How to prove that $n \sqrt{17}$ is irrational?

Question

Prove that $ \sqrt{17}$ is irrational. Subsequently, prove that $n \sqrt{17}$ is irrational too, for any natural number $n \neq 0$. Use the following lemma: Let p be a prime number; if $p | a^2$ then $p | a$ as well. I proved by contradiction that $\sqrt{17}$ is irrational, but I'm not sure how to prove that $n \sqrt{17}$ is irrational. I tried to prove it by contradiction as well, but I'm not sure if that's what I'm supposed to do. Is it easier to prove with induction?

Answer 1

I think the lemma is to show $\sqrt{17}$ is irrational.

Hint for the too part:

• $n$ is rational so $\frac1n$ is rational
• The product of two rational numbers is a rational number
• What is $n\sqrt{17} \times \frac1n$?

Answer 2

Suppose $n\sqrt{17}$ with $n \in \mathbb{N}$, is rational, then $\sqrt{17}n=\frac{p}{q} \to \sqrt{17} = \frac{p}{nq}$ with $p,q \in \mathbb{Q}, n\cdot q \neq 0$. Here is a contradiction since operations between rationals are always rationals, except division by zero. This show us that $\frac{p}{nq} \in \mathbb{Q}$, thus we are done.

Answer 3

Proof by contradiction:

If $\sqrt{17} = \frac{a}{b}$ for $\{ a, b \} \in \mathbb{Z}^+$ where $a$ and $b$ are in reduced form, then

$$17 b^2 = a^2 .$$

The left side has an odd number of prime factors while the right side has an even number of prime factors.

According to the fundamental theorem of arithmetic (unique prime factorization), this equation cannot hold.

Thus the assumption that $\sqrt{17}$ is rational is false. QED.

Likewise suppose $n \sqrt{17} = \frac{a}{b}$.

Now $17 n^2 b^2 = a^2$. Again, the left hand side has an odd number of prime factors, and the right-hand side has an even number of prime factors. This can never happen. Therefore the assumption is false. QED.