Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am now reading the definition of irrational number, which we can describe by the following terms: suppose that we have divided all rational numbers into two classes, a lower class and an upper class, such that every number of the lower class is less then all numbers in the upper class. From the book, it is remarked that we may have three different cases:

1$\text{}$. The lower class can have a greatest number and the upper class no smallest number.

It is clear that if we take number 5 as a division number, in lower class we will have 4,3,2,1 so largest is 4 and in upper just 7,7,7,7,7 no smallest number right yes?

2$\text{}$. The upper class can have a smallest number and lower class no greater number.

It is clear that in case of 5, upper class can be 6,7,8,9,10, so smallest is 6, while lower 4,4,4,4,4,4,4 no greater number and third one.

3$\text{}$. The lower class can have no greatest number and the upper class no smallest number.

I did not understand the method which is used in the book for the third case, namely: if we arrange positive numbers and their squares so that each square number is underneath it's corresponding number, then since the square of a fraction in it's lowest terms is a fraction whose numerator and denominator are perfect squares, we see that there are not rational numbers whose squares are 2,3,5,6,7,8,10,11 and so on. Then the author used an approximation method to show that there are rational numbers which are as near to these numbers as we please and finally he divided so that all negative numbers, 0, and positive numbers whose square is less then 2, and all numbers whose square are more than 2. For rational numbers which are near to numbers which I mentioned are looking like this:

 2, 1.5, 1.42, 1.415, 1.4143
  1, 1.4, 1.41, 1.414, 1.4142

I did not understand why author used these numbers, so please help me, sorry if text is too much big, I couldn't express my question otherwise.

share|improve this question
3  
Which book are you reading? –  lhf Dec 23 '11 at 20:11
add comment

3 Answers 3

up vote 3 down vote accepted

You may have missed the fact that this requires you to split all rational numbers (that is numbers which can be written as $\frac{a}{b}$ with $a$ an integer and $b$ a positive integer where they have no common factor above $1$) into two subsets.

If you divide them into two sets where the lower set is everything negative or whose square is less than or equal to 25, you will find numbers like $-1, 5, \frac{3}{2}, \frac{500}{101}$ in the lower subset and numbers like $6, 999, \frac{5000}{999}$ in the upper subset.

If you divide them into two sets where the lower set is everything negative or whose square is strictly less than 25, you will find numbers like $-1, \frac{3}{2}, \frac{500}{101}$ in the lower subset and numbers like $6, 5, 999, \frac{5000}{999}$ in the upper subset.

If you divide them into two sets where the lower set is everything negative or whose square is less than 2, you will find numbers like $-1, 1, \frac{141421}{100000}$ in the lower subset and numbers like $6, 2, \frac{141421357}{100000000}$ in the upper subset.

share|improve this answer
    
but how to understand that lower class has no largest or smallest elements? –  dato datuashvili Dec 23 '11 at 20:08
    
For that you need to know that $\sqrt{2}$ is irrational –  Henry Dec 23 '11 at 22:06
add comment

These number are approximations of $\sqrt 2$, which is an irrational number. Hence it is an example of case 3.

share|improve this answer
add comment

The way we do it would best be viewed by drawing a line and imagining their are only rationals marked on it.

Then mark a point at root(2) - you know this isn't rational - and then you have seperated the rationals into (where intervals are taken in the rationals):

$(-\infty,0]$ U { $ p > 0 | p \in \mathbb Q , p^2 < 2 $} as a 'lower' set and {$p > 0 | p \in \mathbb Q , p^2 > 2$ }

share|improve this answer
    
Yes - irrational is identified with one [or both of] these sets - supremum ['upper limit'] of the lower set for instance. Note that familiar representation is nothing but this same limiting procedure. 1.41, 1.414, 1.4142 etc. Irrationals are completely non-physical constructs just like complex numbers [in the sense one is as 'real' or as imaginary as the other]. Non-physical in the sense there is nothing corresponding to infinite precision irrationals in real universe. –  Fakrudeen Jul 26 '13 at 10:36
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.