0

Consider the numbers between 1 and 2. The reciprocals of the numbers will be between 0.5 and 1. Now consider the numbers between 2 and 3. The reciprocals of those numbers will lie between 0.333... and 0.5. And the reciprocals of numbers between 3 and 4 will be between 0.25 and 0.333... and so on.
That means the total number of numbers greater than 1 will be equal to the number of numbers between 0 and 1.
That means the distribution of numbers is not equal.

Is this so
or have I done something wrong in interpreting it?

Bill Dubuque
  • 272,048
KeSHAW
  • 7
  • 4
    Two sets are equivalent (i.e. have the same cardinality) if there is a bijection between them. – Wuestenfux Apr 02 '22 at 14:41
  • 1
    What do you mean by "numbers"? – psl2Z Apr 02 '22 at 14:55
  • Welcome to Math.SE! <> It seems there may be two issues being used: How many (real?) numbers lie in an interval, and how long is a real interval. Those are both ways of quantifying in "real life," but for mathematical sets, especially infinite sets, they need not be equivalent. To address the question, that is to resolve any apparent inconsistency in our mathematical understanding, we need to agree on definitions, as the existing comments suggest. – Andrew D. Hwang Apr 02 '22 at 15:37
  • Indeed, there are the same cardinality of numbers between (1,2) as there are $(1,\infty)$, despite (1,2) being a subset of $(1,\infty)$. There is a bijection between $(1,\infty)$ and (0,1) as youve pointed out, and there is also a bijection between (0,1) and (1,2). All three of these sets are equinumerous. – SquishyRhode Apr 02 '22 at 16:11
  • @AndrewD.Hwang Sir, I have studied that mathematics is the subject which is intimately related to nature and I have also observed it. So, isn't it a question arising thing that cardinality of a set calculated in one way is different from calculated in another way, although both methods are correct (at least, as we know it to be). I know it is, but why? (Sorry for misinterpretation, if any because being a high school student, this is the level I know up to). Would be grateful for further clarification. – KeSHAW Apr 05 '22 at 18:39

1 Answers1

1

Given the sets $I_{a,b} = [a, b]$, with $a< b\in \mathbb{R},$

it is clear that each function $f_{a,b,c,d} : I_{a,b} \to I_{c,d}$, defined as

$$f_{a,b,c,d}(x) = (d-c)\frac{x-a}{b-a}+c$$

is invertible (it is just a straight line with non null slope). Hence, $ f_{a,b,c,d}$ is a bijection between $I_{a,b}$ and $I_{c,d}$. In other words, each element in $I_{a,b}$ corresponds to one and only one element of $I_{c,d}$, and vice versa. This means that $I_{a,b}$ and $I_{c,d}$ have the same cardinality.

This work also for the proposed example, i.e.take $a=1$, $b=2$, $c = \frac{1}{b}= \frac{1}{2}$, $d = \frac{1}{a} =1$.

the_candyman
  • 14,064
  • 4
  • 35
  • 62