# Thread: More irrational numbers than rational numbers?

1. I hypothesize that there are more irrational numbers than rational numbers. I have a proof for this but I would first like to know if I am correct.
Are there more irrational numbers than rational numbers given for a set of infinite numbers.
By "more" I mean "more infinitely bigger" if that makes any sense.

2.

3. You're right. Do you know anything about different levels of infinity? Anyway, I'd love to see your proof.

4. I know both the rational and irrational numbers are subsets of the reals, and that the reals have a higher 'cardinality' than either. Similarly that the rationals have a higher cardinality than the integers. Comparing the rational/irrational numbers though I have no idea.

5. The rationals have the same cardinality as the integers!

6. Really? Is that because the rationals are basically made of quotients of integers? I would think that the ratioanls have more cardinality because for any numerator you choose from the integers, you can have an infinite number of denominators....obviously I'm wrong, but where?

P.S. you have whizzed your way through about 4 maths posts in the time it takes me to think about 1!

7. Originally Posted by serpicojr
You're right. Do you know anything about different levels of infinity? Anyway, I'd love to see your proof.
Yeah, I understand the concept of the hierarchy of Infinity.

My proof is actually much longer and complex but I shortened it down to the bare minimum :

U <=> Irrational number

√(2) + n = U where n∈R
Because "n" can be any real number, there will always be a one-to-one relationship between any real number and "U".
But because there are more irrational number (i.e. √(3), √(6)-1, etc...), the amount of irrational numbers is much greater than the amount of rational numbers.

Ask me if you need any clarification.
btw check out this img I made recently:

(8/9 should be colored gray)

This image is unrelated to my previous proof but it shows how limited the rational numbers actually are. The blue colored boxes are the ones that cancel out because they can be simplified.

- Nishant Shukla

8. Originally Posted by bit4bit
Really? Is that because the rationals are basically made of quotients of integers? I would think that the ratioanls have more cardinality because for any numerator you choose from the integers, you can have an infinite number of denominators....obviously I'm wrong, but where?

let [R] = any rational number
let [IR] = any irrational number

[R] = -∞ ..., -1, 0, 1, 2, 3... ∞
[IR] = √2 , √3 , √5 , √6 ...∞
In the above there are clearly more [R] than [IR]
[R] wins
---

[R]/1,2,3,4,5,6......∞ (infinite rational nums)
[IR]/1,2,3,4,5,6......∞ (infinite irrational nums)

now we can ignore the 2 previous facts in determining which is bigger because the one-to-one relationship cancels out.
Its a tie
---

[R]*(-∞...2,3,4,5,6,7.......∞) ∈ [R] (because we already defined [R] above)
however, [IR]*(-∞...2,3,4,5.......∞) = new [IR] number.

therefore in this case, there are clearly A LOT more [IR] than [R]
[IR] wins by a lot

---

[R]+(-∞...2,3,4,5,6,7.......∞) ∈ [R] (because we already defined [R] above)
however, [IR]+(-∞...2,3,4,5.......∞) = new [IR] number.

in this case again there are more [IR] than [R]
[IR] wins by a lot again

---

So far [IR] is winning.

it goes on and there should be more [IR] than [R]

-- Nishant Shukla

9. Originally Posted by bit4bit
Really? Is that because the rationals are basically made of quotients of integers? I would think that the ratioanls have more cardinality because for any numerator you choose from the integers, you can have an infinite number of denominators....obviously I'm wrong, but where?
That's the tricky thing about infinite cardinals--an infinite cardinal times itself is the same cardinal. Here's why the rationals and whole numbers have the same cardinality.

Write out all of the rational numbers as so:

0, 1, -1, 2, -2, 3, -3...
0/2, 1/2, -1/2, 2/2, -2/2, 3/2, -3/2...
0/3, 1/3, -1/3, 2/3, -2/3, 3/3, -3/3...
...

and so on--increasing magnitude of numerator to the right, increasing denominator down. This array certainly contains all of the rational numbers. Now starting at the 0 in the upper left corner, go right to 1, then down the diagonal to 0/2, then down to 0/3, then up the diagonal through 1/2 and -1, right to 2, etc., so that you snake all the way around the array, hitting each number. Traveling along this curve, you can count all of the rational numbers.

P.S. you have whizzed your way through about 4 maths posts in the time it takes me to think about 1!
Don't sweat it--I get paid to do math.

10. Originally Posted by serpicojr
Don't sweat it--I get paid to do math.
Whhhhatt? where!?!
I'm in just tell me who pays you! (unless you were joking)

11. Originally Posted by DivideByZero
U <=> Irrational number

√(2) + n = U where n∈R
Because "n" can be any real number, there will always be a one-to-one relationship between any real number and "U".
Is this supposed to give you a one-to-one correspondence between irrationals and reals? Because if n = 1 - sqrt(2), then sqrt(2) + n = 1. If this isn't what you're saying, then I don't quite understand your argument.

I think the easiest way of seeing that the irrationals are uncountable is to note that the reals are uncountable, the rationals are countable, and the irrationals are the reals minus the rationals. Of course, this assumes can prove the reals are uncountable. Have you seen Cantor's proof of such?

12. Originally Posted by DivideByZero
Whhhhatt? where!?!
I'm in just tell me who pays you! (unless you were joking)
No jokes. I'm a math grad student.

13. Originally Posted by serpicojr
Originally Posted by DivideByZero
Whhhhatt? where!?!
I'm in just tell me who pays you! (unless you were joking)
No jokes. I'm a math grad student.
thats a grand accomplishment. Is there a specific branch of math you are interested in the most? e.g. number theory, applied math, calc etc...

14. Number theory! More specifically, algebraic number theory.

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