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$\begingroup$

Basically, I am asking if this is true.

$$|\{z\mid 0

But I don't know how to prove it.

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    weird, I have always seen a construction involving an enumeration of Q – 2017-01-26
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    $\infty$ is not a cardinal. – 2017-01-26
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    Hint: Take some positive irrational number (like $\pi$) and look at the sequence $(\pi/i)_{i\in\mathbb{N}}$. What can you see? – 2017-01-26

2 Answers 2

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$1/\sqrt{n}$ is irrational for every prime number $n$. I'm leaving uncountably many out but this proves that there are infinitely many.

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    $n=9{}{}{}{}{}{}$ – 2017-01-26
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    What if $n$ is the square of an odd number? – 2017-01-26
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    Haha, oops... fixed. – 2017-01-26
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It's true, and there are many many ways to prove it.

Taking any rational number $q$ such that $0

Or, you could say that for every $n$, there exists an irrational number between $\frac{1}{n+1}$ and $\frac1n$.

Or, you could go decimal. There are infinitely many non-repeating strings of digits from $0$ to $9$

Or, take any irrational number $x$ on $(0,1)$ (for example, $x=\frac{1}{\sqrt 2}$. Then, $x,\frac x2,\frac x3,\dots$ are all irrational and all on $(0,1)$.

Alternatively, the set $(0,1)$ is uncountably infinite, while $(0,1)\cap \mathbb Q$ is countably infinite, so the set $(0,1)\cap\mathbb R-\mathbb Q$ must be uncountably infinite as well.

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    @Garmekain Edited the answer. – 2017-01-26