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Constructing segments with irrational lengths

Since rational numbers can be expressed as fractions (or ratio of two integers), rational numbers can be easily located on the number line using compass and straightedge construction. In principle, a segment of any length can be divided into any number of parts; hence, it is possible to locate any rational number on the number line. Irrational numbers, on the other hand, cannot be expressed as ratio of two integers, so the big question is:

How do we locate irrational numbers on the number line?

The question above is equivalent to: “How do we construct a segment with irrational lengths?”

It is easy to locate some irrational numbers on the number line even with compass and straightedge construction. The irrational $\sqrt{2}$ can be located by constructing square ABCD (Figure 1), getting the diagonal AC, and constructing a circle with radius AC. It follows that the length of AE is $\sqrt{2}$ , and the coordinates point E are $(\sqrt{2},0)$ .

Figure 1

It is also apparent that the easiest way to construct segments with irrational lengths is by constructing diagonals of rectangles. In Figure 2, OB, OC and OD have lengths $\sqrt{2}$ , $\sqrt{5}$ , and $\sqrt{10}$ respectively. More complicated construction is required to construct other irrational lengths, $\sqrt{3}$ for instance, which is the length of EH. » Read more

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Counting the Real Numbers

August 30, 2010 GB Calculus and Analysis, College Mathematics, Set Theory

If we are in a room full of ballroom dancers where each male dancer has a female dancer partner, and no one is left without a partner, we can say that there are as many male as female dancers in the room even without counting. In mathematics, we say that there is a one-to-one correspondence between the set of male dancers and the set of female dancers.

Pairing Infinite Sets

In the A Glimpse at Infinite Sets, we have learned that if we can pair two sets in one-to-one correspondence, we can say that the two sets have the same number of elements. The number of elements of a set is its cardinality. Therefore, the cardinality of the binary numbers {1,0} is 2 and the cardinality of the set of the vowel letters in the English alphabet {a, e, i, o, u} is 5.

The pairing of sets can be extended to compare sets with infinite number of elements or infinite sets. In Figure 1, it is clear that it is possible to pair the set of integers with the set of counting numbers in one-to-one correspondence (can you see why?). Infinite sets whose elements can be paired with the set of counting numbers in one-to-one correspondence is said to be countably infinite.

Figure 1

As a consequence of the analogy above, we can conclude the cardinality of counting numbers is equal to the cardinality of integers (Can you see why?). » Read more

Leave a comment 1=0.999...., aleph null, binary numbers, cantor's diagonalization system, cardinality of real numbers, countably infinite sets, diagonal numbers, infinite sets, non-terminating decimals, one-to-one correspondence, terminating decimals, uncountably infinite sets

Geometric Sequences and Series

June 9, 2010 GB Calculus and Analysis, College Mathematics, High School Algebra, High School Mathematics

Introduction

We have discussed about arithmetic sequences, its characteristics and its connection to linear functions. In this post, we will discuss another type of sequence.

The sequence of numbers 2, 6, 18, 54, 162, … is an example of an geometric sequence. The first term 2 is multiplied by 3 to get the second term, the second term is multiplied by 3 to get the third term, the third term is multiplied by 3 to get the fourth term, and so on. The same number that we multiplied to each term is called the common ratio. Expressing the sequence above in terms of the first term and the common ratio, we have 2, 2(3), 2(3²), 2(3³), …. Hence, a geometric sequence, also known as a geometric progression, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed non-zero number called the common ratio.

The Sierpinski triangle below is an example of a geometric representation of a geometric sequence. The number of blue triangles, the number of white triangles, their areas, and their side lengths form different geometric sequences. It is left to the reader, as an exercise, to find the rules of these geometric sequences.

Figure 1 - The Seriepinski Triangles.

To generalize, if a₁ is its first term and the common ratio is r, then the general form of a geometric sequence is a₁, a₁r, a₁r², a₁r³,…, and the n^th term of the sequence is a₁r^n-1.

A geometric series, on the other hand, is the sum of the terms of a geometric sequence. Given a geometric sequence with terms a₁r, a₁r², a₁r³,…, the sum S_nof the geometric sequence with n terms is the geometric series a₁ + a₁r + a₁r², a₁r³ + … + ar^n-1. Multiplying S_n by -r and adding it to S_n, we have

Hence, the sum of a geometric series with n terms, and $r \neq 1 = \displaystyle\frac{a_1(1-r^n)}{1-r}$ .

Sum of Infinite Geometric Series and a Little Bit of Calculus

Note: This portion is for those who have already taken elementary calculus.

The infinite geometric series $\{a_n\}$ is the the symbol $\sum_{n=1}^\infty a_n = a_1 + a_2 + a_3 + \cdots$ . From above, the sum of a finite geometric series with $n$ terms is $\displaystyle \sum_{k=1}^n \frac{a_1(1-r^n)}{1-r}$ . Hence, to get the sum of the infinite geometric series, we need to get the sum of $\displaystyle \sum_{n=1}^\infty \frac{a_1(1-r^n)}{1-r}$ . However, $\displaystyle \sum_{k=1}^\infty \frac{a_1(1-r^n)}{1-r} = \lim_{n\to \infty} \frac{a_1(1-r^n)}{1-r}$ .

Also, that if $|r| < 1$ , $r^n$ approaches $0$ (try $(\frac{2}{3})^n$ or any other proper fraction and increase the value of $n$ ), thus, $\displaystyle \sum_{n=1}^\infty \frac{a_1(1-r^n)}{1-r} = \lim_{n \to \infty} \frac{a_1(1-r^n)}{1-r} = \frac{a_1}{1-r}$ . Therefore, sum of the infinite series $\displaystyle a_1 + a_2r + a_2r^2 + \cdots = \frac{a_1}{1-r}$ .

One very common infinite series is $\displaystyle \sum_{n=1}^{\infty} \frac{1}{2n} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots$ , or the sum of the areas of the partitions of the square with side length 1 unit shown below. Using the formula above,

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{2n} = \frac{a_1}{1-r} = \frac{\frac{1}{2}}{1-\frac{1}{2}} = 1$ .

Figure 2 - A representation of an infinite geometric series.

This is evident in the diagram because the sum of all the partitions is equal to the area of a square. We say that the series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{2n}$ converges to 1.

8 comments common ratio, geometric progressions, geometric sequence, infinite geometric sequence, sierpinski triangle, sum of infinite geometric series

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