Let

  Φ(x)= 1/√(2π) ∫_{-∞}^{x} e^{-t²/2} dt (x∈ℝ)

denote the standard normal cumulative distribution function.
Put q₁∈ℚ*, q₂∈ℚ and assume

  Φ(q₁)=q₂. (1)

The question is whether (1) forces the “obvious’’ pair

  (q₁ , q₂)=(0 , 1/2).

Short answer. No proof of such a statement is known. In fact we do not even know how to prove that Φ(1), Φ(2) or Φ(1/10) is irrational. The only thing that is known unconditionally is that if q₁≠0 is algebraic, then Φ(q₁) is transcendental if (a very deep but unproved) conjecture such as Schanuel’s conjecture or the refined Lindemann-Weierstrass conjecture is assumed; without those conjectures the problem is completely open.

Below is a summary of what is and is not known.

1. Relation with the error function.
     Φ(x)=½(1+erf(x/√2)),  erf(z)=2/√π ∫_0^{z} e^{–t²} dt .  A rational value Φ(q₁)=q₂ is equivalent to
     erf(q₁/√2)=2q₂–1.
    Thus the question is the same as asking whether the error function can take a rational value at a non–zero algebraic argument.

2. Irrationality and transcendence status.
    • Nobody can presently prove that erf(1) (≈0.84270079…) is irrational, let alone that it is transcendental.
    • More generally, for any non–zero algebraic α we do not know how to decide whether erf(α) or Φ(α√2) is rational, algebraic, or transcendental.
    • The error function is an E–function in the sense of Siegel, so the Siegel–Shidlovsky theorem yields algebraic-independence results for collections of its values together with solutions of its differential equation, but it does not give information about a single value such as Φ(q).

3. What can be proved under extra hypotheses.
    • Lindemann–Weierstrass or, more generally, Schanuel’s conjecture implies that erf(α) (and hence Φ(α√2)) is transcendental for every non-zero algebraic α. Consequently under those conjectures the pair (0,½) is the only rational pair.
    • Various other conditional results (based on, e.g., the theory of periods and 1-motives) reach the same conclusion, but again they rely on statements that are far from being settled.

4. Attempts via elementary expansions.
    One might hope to combine the series representation
     Φ(x)=½+1/√(2π) Σ_{n≥0} (–1)^n x^{2n+1}/(2^n n! (2n+1))
    with the transcendence of e^{x²/2}, but the alternating series makes none of the standard transcendence techniques (Hermite–Lindemann, Mahler, etc.) applicable.

5. Numerical evidence.
    Since Φ is strictly increasing and maps ℝ bijectively onto (0,1), every rational number between 0 and 1 is assumed once. Extensive high-precision computations reveal no obvious rational pairs besides (0,½), but of course numerical searches can never settle the question.

Conclusion. At present there is no known method, even for a single non-zero rational q₁, to prove that Φ(q₁) is irrational, let alone to rule out the possibility that it might be rational. Thus the implication

  Φ(q₁)∈ℚ ⇒ q₁=0, Φ(q₁)=½

is an open problem. All available evidence and all standard conjectures in transcendental number theory predict that the only rational value of Φ at a rational argument is indeed ½ at 0, but this has not been proved.