Mathematics Dictionary

Definition - A real number has a postition on the number lines.
Real numbers :
- Integer - Whole numbers 0, 1, 2, 3, 4, 5, ......
  - Odd number :
    - Integers have no factor of 2.
    - Examples : 1, 3, 5, ... are odd numbers.
  - Even number
    - Integers have factor of 2.
    - 2, 4, 6, ... are even numbers.
- Rational number
  - It can be expressed as ratio of two integers.
  - Example : 0.3 = 3/10.
- Irrational number
  - Defintion : It can't be expressed as ratio of two integers.
  - It has position on the number line. e.g. pi is between 3 and 4.
  - Example : pi = 3.141592...., Sqr(2) = 1.41....
Properties of real numbers
- Positive number : It is on the Right of zero of a number line.
- Negative number : It is on the left of zero of a number line.
Decimal number and fraction
- Repeating decimals - Example : 0.33333... is repeating which is 1/3.
  - Let x = 0.33333....
  - Then 10*x = 3.3333...
  - Hence 10*x - x = 3.
  - Hence x = 1/3.
- Fraction = numerator/denominator.
  - Proper fraction - denominator is greater than numerator.
  - Improper fraction - denominator is less than numerator.

Diagram : Number line

Study | Section 1 : Number line

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AL 03 02. Complex number system
Defintion

Imaginary number : i = Sqr(-1).
- i^2 = -1.
- i^3 = (i^2)*i = -i
- i^4 = (i^3)*i = -i^2 = -(-1) = 1.
- i^5 = (i^4)*i = i
Complex number : z = a + b*i.
- Example : (1 - i)*(1 + i) = 1^2 - i^2 = 1 -(-1) = 2.
- Example : (1 + i)*(1 + i) = 1^2 + 2*1*i + i^2 = 1 + 2*i +(-1) = 2*i.

Complex number coordinate system

It is similar as the rectangular coordinate system.
The x-axis is the real number.
The y-axis is then imaginary number.
z = a + b*i.
- Coordiante at (a,b).
- Make angle with x-axis is A = arctan(b/a).
- Distance to origin is r = Sqr(a^2 + b^2).
- Hence a = r*cos(A) and b = r*sin(A).
Example : z = 1 + 2*i is same as point (1,2).
Example : (1 + i)*(1 + i) = 2*i
- (1 + i) = coordiante (1,1) or (1+i) makes 45 degrees angle with x-axis.
- (1 + i)*(1 + i) = 2*i means makes angle 90 degree with x-axis.

Complex number in polar form

z = a + b*i = r*(cos(A) + i*sin(A)).
Where r = Sqr(a^2 + b^2).
a = r*cos(A) and b = r*sin(A).
A = arctan(b/a)
Example : Convert r = 1/(1+i) into polar form.
- r = 1/(1+i) = (1-i)/((1-i)*(1+i)) = (1-i)/2 = a + b*i.
- Hence a = 1/2 and b = -1/2. r = 1/Sqr(2) = Sqr(2)/2.
- A = arctan(b/a) = arctan(-1) = 225 degrees.
- Hence r = Sqr(2)*(cos(225) + i*sin(225))/2

Conjugate numbers

(a + b*i) and (a - b*i) are conjugate complex numbers
Sum of conjugate numbers = real number. Example (a + b*i) + (a - b*i) = 2*a.
Product of conjugate numbers = real number. Example (a+b*i)*(a-b*i) = a^2 + b^2.
Example : Prove that 1/(1-i) = (1+i)/2
- Change denominator to real number.
- Hence multiply numerator and denominator by (1+i).
- (1+i)*(1-i) = 1^2 - i^2 = 2.
- 1/(1-i) = (1+i)/((1+i)*(1-i)) = (1+i)/2.

Text and examples

Study | Definiotn and application examples of the complex number system.

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AL 03 03. Special properties of real numbers : Factors

Study |
Amicable number pairs
- Example 1. What are amicable number pairs ?
- Example 2. Prove that 220 and 284 are amicable pairs.
Study |
Perfect numbers.

Example 1. What is perfect number ?
Example 2. Prove that 28 is a perfect number.

Properties of numbers

Abundunt numbers : Sum of their factors is greater than number itself.
Deficient numbers : Sum of their factors is less than number itself.
Perfect numbers : Sum of their factors is equal to than number itself.
Prime numbers : Sum of their factors is 1.
Note : Sum of factors is not included number itself.
Example.
- Sum of factors of 27 = 1 + 3 + 9 = 13. Hence 27 is deficient number.
- Sum of factors of 28 = 1 + 2 + 4 + 7 + 14 = 28. Hence 28 is a perfect number.
- Sum of factors of 29 = 1.
- Sum of factors of 30 = 1 + 2 + 3 + 5 + 6 + 10 + 15 = 42. It is abundunt.

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AL 03 04. Magic number patterns

Circle patterns

Use integers 1 to 33 to make magic number circle.

Find the sum of the numbers along each circle.

Find the sum of the numbers along each diagonal (number 9 is not included).

Study | Magic numbers circle pattern is given here.

Square patterns

Use integers 1 to 16 to make magic number circle.
Find the sum of the numbers for each row.
Find the sum of the numbers for each column.
Find the sum of the numbers for each diagonal.
Study | Magic numbers circle pattern is given here.

Square patterns

Use integers 1 to 9 to make magic square.
Use integers 1 to 25 to make magic square.
Use integers 1 to 49 to make magic square.
Use integers 1 to 121 to make magic square.
Reference.
- On PC computer : MD2002 program 01 10.
- On Book : Chapter 11 of Computer mathematics by Dr. K. G. Shih.

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AL 03 05. Numbers arranged in matrix pattern

Use integers to arrange a mattrix pattern

Numbers in first row are arranged in triangular pattern : 1, 3, 6, 10, ....
What is the number at row 1 and coloumn 6.
Find row number and coloumn number for number 99.

Diagram pattern

Study | Number pattern is given here.

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AL 03 06. Number sequences

Numbers in square patterns

Study | How many squares in squares ?.
What are the square number sequences ?
- They are 1, 4, 9, 16, .....
- What is the 10th number ?
Sketch square patterns for 1^2 + 2^2 = 5.
Sketch square patterns for 1^2 + 2^2 + 3^2 = 14.

Numbers in cubic patterns

Study | How many cubes in cubes ?.
What are the cubic number sequences ?
- They are 1, 8, 27, 64, .....
- What is the 10th number ?
Sketch cubic patterns for 1^3 + 2^3 = 9.
Sketch cubic patterns for 1^3 + 2^3 + 3^3 = 36.

Numbers in triangular patterns

Study | What is triangular number sequence ?.
The sequence is 1, 3, 6, 10, 15, .....
Sketch triangular patterns for 1 + 2 + 3 + 4 = 10.
Sketch triangular patterns for 1 + 2 + 3 + 4 + 5 = 15.

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AL 03 07. Perfect square numbers and square free numbers

Definition

Perfect square numbers.
- 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, ...
- Its square roots are integers.
Square free numbers
- 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 13, 15, 17, 18, .....
- Its square roots are not integers.
Simplest form
- For square root - It has only one factor is square free.
  - Example Sqr(180) has factors 4, 5, 9 and 5 is square free.
  - Hence simplest form of Sqr(180) = 36*Sqr(5).
- Fraction - Numerator and denominator has no common factors.
  - 16/20 = 4*4/4*5 = 4/5.

Example : Show that 1.41 < Sqr(2) < 1.42

Method : Every real number other than integers lies between two consecutive integers.
- Sqr(2) between 1 and 2.
- (10*Sqr(2))^2 is between 14^2 and 15^2
  - Hence 10*Sqr(2) is between 14 and 15.
  - Divide by 10, we have Sqr(2) between 1.4 and 1.5.
- (100*Sqr(2))^2 is between 141^2 and 142^2
  - Hence 100*Sqr(2) is between 141 and 141.
  - Divide by 100, we have Sqr(2) between 1.41 and 1.42.

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AL 03 08. Find square root of a real number

Example 1 : Find square root of 2

Method 1 : Use calculator. Sqr(2) = 1.41421356....
Method 2 : Every real number other than integers lies between two consecutive integers.
- Sqr(2) between 1 and 2.
- (10*Sqr(2))^2 is between 14^2 and 15^2
  - Hence 10*Sqr(2) is between 14 and 15.
  - Divide by 10, we have Sqr(2) between 1.4 and 1.5.
- (100*Sqr(2))^2 is between 141^2 and 142^2
  - Hence 100*Sqr(2) is between 141 and 141.
  - Divide by 100, we have Sqr(2) between 1.41 and 1.42.
Method 3 : Approximate method
- 1st approximate : A1 = 2.
- 2nd approximate : A2 = 0.5*(A1 + A1/A1) = 1.5.
- 3nd approximate : A3 = 0.5*(A2 + A1/A2) = 1.41666667
- 4th approximate : A4 = 0.5*(A3 + A1/A3) = 1.41421568
- 5th approximate : A5 = 0.5*(A4 + A1/A4) = 1.41421356
Reference :See chapter 7 in compter methematics by Dr. K. G. Shih

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AL 03 09. Find square root using arithmatic method

Method

Step 1 : Arrange digits of integer number in pairs from right
- Example : 14641 is arranged as 1 46 41
Step 2 : Divide the 1 or 2 digits at left by a and get quotient a.
- Example : Find Sqr(14641)
  - Arranged as 1 46 41.
  - Divide 1 by 1 and 1st digit of quotient is 1 and remainder is 46 41
  - 1 ..... 1st digit of quotient if 1 divide by 1
  - -------
  - 1 46 41
  - 1
  - -------
  - 0 46 41
  - Hence 1st digit of quotient from left is 1.
step 3 : 46 divide by (20*1 + n) where n = 2nd digit of quotient
- Let n = 1 and 46 divide by 21. quotient is 1 and Remainder is 25.
- Let n = 2 and 46 divide by 22. quotient is 2 and Remainder is 2.
- Let n = 3 and 46 divide by 23. quotient is 3 and Remainder is negative.
- The 2nd digit of quotient is 2. Remainder is 241
  - 2 ..... 2nd digit of quotient if 46 divide by 22
  - -------
  - 46 41
  - 44
  - ------- 2 41
  - Hence 1st and 2nd digits of quotient from left is 12
step 4 : 241 divide by (20*12 + n) where n = 3rd digit of quotient
- Let n = 1 and 241 divide by 241. quotient is 1 and Remainder is 0.
- Let n = 2 and 241 divide by 242. quotient is 2 and Remainder is negative.
- The 3rd digit of quotient is 1. and remainder is zero.
  - 1 ..... 3rd digit of quotient if 241 divide by 241
  - -------
  - 241
  - 241
  - ------- 0
  - Hence 1st, 2nd and 3rd digits of quotient from left is 121
Hence Sqr(14641) = 121.
If remainder is not 0, we divide the remainder by (20*121+n).

Example : Find square root of 92.

Divide 92 by 9 and quotient is 9. Remainder is 11.
- 9.591
- ----
- 92
- 81
- ----
- 11 00 Now 11.00 divide by (20*9 + n) where n is the 2nd digit.
- 09 25
- --------
- 01 75 00 divide by (20*95 + m). m = 9
- 01 71 81
- --------
- 00 03 19 00 divide by (20*959 + r). r = 1
- 00 01 91 80
- -----------
- 00 01 27 80 00 divide by (20*9591 + s) s = 6 (5th digit of quotient)

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AL 03 10. Mutiple of an integer whose digits are same

Example

Example : Find Multiple of 49 whose digits are same. 777777 = 49*15873.
Reference : MD2002 program 01 17.

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AL 03 11. Amicable number pairs

Amicable number pairs

Example : Numbers n1 and n2 are amicable pairs. if n1 = 220, find n2.

Study Topic | Amicable number pairs.

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AL 03 12. Perfect numbers

Text and examples

Example : How to find 3rd third perfect number ?.

Study Topic | Perfect numbers.

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AL 03 13. An integer n and n = n1 + n2 + n2, How many arrangement

Note

1 + 2 + 3 = 6 and 2 + 1 + 3 = 6 are 2 arrangement

Projecture a formula

Case 1 : n = 3
- 1 + 1 + 1 = 3
- It has only one arragement. It is the case (n - 2) and has 1 arrangement
Case 2 : n = 4
- 1 + 1 + 2 = 4
  - It has only one arragement if (n - 2) = 2 stands at 3rd place
  - It is the case (n - 2) and has 1 arrangement
- 1 + 2 + 1 = 4
- 2 + 1 + 1 = 4
  - Hence it has 2 arragements
  - This is the case (n - 3) and has 2 arrangements
Case 3 : n = 5
- (n - 2) = 3 : it is (n-2) and only 1 arrangement
  - 1 + 1 + 3 = 5 :
- (n - 2) = 2 : it is (n-3) and only 2 arrangentme
  - 1 + 2 + 2 = 5
  - 2 + 1 + 2 = 5
- (n - 1) = 1 : it is (n-4) and only 3 arrangentme
  - 1 + 3 + 1 = 5
  - 3 + 1 + 1 = 5
  - 2 + 2 + 1 = 5

Conclusion

For (n - 2), it has 1 arrangement
For (n - 3), it has 2 arrangements
For (n - 4), it has 3 arrangements
Hence total arrangements = 1 + 2 + 3 + ..... + (n - 2) = (n-2)*(n-1)/2

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AL 03 14. Prove that 5^(2*n) - 24*n - 1 is divisible by 576

Keyword

Binomial expansion coefficients

Proof

5^(2*n) - 24*n - 1
= (5^2)^n - 24*n - 1
= (1 + 24)^n - 24*n - 1
= C(n,0) + C(n,1)*24 + C(n,2)*(24^2) + ...... + C(n,n)*(24^n) - 24*n - 1
Since c(n,0) = 1 and C(n,1) = n and C(n,r) is integers
Hence 5^(2*n) - 24*n - 1
= C(n,2)*(24^2) + C(n,3)*(24^3) + ...... + C(n,n)*(24^n)
= (24^2)*(C(n,2) + C(n,3)*(24^1) + ..... + C(n,n)*(24^(n-2))
Since (C(n,2) + C(n,3)*(24^1) + ..... + C(n,n)*(24^(n-2)) is integer
Hence 5^(2*n) - 24*n - 1 is divisible by 24^2 or 576

Exercises

1. 3^n - 2*n - 1 is divisible by 4
2. 3^(2*n) - 8*n - 1 is divisible by 64
3. 2^(3*n) - 7*n - 1 is divisible by 49

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AL 03 15. Examples : Perfect numbers

Example 1 : Prove that 496 is a perfect number

Method 1 : Sum of factors
- Factors of 496 are
- 1, 2, 4, 8, 16, 31, 62, 124, 248, 496
- Sum the factors without 496
- Sum = 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 = 496
- Hence 496 is a perfect number
- This is the 3rd perfect number
Method 2 : Use properties
- P1 = 1^3 + 3^3 + 5^3 + 7^3
- = 1 + 27 + 125 + 343
- = 496
- P2 = (2^(n - 1))*(2^n - 1)
- If n = 5, P2 = (2^4)*(31) = 496
- Since P1 = P2, Hence 496 is the 3rd perfect number

Find the 4th perfect number

P1 = 1^3 + 3^3 + 5^3 + 7^3 + .....
- First 4 terms P1 = 1 + 27 + 125 + 343 = 496
- First 5 terms P1 = 496 + 9^3 = 1225
- First 6 terms P1 = 1225 + 11^3 = 2556
- First 7 terms P1 = 2556 + 13^3 = 4753
- First 8 terms P1 = 4753 + 15^3 = 8128
P2 = (2^(n - 1))*(2^n - 1)
- If n = 5, P2 = (2^4)*(31) = 496
- If n = 6, P2 = (2^5)*(61) = 1952
- If n = 7, P2 = (2^6)*(127) = 8128
Since P1 = P2, Hence 8128 is the 4th perfect number

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AL 03 16. Examples : Amicable number pairs

Example 1 : Prove that 220 and 284 are amicable pairs

Factors of 1184 are
1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110, 220
Sum the factors without 220
Sum = 1 + 2 + 4 + 5 + 10 + 11 + 20 + 22 + 44 + 55 + 110 = 284
Factors of 184 are
1, 2, 4, 71, 142, 284
Sum the factors without 284
Sum = 1 + 2 + 4 + 71 + 142 = 220
Hence 220 and 284 are amicable pairs

Example 2 : Prove that 1184 and 1210 are amicable pairs

Factors of 1184 are
1, 2, 4, 8, 16, 32, 37, 74, 148, 296, 592, 1184
Sum factors without 1184
1 + 2 + 4 + 8 + 16 + 32 + 37 + 74 + 148 + 296 + 592 = 1210
Factors of 1210 are
1, 2, 5, 10, 11, 110, 121, 242, 605, 1210
Sum factors without 1210
1 + 2 + 5 + 10 + 11 + 110 + 121 + 242 + 605 = 1184
Hence 1184 and 1210 are amicable pairs

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AL 03 00. Outlines

Defintions

Amicable numbers
- Sum of factors of number 1 = number 2
- Sum of factors of number 2 = number 1
Complex number : It is in the form of z = a + b*i
- Number a and b are real
- The symbol i : i = Sqr(-1), i^2 = -1, i^3 = -i and i^4 = 1
Perfect numbers : Sum of factors = number itself
Real number : It has position on number line
- Rational number : It can be expressed as ratio of two integers
- Rational number : It can be expressed as ratio of two integers
Square free numbers : 1, 2, 3, 5, 7, 10, 11, ....

Properties of numbers

Abundunt numbers : Sum of their factors is greater than number itself.
Deficient numbers : Sum of their factors is less than number itself.
Perfect numbers : Sum of their factors is equal to than number itself.
Prime numbers : Sum of their factors is 1.
Note : Sum of factors is not included number itself.

Number sequences

Triangular pattern : Sum[n*(n+1)/2] = n*(n+1)*(n+2)/6
Square pattern : Sum[n^2] = n*(n+1)*(2*n+1)/6
Cubic pattern : Sum[n^3] = (n*(n+1)/2)^2

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