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I am confused why wouldn't every single point in $\mathbb{R}^2\setminus \mathbb{Q}^2$ be irrational.

It seems that whenever we have a pair of rational value or pair of integer, it is removed. Then the stuff between two removed point has all be irrational. But why can there exist points where only one coordinate is irrational?

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    $\begingroup$ The point $\langle\sqrt2,2\rangle$ is in $\Bbb R^2$, since $\sqrt2$ and $2$ are both real numbers, and it is not in $\Bbb Q^2$, since $\sqrt2$ is not a rational number, so it is in $\Bbb R^2\setminus\Bbb Q^2$. $\endgroup$ – Brian M. Scott Aug 4 '16 at 2:35
  • $\begingroup$ $\mathbb Q^2$ consists of all the points with both coords rational. If only one coord is rational and the other irrational, that point's not in $\mathbb Q^2$. $\endgroup$ – user4894 Aug 4 '16 at 2:37
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The point $\left\langle\sqrt2,2\right\rangle$ is in $\Bbb R^2$, since $\sqrt2$ and $2$ are both real numbers, and it is not in $\Bbb Q^2$, since $\sqrt2$ is not a rational number, so it is in $\Bbb R^2\setminus\Bbb Q^2$. The points that are removed from $\Bbb R^2$ are those with both coordinates rational, so you’ve kept every point that has at least one irrational coordinate. That includes the points with one rational and one irrational coordinate.

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  • $\begingroup$ Is there a set in $\mathbb{R}^2 \backslash \mathbb{Q}^2$ that is path connected? i.e. like a diagonal or something $\endgroup$ – Shamisen Expert Aug 4 '16 at 2:46
  • $\begingroup$ @Unicorn-SharkInvasion: The whole set is path-connected. In fact, you can get from any point to any other point using straight line segments. To get from $\langle\pi,1\rangle$ to $\langle 2,e\rangle$, for instance, move along $x=\pi$ to $\langle\pi,e\rangle$, then along $y=e$ to $\langle 2,e\rangle$. $\endgroup$ – Brian M. Scott Aug 4 '16 at 2:54
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    $\begingroup$ +1. It might be easier to visualize this by first thinking of $\mathbb{Z}^2\setminus\mathbb{E}^2$, where $\mathbb{E}$ is the set of even integers - the bonus is that you can draw (part of) this. (Of course this won't help with the path-connectedness, but it might help you visualize why we're not getting rid of as many points as you think.) $\endgroup$ – Noah Schweber Aug 4 '16 at 3:27

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