1. Aug 6, 2010

### abiyo

Hi everyone,

Just started studying Real Analysis and I have some questions. If I ask anything stupid please excuse me.

1. How do you generate irrational numbers? In other way, are there irrational numbers beyond
pi, e and roots of numbers?

2. What is division? For 5/2, 8/3 or 18/7 I think of it as divisor, remainder problem. But think of dividing an irrational number by another rational number. What does it mean? I am really having a hard time understanding what division really is.

Thanks
Abiyo

2. Aug 6, 2010

1. There are uncountably many irrational numbers, since R is uncountable and Q is countable. (You can only describe countably many of them, though, since there are only countably many strings you can use to describe them with.)

2. If y is not 0, then x/y = z iff x = yz.

3. Aug 6, 2010

### owlpride

As adriank said, there are a *lot* of them. In general real numbers are defined as the limit of a "convergent" sequence of rational numbers, and most of these limits will be irrational. If you want to generate a random real number, just take an arbitrary infinite decimal expansion:

0.348238723509425034657211894590824987542...

If the tail of your expansion is not periodic (e.g. 0.234545454545...), then the infinite decimal represents an irrational number.

4. Aug 7, 2010

### abiyo

1. So I studied how the construction of real numbers and one thing that is really puzzling me is do irrational numbers mean anything? Or are there just abstract artifacts? so for example pi and e have really pretty properties. Are there irrational numbers like pi and e (and not really roots of numbers) that have special
properties in some way?

2. I understand the formal definition of division but what does it mean as a concept? Addition is incremental
operation, multiplication is repeated addition and subtraction is along the same line as addition. Really what do we mean(conceptually) when we divide two irrational numbers?

3. So I also read in my book that the set of natural numbers has the same cardinality as rational numbers.
I know cardinality implies a one-to-one mapping from N to the given set but given that Q is catching up with N(in a literal sense), are they really equal in cardinality? For example for the closed interval [1,2]
we have two natural numbers 1 and 2. The rational numbers in between have a higher cardinality than 2.
But of course we say that the 1-1 map exists if we consider the whole interval of rational and natural numbers? But shouldn't the mapping in some way account in how Q is slowly catching up?

Thanks

Abiyo

5. Aug 8, 2010

### Char. Limit

For 1, pi and e, but not roots of numbers, belong to a class called transcendental numbers. This means that they are not a root of any finite polynomial with rational coefficients. I view transcendence to be a very special property. I believe the Euler-Mascheroni constant might be transcendental as well.

6. Aug 8, 2010

### slider142

All numbers have some special property; pi and e are the most common irrational numbers, because of their relation to periodic and dynamic behavior, respectively.

What does it conceptually mean when you multiply two irrational numbers? Ie., you are not adding 2 to itself pi times, since you cannot count pi occurences of the number two, only a natural number of occurences.
If you backtrack back to division amongst natural numbers, you can see that division can be defined as the number of times a number can be subtracted from another; repeated subtraction to multiplication's repeated addition.
Like that repeated addition definition, this does not generalize easily to irrational numbers without considering some gnarly limiting arguments.
An easier way is to note that each element must of course divide itself once: call the divisor its multiplicative inverse. This generates the full richness of an algebra with division without needing any sort of strange mechanical definition of division; only multiplication of numbers including their multiplicative inverses. Defining division becomes redundant.
The original question is then answered easily: 2/pi is the number x = 2*pi-1. One can either stop here and find a limiting procedure for identifying the decimal expansion of the multiplicative inverse of pi, or multiplying both sides by pi, we see that x is the number satisfying the equation pi*x = 2. This is another easily approximated number, once we know a decimal expansion of pi.

I do not understand what you mean by "catching up". Can you use more rigorous language to define the concept?
Note that cardinality discards any structural information about the set in question; it is concerned only with the elements of sets, not their relationships to each other or any type of topology, ordering, algebraic structure or such that one may place on them. Those structures are studied with other tools. For example, you are concerned that the set of rational numbers that satisfy certain order relations for the natural numbers 1 and 2 is an infinite set, and thus you can associate an infinite set of rationals for each pair of natural numbers.
This is not a problem, for you have not identified which type of infinity it is that is between 1 and 2, just that it is indeed infinite. If it is a countable infinity (which it is), then it is easy to prove that a countable amount of countable infinities is a countable infinity.
Since both sets are infinite, our comparison must come from identifying single elements with single elements and noting whether anything appears to be amiss. Trying to relate infinite sets with finite sets first, not knowing whether those infinite sets can be associated or not, is a path to confusion.

7. Aug 8, 2010

### jgens

I'm not sure what you mean by "catching up", but it's not difficult to prove that the natural numbers and rational numbers have the same cardinality ...

For example, since every rational number can be written in the form m/n where m is an integer and n is a natural number, it's satisfactory to show that there is a bijective function between every pair (m,n) and the natural numbers. Define f:Z X N -> N such that f(m,n) = 2m3n for non-negative m and f(m,n) = 5|m|7n for negative m. Since each natural number's prime factorization is unique, this guarantees that f is bijective, completing the proof.

There might be a mistake or two in my proof (it's late and I won't bother to check it), but the main idea is there. If you do a quick google search on the matter, you'll probably find dozens of other proofs concerning this matter.

8. Aug 8, 2010

It hasn't even been proven that it's irrational.

Regarding 3, when you say N and Q have the same cardinality, that means they are effectively the same when considered as sets (technically speaking: they are isomorphic in the category of sets), so they share exactly the same properties when you consider arbitrary functions from or to them. It's only when you consider some additional structure (say, addition or multiplication, or ordering, or topology) that you can really say that Q is bigger than N.

f is not bijective, but it's injective, which shows |Q| ≤ |N|. And it's obvious that |N| ≤ |Q|, so then |Q| = |N| (by the Bernstein-Schroeder theorem).

9. Aug 8, 2010

### jgens

Figures I would make a mistake. Thanks for calling me on it.

10. Aug 8, 2010

### abiyo

Thanks a lot everyone. I understand things way better now.
Let me go back to studying more.

11. Aug 11, 2010

### abiyo

Along this line

So the continuum hypothesis states that there is no set that has a cardinality between N and R. Can anyone explain why? Out of my tiny knowledge, it is really puzzling why this is so given that we have finite sets, N and higher order infinities. Could anyone explain me why in brief? I am not be competent to understand ZF and stuff.

Secondly there are higher infinities than R. But how many? In other words what is the cardinality of higher infinites
A)Is it countable # of infinities
B)Is it uncountable # of infinities
c)Finite(although I doubt c)

Thanks a lot people. I am learning more and more as I ask in this forum.

Abiyo

12. Aug 11, 2010

There is no "why"; the continuum hypothesis can neither be proven nor disproven from ZFC (the results by Cohen and Gödel, respectively). So CH (or its negation) is an axiom, independent of the other axioms of ZFC.

Next, the class of cardinal numbers is so big, it's not even a set. (Proof: Suppose there was a set K consisting of all cardinal numbers. For each cardinal number α ∈ K, let Xα be a set of cardinality α. The union ⋃{Xα | α ∈ K} of all these sets is another set X, and |X| ≥ |Xα| = α for all α in K, because Xα ⊆ X. Thus |X| is the biggest element of K. Now consider the power set P(X) of X. Observe that |P(X)| > |X|, so because |X| is the biggest element of K, |P(X)| is a cardinal number not in K, contradicting the assumption that K contains all cardinal numbers.)

Last edited: Aug 11, 2010
13. Aug 11, 2010

### abiyo

That makes sense but I am still confused really at the idea of cardinality and set.
For example is there really a cardinal between R and P(R) where R stands for the cardinal of real numbers, P(R) is the cardinality of power set.

When you said there is no such that it contains all cardinals( Your proof makes absolute sense) but it is puzzling why we can't abstractly construct such a set and talk about cardinality?

Sorry if I am asking too much. This is really making me restless.

14. Aug 11, 2010

There is something called a class, which is more general than a set. For instance, one can talk about the class of all cardinal numbers. It just so happens that this class isn't a set; instead, it's a proper class.

Regarding your other question, it can be shown that |P(N)| = |R|. Thus, the nonexistence of a cardinal number between |N| and |P(N)| is the continuum hypothesis. Whether this is the case for all cardinal numbers, not just |N|, is called the generalized continuum hypothesis.

Last edited: Aug 11, 2010
15. Aug 12, 2010

### abiyo

I have cleared my confusions now. My last question for this part

Any recommendations for a readable book in set theory?