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Prove that there exist equal number of irrational numbers between any 2 rational numbers, when the difference between the 2 rational numbers is same.

If the assertion is not true then please prove otherwise.

Bill Dubuque
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Ash_Tag
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    It (perhaps) depends on what you mean by equality. Since the irrational (and rational) numbers are dense in $~\Bbb{R},~$ for any two distinct rational numbers, there will always be an infinite number of irrational numbers between them. Further, infinity is not a number but is instead a symbol of unbounded growth. So, given the distinct rational numbers $~a_1 < ~a_2, ~$ and the positive number $~r,~$ I surmise that you are being asked to demonstrate a bijection between the irrational numbers in $~(a_1, ~a_2),~$ and the irrational numbers in $~(a_1 + r, ~a_2 + r).$ – user2661923 Mar 23 '23 at 16:33
  • As phrased, the question is imprecise: what does it mean for two intervals to contain the "same number" of irrational numbers? The usual way to make this precise is to talk about "cardinality", instead. In terms of cardinality, a stronger statement is true. – Xander Henderson Mar 23 '23 at 17:11
  • See also https://math.stackexchange.com/q/4427283/, https://math.stackexchange.com/q/783020/, https://math.stackexchange.com/q/3305186/, etc. A Google search like all intervals have the same cardinality site:math.stackexchange.com will return more related results. – Xander Henderson Mar 23 '23 at 17:14

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There's a nice proof of a much stronger theorem which is that there exist equal numbers of irrational numbers between any 2 distinct real numbers (even if the difference between the two numbers isn't the same).

To prove this, consider 2 line segments in the cartesian plane s1 and s2. The first is from P1=(s_1, 0) to P2=(e_1, 0), and the other from P3=(s_2, 1) to P4=(e_2, 1). Draw lines L1 and L2 which connect P1 to P3 and P2 to P4 respectively, and label their intersection P5. Now for any point x on s1 we can draw the line connecting P1 to x. This line intersects s2 at exactly one point, and it's fairly simple to prove that this forms a bijection between s1 and s2.

Oscar Smith
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