Ok, let's count them

Number Theory Level 2

In terms of cardinality, are there more rational or irrational numbers?

Since there are infinitely many rational as well as irrational numbers, it cannot be determined Irrational Numbers Both are equal in number Rational Numbers

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Nihar Mahajan by Nihar Mahajan , 20, India

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7 solutions

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John Gilling
Oct 12, 2015

Using knowledge of cardinality and classical sets , the rational numbers are countably infinite, but the real numbers are uncountable. It follows that their set difference, i.e. the irrational numbers, must be uncountable. To see this, a simple contradiction argument does the trick: if the irrational numbers were countably infinite, then the real numbers would have to be countably infinite as well, as the union of two countably infinite sets.

119 Helpful 8 Interesting 5 Brilliant 14 Confused

So? Your proof is not over here is it? Can you prove that the real are uncountable?

Benjamin Katz-Crowther - 5 years, 8 months ago

Oh, my apologies, Benjamin! I didn't realize that a proof of such a fact was expected or required in an informal setting like this.

For those unfamiliar with this result, Cantor's diagonalization proof is one of the easiest ways to see this. Most presentations of the proof prove the uncountability of the open interval ( 0 , 1 ) (0,1) , and use proper inclusion to imply uncountability of the reals.

Hope this helps!

John Gilling - 5 years, 8 months ago

To elaborate on John, take the reals from 0 to 1. Assume these are countable. Then we can make a one-to-one correspondence between them and the natural numbers, and use this to put them into a list, say for example:

  1. 0.8134...
  2. 0.6723...
  3. 0.1567...
  4. 0.7583...

and so on to infinity. Now we construct a real number that is not on this list. We do so in decimal form. Our first digit should be different from our first number's first digit, so that it can't be the same number: so we have 0.9.... Now the second digit should be different from the second number's second digit: so we have 0.98.... We repeat this process to get 0.9874..., and we have a number that cannot be on this list: contradiction, therefore the reals from 0 to 1 are uncountable.

Alex Bean - 5 years, 7 months ago

One can also add that there is a bijection between set of real numbers and (0, 1). Hence the cardinality of (0,1) is same as that of set of real numbers. As proved that (0, 1) is uncountable, it follows that set of real numbers is also uncountable. Hence the argument follows

Anand Chitrao - 5 years, 4 months ago

Can you prove that the number of rational numbers in the interval [1,2] are countably infinite whereas the number of irrational numbers in the same interval are unaccountably infinite?

Devendra Kumar Singh - 5 years, 7 months ago

Look, my explanation is different but very simple… From a single rational number you can make infinite irrational numbers e.g. let that number be 2 So, irrational numbers from 2 are sqrt(2), cbrt(2), … and 2^(any rational or irrational number) will result into an irrational number But can you make all the numbers rational from irrational numbers? Let check for sqrt(2), If you square it you will get rational but on other hand if you cube it, it will not be irrational number Hence it proves that irrational numbers are more than rational numbers Other way around, you can say that both type of numbers are infinite but irrational's infinity is relatively bigger than rational's infinity, this concept is known as relative infinity concept

Himanshu Agrawal - 5 years, 4 months ago

For me The best reply

Nuno Ferreira - 5 years, 4 months ago

Dear Himanshu, note that proof does not require example. However, to disprove one counter example is sufficient.. Picking Sqrt(2) from your citation and squaring it to get 2, we can construct infinitely many rationals such as 2/2,=1, 1/2, 1/3, 1/4..... and so on; another series may be 2/3,2/5, 2/7 etc. Each of the previous rationals could be utilized to construct infinitely many rationals. Further, these rationals when raised to different +ve or negative integral exponents , new rational numbers are constructed. So you can not say irrationals are more than rationals.

Devendra Kumar Singh - 5 years, 3 months ago

Irrational numbers are more because,rational plus irrational is equal to irrational.So the basic irrational are incomplete square roots,and plus all the rational numbers.Real numbers =rational numbers+irrational numbers: irrational numbers and (rational+irrational numbers) also irrational numbers.

Jamil Nizamani - 5 years, 7 months ago

Then, by the same logic, integers would be more than naturals. Which is not so

Anand Chitrao - 5 years, 1 month ago
Wesley Engers
Oct 22, 2015

I do not approve of this image. The real numbers are partitioned by the rational and irrational numbers. There are no real numbers that are both not rational and not irrational.

83 Helpful 0 Interesting 0 Brilliant 0 Confused

The different regions of a Venn diagram are allowed to be empty.

Shawn Plassmann - 5 years, 7 months ago

Yes, they are. From a technical standpoint there is nothing wrong with having nothing in a Venn diagram space, but from an intuitive standpoint it can be misleading.

Wesley Engers - 5 years, 7 months ago

I agree. When I make the diagram in my classes when teaching this topic, I divide the Real numbers oval/rectangle into two pieces by drawing a line down the middle, and then make more ovals on the Rational side.

Najeeb Sheikh - 5 years, 7 months ago

The interesting problem in terms of cardinality would be the number of algebraic numbers.

Marta Reece - 5 years, 4 months ago

The algebraic numbers are countably infinite. Since the real numbers are uncountably infinite, the real transcendental numbers must be uncountably infinite.

Adam Davis - 4 years, 9 months ago

This a very wrong approach on set theory. The regions of a Venn Diagram can be empty and there is absolutely no reason for them not to be. So, as long as there are no elements in the space, the image is perfect.

Nilabha Saha - 4 years, 9 months ago
Lukas Leibfried
Feb 11, 2016

Rational numbers are countably infinite.

Irrational numbers, on the other hand, are uncountably infinite.

Irrational numbers win.

42 Helpful 0 Interesting 0 Brilliant 0 Confused

Here is one way to prove this

Lemma Sum of countably infinite zeroes is zero

Proof (Lemma) Assume sum of n zeroes (n being a a positive integer) and write a sequence Sn which turns out to be 0,0,0,..., i.e., lim as n goes to infinity Sn is zero

Now consider a random population uniformly distributed in [a,b]. Let p be the probability that a sample drawn from this population is rational. p would then be a countably infinite sum of zeroes and is hence zero

Therefore Pr of picking an irrational is 1 indicating that it is the bigger set

11 Helpful 0 Interesting 0 Brilliant 1 Confused

how can it be inferred from sum? the question is asked for numbers.

Jamil Nizamani - 5 years, 7 months ago

It is rather subtle and uses measure theory. The point is that the probability of, say, x being in the interval [a,b] is clearly 1. We can "divide" this probability into the probabilily that p taking a rational or irrational value in [a,b]. It is clear that this is exclusive and exhaustive. He argued that the first evaluation (i.e. that the probabiltly that p( Q \mathbb{Q} )=0) and so, must have that p( R ∖ Q \mathbb{R}\setminus \mathbb{Q} )=1 .

Patrick Bourg - 5 years, 4 months ago
Karan Gujar
Feb 13, 2016

"Between any two rational numbers there are infinite irrational numbers but between 2 irrational numbers there is atleast one rational number"

9 Helpful 0 Interesting 0 Brilliant 0 Confused
Olivier Poulard
Feb 10, 2016

Maybe a simpler way to demonstrate: one can make an infinite number of irrational numbers out of rational numbers, not the other way around.

8 Helpful 0 Interesting 0 Brilliant 1 Confused

I've given the proof a try by adapting cantor's diagonal argument:

Take the following rational-->irrational function F, given for base 10:

For every non-repeating rational number r you take the irrational number obtained by adding decimals increased according to an arithmetic progression. If the last two digits of r are not in sequence, add a d=1 progression:

For r=0.5, you have F(r)=0.567891011121314...

For r=1.14, you have F(r)=1.1456789...

For r=773, you have F(r)=773.45678...

If the last digit of r is the result of adding 1 to the penultimate, you instead add a d=2 progression:

r=1.23, F(r)=1.23579111315...

r=0.56 F(r)=0.568101214...

r=0.12345, F(r)=0.12345791113...

r=23.4, F(r)=23.468101214...

For numbers like 0.33333..., you just add the d=1 progression between the repeating decimals:

r=0.3333..., F(r)=0.30313233343536...

r=0.472472472..., F(r)=0.4720472147224723...

If I prove that F is injective, and find a set of infinite irrational numbers that are not contained in its image, then the irrational number set is larger than the rational.

To prove F is injective, I describe how every F(r) can be linked to a single r by the following process:

1: If the decimals form a progression after a certain point in the number, then remove all but the first digit forming that progression.

2: If the decimals do not form a progression, it's a progression intertwined with a repeating series. Find the progression decimals and remove them. (There are formal ways to do that)

Hence, F is injective, so every rational number r has a corresponding irrational number x=F(r).

Still, there are infinite irrational numbers not contained in the image of F, like the set {0.101001000100001..., 0.202002000200002..., ...}.

Hence, the set of the irrational numbers is larger than the set of the rational numbers.

Is this right? Maybe unnecessarily bulky? I'm not a mathematician, just an enthusiast, so forgive any slips.

4 Helpful 0 Interesting 0 Brilliant 0 Confused

By definition, a real number R=0.---(arbitrarily large no. M of decimal places)---- with M<infinity, and next arbitrarily chosen higher real number with a higher M but that is also < infinity, has infinite number of irrational numbers between them.... Simply put: Set of irrational no.s is a infinite set of uncountable no.of numbers between each pair in a countably infinite set of rational numbers...

But since M can be as big as arbitrarily chosen, I think, the above paragraph is meaningless ;) or is it!? :D

Vikram Kumar Vuyyuru - 5 years, 7 months ago

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