## Task 4 - 221214

Determine all real numbers $x$ that have the property that of the following statements (A) to (F), four are true and two are false!

(A) $x$ is a positive rational number.

(B) $x$ is a natural number, or $x$ can be represented in the form $x=\frac{1}{g}$ with an integer $g \neq 0$.

(C) $x^{2}$ is an integer, but $x$ itself is not an integer.

(D) $7 < x^{2} < 9$.

(E) $x$ is a positive real number, but not a natural number.

(F) If $x$ is rational, then $x$ is an integer.

Hint: A statement of the form "If $p$, then $q$ " is true if and only if the statement "(not $p$) or $q$ " is true. A statement of the form " $u$ or $v$ " is true if and only if at least one of the two sub-statements $u$ and $v$ is true.
We start with (C) and (D).

They cannot both be false, as otherwise all the others would have to be true. This is not possible, since from (A) and (F) it follows that $x$ is a natural number, which contradicts (E).

If both are true, then $x=2 \sqrt{2}$ or $x=-2 \sqrt{2}$. For $x=2 \sqrt{2}$, (A) and (B) are false, (E) and (F) are true. For $x=-2 \sqrt{2}$, (A), (B), and (E) are false. Therefore, $x=2 \sqrt{2}$ satisfies our conditions.

If (C) is true but (D) is not, then (A), (B), and (D) are false.

If (C) is false and (D) is true, $x$ cannot be a natural number. Therefore, one of the two statements (A) or (F) is false. Since exactly 2 statements are false, (B) and (E) must be true.

If (A) is true and (F) is false, it follows from (B) that $x$ is a positive rational number of the form $1 / g$ with $g \in \mathbb{Z}$, and thus has an absolute value less than or equal to one. This is not possible with (D).

Therefore, (B), (D), (E), and (F) are true, and (A) and (C) are false, or $x=2 \sqrt{2}$ must hold.

From (B) it follows that $x$ is rational, and from (F) it follows that $x$ is natural. This contradicts (D).

Therefore, $x=2 \sqrt{2}$ is the only solution.