Answer to Question #115952 in Calculus for Pappu Kumar Gupta

Question

Prove or disprove that 2 +(11)^1/2 , beloong to Q

Answer 1

Let's show that $2 + \sqrt{11} \notin Q$ . We will use the fact, that if $r \in Q$ , then $r = \frac{p}{q}, q\neq0,p,q \in Z$, and $gcd(p,q) = 1$ . Let's assume $2 + \sqrt{11} \in Q$ . Then $2 + \sqrt{11} = \frac{p_1}{q_1}, q_1 \neq 0, p_1,q_1 \in Z, gcd(p_1, q_1) =1.$ Hence, $\sqrt{11} = \frac{p_1 - 2q_1}{q_1}.$ Now, we notice that $q_1$ still $\neq 0$ , $q_1, p_1 - 2q_1 \in Z$ and $gcd(p_1 - 2q_1, q_1) =1$. The last equation is true since, using the property of gcd, namely: $gcd(a,b) = gcd(b, a \mod b)$ for $a = p_1 - 2q_1, b = q_1,$ we obtain, that:

$gcd(p_1 - 2q_1, q_1) = gcd(q_1, p_1 \mod q_1) = gcd(p_1, q_1) = 1.$ Hence, $\sqrt{11} \in Q$. Let's now assume that:

$\sqrt{11} = \frac{p}{q}, q \neq 0, p,q \in Z, gcd(p,q) = 1.$ Then, $11 = \frac{p^2}{q^2}$, and $p^2 = 11q^2$ . We obtain that $p^2$ is divisible by 11, and since 11 is prime, $p$ is also divisible by 11. Then $p = 11p_0, p_0 \in Z$ . So:$11q^2 = p^2 = 11^2{p_0}^2$ , therefore $q^2 = 11{p_0}^2$ . So, $q^2$ is also divisible by 11, and since 11 is still prime, so is $q$ . Hence, $q = 11q_0, q_0 \in Z$. But this means that $gcd(p,q) = gcd(11p_0, 11q_0) \ge 11$ . So, we obtained the contradiction ($gcd(p,q) = 1$) with the assumption $2 + \sqrt{11} \in Q$ . Thus, $2 + \sqrt{11} \notin Q$ .