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2006-10-06 04:20:44 · 15 answers · asked by ro 2 in Science & Mathematics Mathematics

Aparently once has to be bigger

2006-10-06 04:26:54 · update #1

Apparently one has to be larger*

2006-10-06 04:27:47 · update #2

15 answers

@Draco: your argument does prove much; there are also infinitely many rational numbers between any two irrational numbers!

The set of irrational numbers.

It is easy prove that it is AT LEAST as large as the set of the rational numbers. For every rational number x, multiply it with pi and you get an irrational number pi x.

But the fact that there are other irrational numbers apart from the rational multiples of pi does not mean that the set of irrational numbers is strictly greater. (This is a strange thing about infinite sets. For instance, the set of perfect squares is not smaller than the set of all counting numbers!)

A complete answer: the set of rational numbers is COUNTABLE, i.e. of the same size as the set of counting numbers. The set of real numbers is UNCOUNTABLE, i.e. strictly greater. You get the irrational numbers when you remove the rational numbers from the set of real numbers; removing a COUNTABLE number of elements from an UNCOUNTABLE does not make that set smaller, so you end up with an UNCOUNTABLE set of irrational numbers, strictly greater than the COUNTABLE set of rational numbers.

Prove that real numbers are UNCOUNTABLE: the famous "diagonal argument" by Cantor.

Prove that (positive) rational numbers are COUNTABLE: write them in a table with numerators horziontal and denominators vertical. Now zigzag through the table and label all the (reduced) fractions you encounter with a natural number. It is clear that you will encounter any fraction in a finite number of steps, so every rational number gets assigned a natural number, proving that the set of rational numbers is not bigger than the set of natural numbers.

The zigzag pattern looks like this:
1/1; 2/1; 1/2; 1/3; (2/2); 3/1; 4/1; 3/2; 2/3; 1/4; 1/5; (2/4); (3/3); (4/2); 5/1; 6/1; 5/2; etc.

2006-10-06 10:51:08 · answer #1 · answered by dutch_prof 4 · 1 0

They are both infinite, but the number of irrational numbers is a larger kind of infinity.

The set of rational numbers is countable infinite: You can make an infinitely long list of all rational numbers without leaving out one of them. See below for a possible approach. Of course you would never get the list finished, but any rational number would appear on the list at some point given enough time.

It is impossible to do the same thing for the set of real numbers. There is no procedure by which you could make a list of all real numbers without leaving out some of them. For this reason, we say that the set of real numbers is uncountable infinite. The set of irrational numbers is simply the (uncountable) set of real numbers, with the (countable) subset of rational numbers left out, therefore this set is also uncountable infinite.

(Note that irrational numbers are a subset of the real numbers, so arguments involving imaginary numbers do not apply here.)

Possible method to list all rationals:
- start with just 0, 1 and 2
- add all fractions with denominator 2 and numerator smaller than 2
- add the number 3
- add all fractions with numerator 3 and denominator smaller than 3
- add all fractions with denominator 3 and numerator smaller than 3
- add the number 4
- add all fractions with numerator 4 and denominator smaller than 4
- etc etc
You would count some numbers more than once, but that does not really matter.
You will have to fit in the negative numbers, which is not difficult.

2006-10-06 04:57:29 · answer #2 · answered by Joris 2 · 0 0

There are several ways of measuring 'size' of a set of real numbers. In all the ways, the set of irrationals is larger than the set of rationals.

For example, the most crude method ofmeasuring size is 'cardinality'. A set that can be paired up with the collection of all natural numbers is called 'countably infinite'. While both are infinite, the set of rational numbers is 'countably infinite', while the set of irrational numbers is 'uncountbly infinite'. This means that the set of irrationals is much, much bigger.

Another way of measring size is the 'measure' of the set. This is a generalization of the concept of length and is developed in analysis classes. In this way, the set of rationals has measure zero, while the set of irrationals has infinite measure. More specifically, the collection of rationals between 0 and 1 has zero measure and the collection of irrationals between 0 and 1 has measure 1.

Another way to measure 'size' is through the concepts in topology. There is a concept of 'category' which says some things about the size of a set. In this system, the set of rationals has 'category 1' (which means they are 'small') while the set of irrationals has 'category 2' (which means they are 'large').

2006-10-06 04:46:12 · answer #3 · answered by mathematician 7 · 3 0

It is true that there are infinite rational numbers and also infinite irrational numbers. But there are infinitely many more irrational than rational numbers. Rational numbers can be expressed as a fraction m/n where both m and n are integers. However, there is an infinite number of irrational numbers between each rational number. Examples of irrational numbers are pi, e, sqrt(2), etc.

2006-10-06 05:30:04 · answer #4 · answered by Draco Moonbeam 3 · 0 0

There are more irrational numbers than rational numbers, but there are as many rational numbers as integers. See link below.

The basic idea is that it's possible to pair every integer with a rational number. But when you try and pair rational numbers with irrational numbers (usually shown by trying to pair integers with real numbers), you can always show at least one number in the latter set that is not paired. So the irrationals (or reals) have more elements. Counterintuitive but true.

2006-10-06 04:51:32 · answer #5 · answered by Joe C 3 · 0 0

There are more irrational or imaginary numbers. I think infinately more. Because there are different degrees of infinity, i believe you can say there are more irrational numbers.

Think of the rational numbers as a 1 dimentional line. They go to +infinity to -infinity so there are a lot of them, but irrational numbers are that line with another dimention added on. Like going from a line to a piece of paper. It has an x and y direction. Or real and imaginary. This makes irrational numbers more "dense" i suppose and so at the real number 1 there are infinite irrational numbers along the y direction, and at the number 2 there are infinate numbers along the y direction, and so on.

Lots and lots.

2006-10-06 04:27:38 · answer #6 · answered by Bash_03 2 · 2 2

Irrational numbers, their universe of acceptability is larger
For example:

sqrt(-1) or in fact sqrt of any negative number would not fall within the scope of rational numbers but it falls within the scope of irrational numbers, thus the set of irrational numbers is larger.

2006-10-06 04:30:52 · answer #7 · answered by bostoncity_guy 2 · 0 2

Bash is right, there are degrees of infinite sets.

As another example, there are more real numbers than whole numbers, right? For every two consecutive whole numbers (like, say, 0 and 1) there are an infinite number of real numbers between them.

2006-10-06 04:38:59 · answer #8 · answered by Sheik Yerbouti 4 · 0 2

Both are infinite order. Both are same order.
The reason. for every rational number say n we can find one irrational number. if n is rational n+e is irrational. For every irrational one can find rational. so mapping is 1 to 1. so same order

2006-10-06 04:52:23 · answer #9 · answered by Mein Hoon Na 7 · 0 4

Since both sets are theoretically infinite, they're the same, right?

2006-10-06 04:29:32 · answer #10 · answered by Anonymous · 0 2

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