1

I’m trying to prove that $\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}$ is irrational.

I have already proved that $\sqrt{2}+\sqrt{3}$ is irrational. Should I use a similar approach as below or is there a different way?

Proof: $\sqrt{2}+\sqrt{3}$ is irrational Assume $\sqrt{2}+\sqrt{3}$ is rational so $\sqrt{2}+\sqrt{3}=p/q$ where $p$ and $z$ are integers, $q\neq 0$, $\gcd(p,q)=1$ $r^2=5+2\sqrt{6}$ $r^2-5=2\sqrt{6}$ then I squared both sides… $r^4-10r^2+25=24$ $r^4-10r^2+1=0$

$r=p/q$ so $p^4/q^4-10p^2/q^2+1=0$ then I got a common denominator $p^4-10p^2q^2+q^4=0$ $p^4=10p^2q^2-q^4$

I found that $q=1$ or $-1$ and then I plugged it back into $r^4-10r^2+1=0$ and found that it can’t $=-1$ so there is a contradiction.

Tomasz Lenarcik
  • 471
user2553807
  • 1,245
  • 25
  • 46

3 Answers3

1

This is covered by the proof that $\mathbb Q(\sqrt p_1, \sqrt p_2,\cdots, \sqrt p_n)$ is degree $2^n$ over $\mathbb Q$, where $p_1$, etc. are distinct primes. The proof is by induction, using the same method of proof as for two primes.

Tomasz Lenarcik
  • 471
Bill Kleinhans
  • 1,458
0

You have a shorter proof: if $\sqrt{2}+\sqrt{3}=\frac{p}{q}$, where $p\in\mathbb Z$ and $q\in\mathbb N$, $q\neq 0$, then $5+\sqrt{6}=\frac{p^2}{q^2}$. So, $\sqrt{6}=\frac{p^2-5q^2}{q^2}$ is rational, which is known to be false.

For $\sqrt{n}$, with $n\in\mathbb N$, you have the following alternative:

1) $\sqrt{n}$ is an integer (it happens when $n$ is a perfect square)

2) $\sqrt{n}$ is irrational.

Taladris
  • 11,339
  • 5
  • 32
  • 58
0

First, I would try proving something more general in nature, e.g. that the set $\{1,\sqrt{p_1},\ldots,\sqrt{p_k}\}$, where $p_1,\ldots,p_k$ are different primes is linearly independent over $\mathbb{Q}$. This should give you enough clues on how to approach your problem.

Tomasz Lenarcik
  • 471