Transcript Slide 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
Real number = integer-part + fractional-part
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”).
 4   1 2 2  2  1  2i
2  2  22  4
 4  not realsince - 4  - 2  -2 or  2  2
4 + 2i
real part
 
4 2  2
all have to be
either -2 or +2
imaginary part
2
2
1
Surds: an irrational root such as √3
2
2
fractional index
lacking sense : IRRATIONAL; absurd
Radicals become easier if you think of them in
1
terms of indices. Think 4 2 instead of 4
http://www.itc.csmd.edu/tec/GGobi/index.htm
Rational Numbers VS. Irrational Number
100.3; 1/6 = .16666; 2/7 = .285714285714
Approximating irrational number by
rational numbers: number theory
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
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).
integer
any of the natural numbers, the negatives of these numbers, or zero
natural
numbers
the number 1 or any number (as 3, 12, 432) obtained by adding 1 to
it one or more times : a positive integer
real number
one of the numbers that have no imaginary parts and comprise the
rationals and the irrationals
imaginary
number
a complex number (as 2 + 3i) in which the coefficient of the
imaginary unit is not zero
irrational
number
a number that can be expressed as an infinite decimal with no set of
consecutive digits repeating itself indefinitely and that cannot be
expressed as the quotient of two integers
Rational
Numbers
an integer or the quotient of an integer divided by a nonzero integer
Radical
Root; Foundation
Page 18 – Example 10b
3
16x  3 54x4
x  x3
16  2  8  2  2  2  2  2  2
3
54  2  27  2  33
2  2 x 
3
1
3

 2  33  x  x 3
22 x  3  3x2 x 
1
2 x  3 2  3x 
3
2 x 2  3x 
1
1
3

1
3
8 x 9 = 4 x 2 x 9 = 4 x 18  18 – 4 = 14
8 x 2  14x  9  8 x 2  (18  4) x  9
8 x 2  18x  4 x  9  8 x 2  4 x  18x  9
4 x(2 x  1)  9(2 x  1)  (2 x  1)(4 x  9)
Page 29 example 9
1 x 25 = 1 x 5 x 5  5 + 5 = 10
16 x 1 = 4 x 4  4 + 4 = 8
x 2  10x  25
x  (5  5) x  25
2
x 2  5 x  5 x  25
xx  5  5x  5
x  5x  5
x  52
16x 2  8 x  1
16x 2  (4  4) x  1
16x 2  4 x  4 x  1
4 x4 x  1  1(4 x  1)
4 x  14 x  1
4 x  12
Page 30 example 12, 13
1 x 12 = 3 x 4  3 + 4 = 7
x 2  7 x  12
x 2  3  4 x  12
x 2  3 x  4 x  12
x x  3  4 x  3
x  3x  4
2 x 15 = 2 x 3 x 5 = 6 x 5  6 - 5 = 1
2 x 2  x  15
2 x 2  (6  5) x  15
2 x 2  6 x  5 x  15
2 xx  3  5x  3
x  32 x  5
Page 40 example 7
Page 39 example 6
x
2

x  3 3x  4
x3x  4   2 x  3
x  33x  4
3x 2  4 x  2 x  6
x  33x  4
3x 2  2 x  6
x  33x  4
3
2 x3
x 1x  1
  2
x 1 x x 1
3 x  x  1  2x  1 x  1  x  3x
x  1x x  1
3x 2  3x  2 x 2  2  x 2  3x
x  1x x  1
2x2  6x  2
x  1x x  1
Page 42 – Example 9
x1  2 x 
3
2
 1  2 x 
1
x
2

1
4  x 
2
1  2 x 1 1  2 x  2 1  2 x  2
1
1 3
1 
2 2
1
1  2 x  2
1
1
1  2 x  2
1
2
 x


1
 1  2 x  


2
2
1
2
2
1
2
2
 x  11  2 x  
 1  2 x  


Page 42 –
Example 10
x2

4  x 
4  x   4  x 
4  x 
4  x   x
4  x 
1
1
1
2
2
 x2
Combine the
numerator
terms
2
2
2
1
2
4
4  x 
2
1 x
1  2 x  2
1
1
x2
4  x 
1
2
2
4  x 
3
2
a
b  a
c
bc
1
2

 x2 4  x2
4  x2
4
4  x  4  x 
2
1
2
2
4
4  x 
2
3
2
1
1 3

2 2

2
1
2
Climbing the mountain on a straight slope
8
10, 7
7
6
On the 2nd day
You climbed 4
miles vertically
5
4
4, 3
3
2
1
0
0
2nd
2
day
start point
4
6
8
10
On the 2nd day,
You covered 3
miles horizontally
12
3rd day end
point
y2  y1
verticalheight
7 3 4 2
gradient( slope) 

 
horizonalheight 10  4 6 3
x2  x1
How far did we walk ?
Slope of a Straight Line
tangent
y2  y1
m
x2  x1
 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 ?
Given two point on a line, what is the distance
between the two points?
8
B
7
Vertical distance
10, 7
6
5
4
A
4
3
C
(4, 3)
2
(10,3)
6
1
0
0
2
4
6
8
10
12
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
Horizontal distance
We can find AB using the Pythagoras’ theorem
AB  (10  4) 2  (7  3) 2  6 2  4 2
AB  ( x2  x1 ) 2  ( y2  y1 ) 2
1
1





x

x
,
y

y
2
1
2 
Mid-point of line segment AB   2 1
2

Equation of a line
y2  y1
m
 y2  y1  mx2  x1   y2  mx2  x1   y1
x2  x1
y  mx  c
slope
x and y are variables -- various
points along the line
Slope of a line
joining points (0,c)
and (x,y)
y-intercept
yc
m
x0
x can’t be 0
Point (0,c) lies on y axis
Properties of
y  mx  c
Set y = 0 to find
the x-intercept
Set x = 0 to find
the y-intercept
When m = 0
x
c
m
no incline, line is
parallel to x-axis.
No x intercept
M can’t be 0
No gradient (undefined),
straight up, perpendicular to
the x axis. No y intercept.
Parallel to y axis. x = k.
y = c
y=c
c is y-intercept
Ex: (1,2), (-1,2), (5,2)…
2.5
5
4
2
(-1, 2)
1.5
(3, 4)
3
2
(5, 2)
(2, 2)
1
(3, 2)
0.5
1
(3, 0)
0
0
1
2
3
4
0
-2
0
2
4
6
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
4
3
B(5, 3)
2
A(1, 1)
1
C(3, 0)
0
-2
0
2
4
-1
D(-1, -2)
-1,-2-2
-3
Matlab: plot([-1,1,5,3,-1],[-2,1,3,0,-2])
6
Example: given gradient, and a point on the line, find the line’s equation
Slope of the line is given
P(x,y)
m
y2  y1
x2
2
x2  x1
y 1
The lines passes through (2,1)
 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
Step1: given two points,
it is easy to find m
m
y2  y1
x2  x1
Step2: once m is known, use the
same equation and one of the
points to find the equation
m
y2  y1
x2  x1
Try this: (3,4), (-1,2)  2y = x + 5