1
$\begingroup$

Theorem: There exists infinitely many irrationals in a range of reals.

Proof: Let the range be $R=(a,b)$ where $a

Suppose some numbers $n_1\ldots n_k\in R$ where each number is greater than the one before, then $a

\begin{align} a&

So

$$ a+\sum\limits_{i=1}^k n_i

Because $k$ can be arbitrarily big, and the truth values of $a

Because there are infinitely many numbers in a range where $a

My question is, is this argument valid? Any not-so-right steps?

  • 0
    **PD:** I would like ti know how the vertical dots should be used in this case with MathJax, as I don't see them right. – 2017-02-24
  • 0
    In latex it is \vdots – 2017-02-24
  • 0
    @joeb I used them like that, but they are not centered. – 2017-02-24
  • 0
    I don't follow the idea of the proof. You seem to be pulling $k$ numbers out of a hat, then somehow "prove" the hypothesis that $a \lt b$. – 2017-02-24
  • 0
    $k$ is just the number of numbers between $a$ and $b$. At first finite, but the conclusion is that $k$ can be infinity because it cancels when grouping all the inequalities. – 2017-02-24
  • 0
    Yes you do seem to be arguing in the wrong direction. Another thing about your arguement, what happens if one of the $n_i = 0$. Then $a \prod_i n_i < b \prod_i n_i$ does not imply $a < b$. It doesn't matter anyway, because you don't need anything to imply $a – 2017-02-24
  • 0
    What you've written makes no sense to me. Nothing in your argument says anything at all about irrational numbers. I suggest you start over. Can you find ONE irrational number between $0$ and $1$? When you have done that, find ONE irrational number between $a$ and $b$. Then think about how you could know you could find more. – 2017-02-24
  • 0
    @joeb That shows there are an infinite number of rationals in the interval. The question was to prove there are an infinite number of **irrationals**. – 2017-02-24
  • 0
    @MarkFischler But the fact that the distance between two irrationals is finite and there are infinitely many reals between a range doesn't imply that there are infinitely many irrationals within that range? – 2017-02-24
  • 0
    Wouldn't it be easier to, say, construct a bijection (or injections both ways would be sufficient due to Cantor-Schroeder-Bernstein) between an arbitrary range and all of $\mathbb{R}$? That tells you the range is uncountable, and you presumably know that rationals are countable, thus this uncountable set of reals minus a countable set of rationals in the range gives you an uncountable (infinite) set. – 2017-02-24
  • 0
    @joeb That is exactly what I've done, just in another words. – 2017-02-24

1 Answers 1

1

No, your proof misses the point. Aside from the fact that you start with an assumption about the existence of a set of ordered intermediate values of arbitrary size, and spend most of your time proving that $a

I would start by proving that between any two rational numbers $q_1$ and $q_2$ with $q_1

Then I would use the fact that at least one rational number must exist between any two unequal reals (proved by using a denominator bigger than twice the reciprocal of the distance between the reals). Apply that to $a$ and $b$, getting a rational $q_x$, and apply it again between $a$ and $q_x$ getting another rational $q_y$.

Now for any given $N$ you can break up the interval between $q_x$ and $q_y$ into $N+2$ rational sub-intervals and by the first part of the proof you can slot $N+1>N$ irrationals, one in each interval. So for any arbitrary $N$ there are more than $N$ irrationals in the interval.