB = 2 which is rational and hence would satisfy the theorem. (\sqrt{2}$$\sqrt{2}$)$\sqrt{2}$ = $\sqrt{2}$$2$ = 2