How to prove $(1+\sqrt{2})^n$ is an irrational number for every $n \in \mathbb{N}$?

Question

Let $k,m \in \mathbb{N}$, $a = k + m\sqrt{2}$

I have proved that $a = k + m\sqrt{2}$ is an irrational number.

Now, I am asked to prove that to every $n \in \mathbb{N}$, $(1+\sqrt{2})^n$ is an irrational number. And I can use what I have proved before (about $a = k + m\sqrt{2}$ being irrational).

I was advised to prove it with mathematical induction.

The base

We check the arguement for n=1: $(1+\sqrt{2})^1 = 1+\sqrt{2}$

We define $k,m=1$ , and 1 is a natural, so $k,m \in \mathbb{N}$

so $(1+\sqrt{2})^1 = 1+\sqrt{2} = k + m\sqrt{2}$ , Which I already proved to be irrational.

Inductive step

We assume $(1+\sqrt{2})^n$ is an irrational number. We need to prove $(1+\sqrt{2})^{n+1}$ is also irrational.

So, $(1+\sqrt{2})^{n+1}$ = $(1+\sqrt{2})^n \cdot (1+\sqrt{2})^1$ Because of exponent rules.

We proved in the base step that $(1+\sqrt{2})^1$ is irrational, and we assumed in the inductive step that $(1+\sqrt{2})^n$ is an irrational.

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But how does it help me forward?

I cannot state that multiplication of 2 irrationals is an irrational. It is simply not true. So it is kind of confusing what I should do next.

Answer 1

In your inductive step, assume that $\exists k, m > 0 : (1+\sqrt{2})^n = k + m\sqrt2$. Then:

$$(1+\sqrt{2})^{n+1}$$ $$= (k + m\sqrt2)(1 + \sqrt 2)$$ $$= k + k\sqrt 2 + m\sqrt2 + 2m$$ $$= (k + 2m) + (k + m)\sqrt2$$

Since $k + 2m$ and $k + m$ are both positive integers, this expression fits the form that you have already proven to be irrational.

Answer 2

HINT: [Don't use induction.] Note that $(1+\sqrt{2})^{n}$ satisfies $(1+\sqrt{2})^{n} = a+ m\sqrt{2}$, where $a$ satisfies $$a = \sum_{k \in \mathbb{Z};\ 0 \le 2k \le n} {n \choose {2k}}2^k,$$ and where $m$ satisfies $$m = \sum_{k \in \mathbb{Z}; \ 1 \le 2k+1 \le n} {n \choose {2k+1}}2^k.$$

[Make sure you see why this is.] However, ${n \choose {2k+1}}$ is positive and integral for every such $k$. So then, what can you conclude about ${n \choose {2k+1}}2^k$ being positive and integral? And thus $m$ as above, isn't it always nonzero and integral?

Answer 3

Here's a very different approach that doesn't necessarily use induction.

• Show that $ ( 1 + \sqrt{2})^n + (1 - \sqrt{2})^n$ is an integer $A$.
  • If we absolutely wanted, we can show this via induction.
• Show that $ ( 1 + \sqrt{2})^n \times (1 - \sqrt{2})^n$ is the integer 1.
• Hence, $( 1 + \sqrt{2})^n , (1 - \sqrt{2})^n$ are the roots of $ x^2 - Ax + 1 = 0$
• By the rational root theorem, the only possible rational roots are $ \pm \frac{1}{1}$.
• Show that $( 1 + \sqrt{2})^n > 1$, hence it is not rational.