# Sme question about irrational numbers

• Tann
In summary, the conversation discusses Cantor's second diagonal proof and how it shows that there are more irrational numbers than rational numbers. The cardinality of irrational numbers is shown to be 2^aleph_0, which is the same as the power of the continuum, or the cardinality of the real numbers. However, a question is raised about whether the set of all irrational numbers truly has the power of the continuum, as it is shown that sets of irrational numbers can be disjoint and still have the same cardinality. The conversation also includes examples using S1 and S2 to demonstrate this concept.

#### Tann

Some question about irrational numbers

Our teacher showed us Cantor's second diagonal proof.

He said that by this proof we can show that there are more irrational numbers
than rational numbers.

He also said that the cardinality of natural numbers or rational numbers has a magnitude called aleph_0, where the cardinality of irrational numbers has a magnitude of 2^aleph_0.

He said that any set that its cardinal = 2^aleph_0 has the power of the continuum, so we can conclude that the set of all irrational numbers has the power of the continuum.

I thought about it and I have this question:

If we have two sets of irrational numbers (let us call them S1 and S2) such that:

S1={x: x < 1}

S2={x: x > 1}

then in this case the boundary point between S1 and S2 is the natural number 1.

In this case S1 and S1 are disjoint sets, because the boundary point is a natural number.

But then we can conclude that the irrational numbers do not have the power of the continuum, even if the magnitude of their cardinal is 2^aleph_0.

I think that from what I have shown, the irrational numbers are uncountable (by Cantor’s
Second diagonal method) but because S1 and S2 are disjoint sets, the set of all irrational numbers does not have the power of the continuum.

So, my question is:

What is the cardinality of the set of all irrational numbers, and how it is related to the continuum hypothesis?

Last edited:
Tann said:
But then we can conclude that the irrational numbers do not have the power of the continuum, even if the magnitude of their cardinal is 2^aleph_0.

I do not see how this statement follows from your previous statements.

"Power" means cardinality. "The continuum" means the reals. "Power of the continuum" means cardinality of the reals.

Since the reals have cardinality $2^{\aleph_0}$, anything with that cardinality has the power of the continuum.

The cardinality of the irrationals follows from this fact about cardinal arithmetic:
if α and β are infinite cardinals, then α + β = max {α, β}

hypermorphism said:
I do not see how this statement follows from your previous statements.
There is no common member to S1 and S2 because the boundrey point is not an irrational number but a natural number.

In this case we can conclude that the set of all irrational numbers does not have the power of the continuum.
Hurkyl said:
"Power" means cardinality. "The continuum" means the reals. "Power of the continuum" means cardinality of the reals.
Thank you for your better version.
Hurkyl said:
Since the reals have cardinality , anything with that cardinality has the power of the continuum.

The cardinality of the irrationals follows from this fact about cardinal arithmetic:
if α and β are infinite cardinals, then α + β = max {α, β}

Is the cardinality of all irrational numbers is 2^aleph_0?

erm, you are very mistaken. in fact there is ablsoutely no connection between the premiss and conclusion with those two sets S1 and S2.

just because I can write a set S, as the union of two disjoint sets, A and B, does not mean that A and B must have different cardinality from S. (infinite) cardinals do not behave like that.

Do you know the criteria for sets having the same cardinality? One of the equivalent ones is that there is a bijection between them. So, N, the natruals is the disjoint union of the odd and even elements yet all have the same cardinality *by definition*. Prove this. And accept it is the definition.

Yes the set of all irrationals has the cardinality of the continuum.

I don't see your reasoning behind your claim that the irrationals don't have the cardinality of the continuum, which is wrong anyway.

Tann said:
There is no common member to S1 and S2 because the boundrey point is not an irrational number but a natural number.

In this case we can conclude that the set of all irrational numbers does not have the power of the continuum.
In addition to the above replies, not that the set of all even natural numbers and the set of all odd natural numbers are disjoint but have the same cardinality because there exists a bijection between them.
If you have a set of 3 distinct apples and another set of 3 distinct oranges, you don't claim that they have a different amount of elements simply because the nature of the elements are different (Ie., by claiming the set of oranges doesn't include any apples).

Last edited:
matt grime said:
Do you know the criteria for sets having the same cardinality? One of the equivalent ones is that there is a bijection between them. So, N, the natruals is the disjoint union of the odd and even elements yet all have the same cardinality *by definition*. Prove this. And accept it is the definition.
I do not say that S1 and S2 do not have the same cardinality.

I say that the irrational numbers are uncountable but they do not have the power of the continuum.

Our teacher gave this example to show that rational numbers do not have the power of the continuum.

He wrote:

S1={x: x2 < 2}

S2={x: x2 > 2}

where S1 and S2 are all of R members.

In this case the boundary point is the square root of 2, where in the left side of it we have S1 and in the right side of it we have S2, where S1 and S2 are disjoint sets of rational numbers.

The same thing holds for:

S1={x: x < 1}

S2={x: x > 1}

where S1 and S2 are all of irrational members.

In this case the boundary point is the natural number 1, where in the left side of it we have S1 and in the right side of it we have S2, where S1 and S2 are disjoint sets of irrational numbers.

It means that the collection of all irrational numbers are uncountable (by Cantor's second diagonal proof) but they do not have the power of the continuum, because the boundary point is the natural number 1, that cannot give us the ability to define an irrational number as a boundary point of both S1 and S2.

Now, we can conclude that the set of all irrational numbers are uncountable but they do not have the power of the continuum.

Last edited:
Your teacher's example has nothing to do with cardinality. It's a lesson about connectedness -- the rationals (and the irrationals) are not a connected set. Or, loosely speaking, they do not form a continuum.

Something doesn't have to be a continuum to have the same cardinality as R.

Hurkyl said:
Your teacher's example has nothing to do with cardinality. It's a lesson about connectedness -- the rationals (and the irrationals) are not a connected set. Or, loosely speaking, they do not form a continuum.

Something doesn't have to be a continuum to have the same cardinality as R.

But our teacher told us that the cardinality of all irrational numbers is 2^aleph_0 and 2^aleph_0 = the power of the continuum.

Now you say that the set of all irrational numbers do not form a continuum.

So please let me ask you again: what is the cardinality of the set of all irrational numbers?

I reiterate:

Something doesn't have to be a continuum to have the same cardinality as R.

I do not see how someone can come to this conclusion.

If there is an unclosed hole between S1 and S2 they cannot have the cardinality 2^aleph0 = |R| = c

Last edited:
Cardinality merely talks about a sets. It has absolutely nothing to do with any topological or geometric ideas.

Hurkyl said:
Cardinality merely talks about a sets.
Ho!??, how can you say that there is no connection beween some collection and its cardinality?

So I think that the solution is to say that if 2^aleph_0 is the cardinality of the set of all irrational numbers, then 2^aleph0 < |R| = c

You can say it all you like, but that doesn't change the fact that $|\mathbb{R}| = 2^{\aleph_0}$.

tann your mixing up didfferent properties i.e. not all sunbstes of the rational numbers or irartional numbers bounded form above have a least upper bound (which is in the set they are subsets of) whears every subset of the real numbers which is bounded form above has a leats upper bound which is a rela number. This proeprty of the rational and irrational numbers does not dictate their cardinality.

Hurkyl said:
You can say it all you like, but that doesn't change the fact that $|\mathbb{R}| = 2^{\aleph_0}$.
Sorry but I came here to share my thoughts with others and learn somthing from professionals.

I do not see that you give any proof to your "facts", so please be professional and give us a proof that clearly show us that the cardinality of the set of all irrational numbers has the power of the continuum.

And if you can, then please explain in simple words evry part of this proof.

Last edited:
Perhaps what is confusing you is that we can 'take away' elements of an iifnite set a dn still have a set with the same cardinality. This is just the dfeitnion thoguh of an inifte set i.e. a set which has proper subsets with the same cardinality as itself.

I already told you the fact that for infinite cardinals, $\alpha + \beta = \max \{\alpha, \beta\}$.

Now, the cardinality of the reals equals the cardinality of the rationals plus the cardinality of the irrationals. Apply the aforementioned fact. :tongue2:

Perhaps what is confusing you is that we can 'take away' elements of an iifnite set a dn still have a set with the same cardinality. This is just the dfeitnion thoguh of an inifte set i.e. a set which has proper subsets with the same cardinality as itself.
I know the properties of the tarsfinite universe, but it says nothing about the cardinality of the set of all irrational numbers and the power of the continuum, and you did not give until now even a single proof that the set of irrational numbers has the power of the continuum.

Hurkyl said:
I already told you the fact that for infinite cardinals, $\alpha + \beta = \max \{\alpha, \beta\}$.

Now, the cardinality of the reals equals the cardinality of the rationals plus the cardinality of the irrationals. Apply the aforementioned fact. :tongue2:
All I see here is a collection of symbols, that does not prove anything.

So please someone give a proof?

In N we can show that the is a bijection between N and a proper subset of N.

Please show me how can you prove the same thing between R and the set of all irratioanal numbers?

Last edited:
Well if we accpet a few things, for example:

a) That the cardianlity of Q is |N|

b) R is the union of Q and the rationals

If the cardianlity of the irrationals is less than |R| then the cradianlity of R must be less tahn |R|, if the cardianlity of the irrationals is greater than |R| then the cardinality of R must be greater than |R|, obviously neither is the case, so the irratipoansl have the cardianltiy |R|.

jcsd said:
If the cardianlity of the irrationals is less than |R| then the cradianlity of R must be less tahn |R|

How you come to this conclusion?

In N we can show that there is a bijection between N and a proper subset of N.

Please show me how can we prove the same thing between R and the set of all irrational numbers?

Last edited:
We can prove there exists a bijection between R and the set of irrational numbers by proving their cardinalities are equal.

We can prove their cardinalities are equal using the cardinal-arithmetic fact I showed you -- adding infinite sets does not yield a bigger set. This is by far the simplest way, and exemplifies a vital fact about cardinalities.

It's far more important that you understand this approach than it is for you to see an explicit bijection.

From the fact that the cardiamlity of the union of two infinite sets is always the cardinality of the 'larger' set.

f=
x + 10 if x is a postive irrational
1/x + pi + sqrt(2) if x is a postive rational >1
1+ pi + sqrt(2) if x=1
x + pi if x is a postive rational <1
pi + sqrt(2) if x = 0
x - 10 if x is a neative irrational
1/x - pi - sqrt(2) if x is a negative rational < -1
-1 - pi - sqrt(2) if x = -1
x - pi if x is a negative rational > -1

is a bijection bwteen the reals and a subset of the irrationals, as the irrationals is a subsert of the reals we know that the cardinality can't be greater than |R|, but as there exists a bijection between a susbet of the irratioanls and the reals it must be |R|.

Last edited:
Clever -- much more concise that what I would have come up with (though mine is an actual bijection between R and R-Q). I think you made one or more typos in the last line of your function, though.

However, there is a minor technicality -- we don't know (1/pi - 1/e) is an irrational number.

jcsd said:
From the fact that the cardiamlity of the union of two infinite sets is always the cardinality of the 'larger' set.
Hey people, I do not see any fact here, all I see until now is that you try to force on me your points of view, whitout giving even a single logical proof about these "facts".

So where are the logical and rigorous proofs behind these "facts"?

jcsd said:
f=
x + 2 if x is a postive irrational
1/x + 1/pi if x is a postive rational >1
x + 1/e if x is a postive rational <1
1/pi if x = 0
x -2 if x is a neative irrational
1/x - 1/pi if x is a negative rational < -1
x + 1/e if x is a negtaive irrational > -1
Nice try but I do not see any proof here because I do not see here the set of all irrational numbers (you used only pi and e to get some result).

Last edited:
I've been somewhat suspicious for a while now that you have no interest in understanding why the cardinality of the irrationals is the same as that of the reals, but instead desire to take potshots at people who try to explain why. Such an attitude is not welcome here.

Surely Tann you accept that a set of gretare cardinality cannot be a subset of a set of lesser cardianlity.

Last edited:
Hurkyl said:
I've been somewhat suspicious for a while now that you have no interest in understanding why the cardinality of the irrationals is the same as that of the reals, but instead desire to take potshots at people who try to explain why. Such an attitude is not welcome here.
Sorry, you are right, and thank you for this post, I'll hold my horses.

Last edited:
jcsd said:
Surely Tann you except that a set of gretare cardinality cannot be a subset of a set of lesser cardianlity.

What I say is this.

As much as I know, we have no way to use a 1-1 mapping between any two sets which are uncountable.

We can make an extension in our mind from N to R but (as much as I know) there is no proof that a 1-1 mapping holds in R as it holds in N or Q.

If I am wrong then please show the proof that clearly shows that the 1-1 mapping holds also in uncountable sets.

Last edited:
Tann, do you even know what a cardinal number is? I don't think you do. And I'm writing as one of the professionals you came here to learn from (your words). You have not shown the slightest inclination to LEARN anything. You have not learned when two sets have the same cardinality, not learned that connectedness and cardinalilty are totally unrelated, and even in your own posts said "[your teacher says] it has the power [cardinality] of the continuum" followed by "but it isn't a continuum" well, look at the first phrase. Does it say it *is* the continuum? Do you think the continuum is uniqely determined merely by its cardinality? If you knew what the *definition* of cardinality is you would see that that is completely and trivially false.

Tann said:
What I say is this.

As much as I know, we have no way to use a 1-1 mapping between any two sets which are uncountable.

We can make an extension in our mind from N to R but (as much as I know) there is no proof that a 1-1 mapping holds in R as it holds in N or Q.

If I am wrong then please show the proof that clearly shows that the 1-1 mapping holds also in uncountable sets.

Well I've already shown you a one-to-one function between two uncountable sets, but here's a simple one:

the set A which is set the set of all real numbers between 1 and o is uncountable and the set B which is the ste of all rela numbers between 1 and 2 is uncountable and f:A-->B f(x) = x+1 is a bijection between them. Or even smple f:R-->R f(x) = x is a bijection.

What on Earth does it even mean for "a mapping to hold"?

Are you sure you're not actually Doron Shadmi? No, you're more coherent than him.

Oh, these "points of view" we're forcing on you are either definitions, or are trivial consequences of the definitions.

matt grime said:
Tann, do you even know what a cardinal number is? I don't think you do. And I'm writing as one of the professionals you came here to learn from (your words). You have not shown the slightest inclination to LEARN anything. You have not learned when two sets have the same cardinality, not learned that connectedness and cardinalilty are totally unrelated, and even in your own posts said "[your teacher says] it has the power [cardinality] of the continuum" followed by "but it isn't a continuum" well, look at the first phrase. Does it say it *is* the continuum? Do you think the continuum is uniqely determined merely by its cardinality? If you knew what the *definition* of cardinality is you would see that that is completely and trivially false.
Well a cardinal is a size of a set and it can be defined by the 1-1 mapping technique.

Please show that this technique holds also in uncountable sets.

Last edited:
Define "holds". And cardinals are defined by bijections, not , usually, injections. SO, what is a cardinal number then? Approximately.

And, to use your own definition. What has connectedness to do with size?