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Prove there is an irrational number between any two real numbers.

Let x, y \in \mathbb{R} be given with x < y. Prove that there exists an irrational number z such that x < z < y.


Note: To do this problem, I think we need to assume the existence of an irrational number. We will prove the existence of such a number (the \sqrt{2}) in I.3.12, Exercise #12.

Proof. Since the rationals are dense in the reals I.3.12, Exercise #6, we know that for x, y \in \mathbb{R} with x<y there exist r,s  \in \mathbb{Q} such that

    \[ x < r < s < y. \]

Now, assume the existence of an irrational number, say w (see note preceding the proof about this). Since w \in \mathbb{R} we know -w \in \mathbb{R} and from the order axioms exactly one of w or -w is positive (w is nonzero since 0 \in \mathbb{Q}). Without loss of generality, let w > 0. Then, since s-r > 0, we know there exists an integer n such that

    \[ n (s-r) > w \implies s > \frac{w}{n} + r \]

Also, since n,w > 0, we have \frac{w}{n} > 0; thus, r < r + \frac{w}{n}.
Then, by I.3.12, Exercise #7 we have \frac{w}{n} irrational and hence r+\frac{w}{n} irrational.
Thus, letting z = r + \frac{w}{n}, we have x < z < y with z irrational. \qquad \blacksquare

2 comments

  1. Camilo says:

    I think it is not necessary to put those x and y since the proof only argues over s and r, and as you see, the conclusion is that the irrational number is between s and r, these two numbers can be any real, so the proof is already complete.

    • Bobby says:

      You need the rational numbers because if you just used x for eg, x can be irrational, and an irrational plus an irrational can be rational so the proof wouldn’t work in that case

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