Is $\sqrt 3 + \sqrt 2$ a surd?

Question

Is $\sqrt 3 + \sqrt 2$ a surd?

By definition an irrational root of a rational is called a surd.

Easy to observe that $\sqrt 3 + \sqrt 2$ is irrational. If not, $\sqrt 3 - \sqrt 2$ being the reciprocal of $\sqrt 3 + \sqrt 2$, is also rational, implying $(\sqrt 3 + \sqrt 2)+(\sqrt 3 - \sqrt 2)=2\sqrt 3$ is also a rational, which is not true.

Now the question is does there exist any $n\in \mathbb N$ such that $(\sqrt 3 + \sqrt 2)^{n}\in \mathbb Q$?

It is clear to me $|\mathbb Q(\sqrt 3 + \sqrt 2):\mathbb Q|=4$. But it is not really helping me. Looking for some help. Thanks.

Answer 1

No, $\sqrt{3} + \sqrt{2}$ is not a surd.

As you have correctly observed, ${\Big\lvert \mathbb{Q} {\left( \sqrt{3} + \sqrt{2} \right)} : \mathbb{Q} \Big\rvert} = 4$. This allows us to know for a fact that if we can find a quartic polynomial over $\mathbb{Q}$ that $\sqrt{3} + \sqrt{2}$ is a root of, then this is the minimal polynomial (up to multiplication by a constant, of course). Note that ${\left( \sqrt{3} + \sqrt{2} \right)}^4 = {\left( 5 + 2 \sqrt{6} \right)}^2 = 49 + 20 \sqrt{6}$; this is also $10 {\left( \sqrt{3} + \sqrt{2} \right)}^2 - 1 = 10 {\left( 5 + 2\sqrt{6} \right)} - 1 = 50 - 1 + 20 \sqrt{6}$. As such, $\sqrt{3} + \sqrt{2}$ is a root of $x^4 - 10 x^2 + 1$, which is the minimal polynomial. But it is then easy to see that this cannot be a factor of any $a x^n - b$, so no power of $\sqrt{3} + \sqrt{2}$ is in $\mathbb{Q}$.

In particular, if $\sqrt{3} + \sqrt{2}$ were a surd, then ${\left( \sqrt{3} + \sqrt{2} \right)}^2 = 5 + 2 \sqrt{6}$ would be as well. But the minimal polynomial of the latter is $x^2 - 10 x + 1$, which is not a multiple of a cyclotomic polynomial and thus $5 + 2 \sqrt{6}$ cannot be a surd.

Answer 2

HINT.-$\displaystyle(\sqrt3+\sqrt2)^n=\sum_{k=0}^n\binom{n}{k}(\sqrt3)^{n-k}(\sqrt2)^k$ so

► if $n$ is odd then $\displaystyle(\sqrt3+\sqrt2)^n=\sum_{k=0}^n a_k\theta_k$ where $a_k$ is rational and $\theta_k$ is irrational.

►if $n$ is even then $\displaystyle(\sqrt3+\sqrt2)^n=\sum_{k\space even} b_k+\sum_{k\space odd} a_k\theta_k$ where $a_k$, $b_k$ are rational, $\theta_k$ is irrational and all the numbers are positive.

Thus never $\displaystyle(\sqrt3+\sqrt2)^n$ is rational.

Answer 3

Is there any $n$ such that $(\sqrt{2}+\sqrt{3})^n \in \mathbb Q$?
No. These four are linearly independent over $\mathbb Q$: $$ 1,\quad\sqrt{2},\quad\sqrt{3},\quad \sqrt{6} $$ If we write $$ (\sqrt{2}+\sqrt{3})^n = a_n + b_n\sqrt{2}+c_n\sqrt{3}+d_n\sqrt{6} $$ we may obtain recurrence \begin{align} a_{n+1} &= 2b_n+3c_n \\ b_{n+1} &= a_n+3d_n \\ c_{n+1} &= a_n+2d_n \\ d_{n+1} &= b_n+c_n \end{align} We could solve this. But it is simpler to note by induction that $a_n,b_n,c_n,d_n \ge 0$ and $(c_{n+1}+d_{n+1}) \ge (c_n+d_n)$, so every power $(\sqrt{2}+\sqrt{3})^n$ has either $c_n \ne 0$ or $d_n \ne 0$, and is therefore irrational.

Answer 4

We have to show that, if $n\in\mathbb{Z}^+,$ then $\left(\sqrt{3} + \sqrt{2}\right)^n\not\in\mathbb{Q}. $

Using NDB's hint in the comments, first we have that:

$$ \frac{1}{ \left(\sqrt{3} + \sqrt{2}\right)^n } = \frac{1}{ \left(\sqrt{3} + \sqrt{2}\right)^n } \cdot \frac{ \left(\sqrt{3} - \sqrt{2}\right)^n }{ \left(\sqrt{3} - \sqrt{2}\right)^n } = \frac{ \left(\sqrt{3} - \sqrt{2}\right)^n }{ \left(\left(\sqrt{3} + \sqrt{2}\right)\left(\sqrt{3} - \sqrt{2}\right)\right)^n }$$

$$ = \frac{ \left(\sqrt{3} - \sqrt{2}\right)^n }{ \left(3-2\right)^n } = \left(\sqrt{3} - \sqrt{2}\right)^n. $$

Now suppose by way of contradiction, that $\left(\sqrt{3} + \sqrt{2}\right)^n\in\mathbb{Q}.$

Since $\left(\sqrt{3} + \sqrt{2}\right)^n\in\mathbb{Q}, \frac{1}{ \left(\sqrt{3} + \sqrt{2}\right)^n } = \left(\sqrt{3} - \sqrt{2}\right)^n$ is also rational. Therefore their difference, $\left(\sqrt{3} + \sqrt{2}\right)^n - \left(\sqrt{3} - \sqrt{2}\right)^n,$ is also rational.

You can check using the Binomial Theorem that:

$ \left(\sqrt{3} + \sqrt{2}\right)^n - \left(\sqrt{3} - \sqrt{2}\right)^n= \begin{cases} a\sqrt{2}\ \text{ for some } a\in\mathbb{N},&\text{if}\, n \text{ is odd}\\ b\sqrt{6}\ \text{ for some } b\in\mathbb{N},&\text{if}\, n \text{ is even}.\\ \end{cases} $

In either case, $\left(\sqrt{3} + \sqrt{2}\right)^n - \left(\sqrt{3} - \sqrt{2}\right)^n$ is irrational. contradicting the fact that $\left(\sqrt{3} + \sqrt{2}\right)^n - \left(\sqrt{3} - \sqrt{2}\right)^n$ is rational.