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natural numbers (a positive integer)

In the early days, one orange, 3 oranges, ..etc. any of the natural numbers, the negatives of these numbers, or zero. natural numbers (a positive integer). Root, Radix, Radicals. Real Number VS. Imaginary (complex) Number.

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natural numbers (a positive integer)

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  1. In the early days, one orange, 3 oranges, ..etc. any of the natural numbers, the negatives of these numbers, or zero natural numbers (a positive integer) Root, Radix, Radicals Real Number VS. Imaginary (complex) Number Ninth-century Arab writers called one of the equal factors of a number a root, and their medieval translators used the Latin word radix (“root,” adjective “radical”). Real number = integer-part + fractional-part Surds: an irrational root such as √3 lacking sense : IRRATIONAL; absurd fractional index 4 + 2i all have to be either -2 or +2 Radicals become easier if you think of them in terms of indices. Think instead of real part imaginary part http://www.itc.csmd.edu/tec/GGobi/index.htm Rational Numbers VS. Irrational Number 100.3; 1/6 = .16666; 2/7 = .285714285714 Number that can’t be expressed as p/q. Not a quotient of two integers 2½ = 1.4142135623730950488016887242097….  (3.1416…) http://mathworld.wolfram.com/Pi.html Approximating irrational number by rational numbers: number theory

  2. How do you represent large multiples such as 2x2x2x2  takes too much space to print 2x2x2x2 = 24 the birth of exponential notation (base, exponent or index (indices)) Now we need a set of rules to figure out what things such as is 22 x 23 Or 23 x 32 Properties of exponents Logarithms: Math based on the exponents themselves, invented in the early 17th century to speed up calculations. Also from the result of the study of arithmetic and geometric series. (study tip: the exponent is the logarithm).

  3. Page 18 – Example 10b

  4. 8 x 9 = 4 x 2 x 9 = 4 x 18  18 – 4 = 14

  5. Page 29 example 9 1 x 25 = 1 x 5 x 5  5 + 5 = 10 16 x 1 = 4 x 4  4 + 4 = 8 Page 30 example 12, 13 1 x 12 = 3 x 4  3 + 4 = 7 2 x 15 = 2 x 3 x 5 = 6 x 5  6 - 5 = 1

  6. Page 40 example 7 Page 39 example 6

  7. Page 42 – Example 9 Page 42 – Example 10 Combine the numerator terms

  8. Climbing the mountain on a straight slope On the 2nd day You climbed 4 miles vertically On the 2nd day, You covered 3 miles horizontally 3rd day end point 2nd day start point How far did we walk ?

  9. Slope of a Straight Line tangent  Equation of a straight line • We need to know two points (locations) • The second day’s starting location and ending location • Can you identify the right-triangle in the previous slide? • Can you identify the right-angle? • It is customary to denote the slope of a straight line by “m” Now that we know there is a right-triangle, how far did we walk ?

  10. Given two point on a line, what is the distance between the two points? Vertical distance B 4 A We were only given points A and B. Using A and B we could simply figure out point C. Point C is same height as point A but it is (10 – 4) or 6 units away from A C (10,3) 6 Horizontal distance We can find AB using the Pythagoras’ theorem Mid-point of line segment AB 

  11. Equation of a line slope y-intercept x and y are variables -- various points along the line Slope of a line joining points (0,c) and (x,y) x can’t be 0 Point (0,c) lies on y axis

  12. Properties of Set y = 0 to find the x-intercept Set x = 0 to find the y-intercept When m = 0 no incline, line is parallel to x-axis. No x intercept y = c M can’t be 0 y = c No gradient (undefined), straight up, perpendicular to the x axis. No y intercept. Parallel to y axis. x = k. c is y-intercept Ex: (1,2), (-1,2), (5,2)…

  13. Typical problems involving straight lines? • Find whether 4 points form a parallelogram. • Method1: Calculate the distances between them to see if AB = DC and CB = DA • Method2: Using mid-points • If the mid-points of the diagonals AC and BD bisect each other then ABCD is a parallelogram • Mehtod3: Using gradients • If the gradients of AB and DC are same Matlab: plot([-1,1,5,3,-1],[-2,1,3,0,-2])

  14. Example: given gradient, and a point on the line, find the line’s equation Slope of the line is given The lines passes through (2,1) P(x,y)  y=2x-3 A(2,1) Try this: (-2,3); m = -1 y = -x + 1 Example: given two points on a line, find the line’s equation Step2: once m is known, use the same equation and one of the points to find the equation Step1: given two points, it is easy to find m Try this: (3,4), (-1,2)  2y = x + 5

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