Write Φ(x) for the standard‐normal CDF

 Φ(x) = (2π)^(–1/2) ∫_{–∞}^x e^{–t²/2} dt.

We are asking:

“If q₁,q₂∈ℚ satisfy Φ(q₁)=q₂, must it be that q₁=0 and q₂=½?”

Remark. Φ can be written in terms of the “error‐function”
 erf z = (2/√π)∫₀^z e^{–t²} dt
via
 Φ(x) = ½ [1+erf (x/√2)].

Thus the question is equivalent to asking whether erf (α) can be rational for any nonzero rational α. In fact:

• If q₁=0 then Φ(0)=½; that gives the obvious solution (q₁,q₂)=(0,½).
• No other rational examples are known.

From the point of view of transcendence‐theory one would like to prove:

“For any nonzero algebraic α, erf (α) is transcendental, hence in particular cannot be a rational number. ”

Unfortunately the standard transcendence‐methods (e.g. Lindemann–Weierstrass or the Siegel–Shidlovskiĭ theorems) do not apply to erf (z). Concretely:

– erf (z) is not an “E–function” in Siegel’s sense, nor a “G–function” in the usual theory, so one cannot invoke the theorems that force transcendence of values at algebraic points.
– No unconditional result is known which says “erf (algebraic) ≠ algebraic (unless zero).”

As a consequence the assertion

“The only rational–rational point on x↦Φ(x) is x=0 with Φ(0)=½”

remains an open problem in transcendence theory. Equivalently:

Open problem. Do there exist any q₁∈ℚ{0} with Φ(q₁)∈ℚ ?

No nontrivial example is known, and no proof has yet been found that excludes such an example. In particular one cannot at present prove unconditionally that “q₁=0, q₂=½” is the only solution in rationals.