# Mathematics

### Saturday, July 14, 2012

#### Fibonacci Numbers

Fibonacci Numbers Fibonacci numbers are numbers in the Fibonacci sequence. The Fibonacci starts out as follows: 1,1,2,3,5,8,13,21,34,... After the second term, each term is the sum of the two preceeding it. Use the following convention to name the Fibonacci numbers. The Fibonacci sequence is... Sign in to see full entry.

### Monday, July 2, 2012

#### Square root of 2 is not the ratio of integers. It is irrational.

Rational numbers are the RATIO of integers. That is why they are called rational numbers. Irrational numbers are not the ratio of integers. That is why they are called irrational numbers. Since sqrt(2) was proven to be not rational, it was proven to not be the ratio of integers. (a+b)^2 + (a-b)^2 =... Sign in to see full entry.

### Sunday, July 1, 2012

#### 0! and 0^0

1! = 1 2! = 1 * 2 3! = 1 * 2 * 3 etc So how would we decide what 0! should be? Reverse the process. 3! = 6 2! = 3!/3 = 6/3 = 2 1! = 2!/2 = 2/1 = 1 0! = 1!/1 = 1/1 = 1 (-1)! = 0!/0 = 1/0. oops. (-1)! does not exist. thus (-2)! does not exist. etc We can find a factorial for all numbers except for the... Sign in to see full entry.

### Monday, May 28, 2012

#### Arithmetic Pattern

12*12 = 144 144 - 1 = 143 = 11 * 13 143 - 3 = 140 = 10 * 14 140 - 5 = 135 = 9 * 15 135 - 7 = 128 = 8 * 16 128 - 9 = 119 = 7 * 17 119 - 11 = 108 = 6 * 18 108 - 13 = 95 = 5 * 19 95 - 15 = 80 = 4 * 20 80 - 17 = 63 = 3 * 21 63 - 19 = 44 = 2 * 22 44 - 21 = 23 = 1 * 23 23 - 23 = 0 = 0 * 24 Sign in to see full entry.

### Friday, May 18, 2012

#### The Golden Rectangle

The ancient Greeks expressed the opinion that the rectangle with adjacent sides in a particular ratio was the most beautiful of the possible rectangles. To construct this rectangle, they started with a rectangle whose length was twice its width. They drew in the diagonal of this 1 by 2 rectangle.... Sign in to see full entry.

### Tuesday, May 8, 2012

#### Arithmetic Exercise

Ask a friend to: Think of a positive number. Example: 2 Add 3. 2+3 = 5 Multiply the result by itself. 5 * 5 = 25 subtract 9. 25 - 9 = 16. divide by original number. 16 / 2 = 8 subtract 6. 8 - 6 = 2. Then you can tell your friend, "If you did all the arithmetic correctly, the result is your original... Sign in to see full entry.

### Sunday, April 15, 2012

#### puzzle of sum to composite

Prove that every positive composite integer is the sum of 4 positive integers such that the product of the four integers is a square integer. Sign in to see full entry.

### Sunday, April 8, 2012

#### Every product of two numbers is the difference of two squares.

Every product is the difference of two squares. 3 * 5 = 15 (5-3)/2 = 2/2 = 1 (5+3)/2 = 8/2 = 4 4^2 - 1^2 = 4*4 - 1*1 = 16 - 1 = 15 3 * 7 = 21 (7-3)/2 = 4/2 = 2 (7+3)/2 = 10/2 = 5 5^2 - 2^2 = 25 - 4 = 21 5 * 7 = 35 (7-5)/2 = 1 (7+5)/2 = 6 6^2 - 1^2 = 35 3*11 = 33 (11-3)/2 = 4 (11+3)/2 = 7 7^2 - 4^2 =... Sign in to see full entry.

### Saturday, March 24, 2012

#### Factoring numbers of the form (h^n + a1 h^m + a0)(h^n + b1 h^m + b0)

How would you determine if a given integer were of the form (h^n + a1 h^m + a0)(h^n + b1 h^m + b0), where 2 * m > n? (h^n + a1 h^m + a0)(h^n + b1 h^m + b0) = h^(2n) + (a1+b1) h^(n+m) + a1 b1 h^(2m) + (a0+b0) h^n + (a1 b0 + b1 a0) h^m + a0 b0 = z Choose values for m and n. you can try n =... Sign in to see full entry.

#### Factoring numbers of the form (h^n + a)*(h^m + b)

Supposse that you wished to factor some integer. A very very small percentag of integers are of the form (h^n + a)*(h^m + b) = h^(n+m) + b * h^n + a * h^m + a0*b0. (assume that m How might we determine whether or not a particlar integer is of that form. For example, could we solve h^(n+m) + b * h^n... Sign in to see full entry.