MAT01A1 Numbers, Inequalities and Absolute Values (Appendix A) Dr Craig 8 February 2017
Leftovers from yesterday: lim n i=1 3 = lim n n 3 = lim n n n 3 i ) 2 ] + 1 n[( n ( n i 2 n n + 2 i=1 i=1 3 = lim n n ) 1 ( 1 n(n + 1)(2n + 1) n2 6 ) = lim n ( 2n 2 + 3n + 2 2n 2 + 3 ( = lim 1 + 3 n 2n + 1 ) n + 3 2 = 4 n i=1 [ i 2 n 2 + 1 ] ) + n
Introduction Who: Dr Craig What: Lecturer & course coordinator for MAT01A1 Where: C-Ring 508 acraig@uj.ac.za Web: http://andrewcraigmaths.wordpress.com
Important information Course code: MAT01A1 NOT: MAT1A1E, MAT1A3E, MATE0A1, MAEB0A1, MAA00A1, MAT00A1, MAFT0A1 Learning Guide: available on Blackboard. Please check Blackboard twice a week. Student email: check this email account twice per week or set up forwarding to an address that you check frequently.
Important information Lecture times: Tuesday 08h50 10h25 Wednesdays 17h10 18h45 Lecture venues: C-LES 102, C-LES 103 Tutorials: Tuesday afternoons (only!) 13h50 15h25: D-LES 104 or D-LES 106 OR 15h30 17h05: C-LES 203 or D1 LAB 408
Other announcements No tuts for MAT01A1 on Wednesdays. If you see this on your timetable, it is an error. (To move your Chem. prac., email Mr Kgatshe ckgatshe@uj.ac.za) CSC02A2 students. Email Dr Craig regarding tutorial clash. Maths Learning Centre in C-Ring 512: 10h30 14h35 Mondays 08h00 15h30 Tuesday to Thursday 08h00 12h55 Fridays
Lecturers Consultation Hours Monday: 10h30 11h30 Ms Richardson (C-503) Wednesday: 14h30 16h00 Ms Richardson (C-503) Thursday: 11h00 13h00 Dr Craig (C-508) 13h30 14h00 Ms Richardson (C-503) Friday: 11h30 13h00 Dr Craig (C-508)
Today s lecture: Number systems Set notation Inequalities Absolute value
Number systems The integers are the set of all positive and negative whole numbers, and zero:..., 3, 2, 1, 0, 1, 2, 3, 4,... The set of integers is denoted by Z. From the integers we can construct the rational numbers. These are all the ratios of integers. That is, any rational number r can be written as r = m n where m, n Z, n 0.
The following are examples of rational numbers (elements of Q): 1 3 22 7 6 7 3 = 24 8 = 3 1 1.72 = 172 10 = 86 5 1 = 1 1
Number systems Some numbers cannot be written as m n for m, n Z. These are the irrational numbers. 2 3 9 π e log 10 2 Combining the rational and irrational numbers gives us the set of real numbers, denoted R. Every x R has a decimal expansion. For rationals, the decimal will start to repeat at some point. For example: 1 3 = 0.33333... = 0.3 1 7 = 0.142857
The real numbers Q: Why the name? A: To distinguish them from imaginary numbers (explained on 15 Feb.) Fun fact: there are as many integers as there are positive whole numbers. In fact, there are as many rational numbers as there are positive whole numbers. However, there are more real numbers than rationals. Read more: The Pea and the Sun, by Leonard Wapner
The real numbers The real numbers are totally ordered. We can compare any two real numbers and say whether the first one is bigger than the second one, whether the second is bigger than the first, or whether they are equal. The following are examples of true inequalities: 7 < 7.4 < 7.5 π < 3 2 < 2 2 2 2 2 10 < 100
Important: there is a big difference between and <, and also between and >. In order to score a distinction for MAT01A1 (or any module at UJ), you must have a final mark 75. You will not get exam entrance if your semester mark is < 40%.
Set notation A set is a collection of objects. If S is a set, we write a S to say that a is an element of S. We can also write a / S to mean that a is not an element of S. Example: 3 Z but π / Z. Example of set-builder notation: A = {1, 2, 3, 4, 5, 6} = {x Z 0 < x < 7 }
Now let us look at Table 1 on page A4 of the textbook. This shows how different intervals can be written using interval notation, set-builder notation, and how they can be drawn on the real number line. Intervals For a, b R, (a, b) = { x R a < x < b } whereas [a, b] = { x R a x b }.
Understanding inequalities graphically Consider the following functions: f(x) = x 2 1 g(x) = (x 1) 2 and the inequality g(x) < f(x). Solving for x (yesterday s tut question): x 2 + 3x 18 = 0 and x 2 + 3x 18 0
Inequalities Let a, b, c R. 1. If a < b, then a + c < b + c. 2. If a < b and c < d, then a + c < b + d. 3. If a < b and c > 0, then ac < bc. 4. If a < b and c < 0, then ac > bc. 5. If 0 < a < b, then 1 a > 1 b. Very important: note that for (3) and (4) we must know the sign of c. We cannot multiply or divide by a term if we do not know its sign!
Solving inequalities We will often make use of a number line or a table to determine the sign of the function on particular intervals. We use critical values (where a function is 0 or undefined) to determine the intervals. Examples: 1. Solve for x: 1 + x < 7x + 5 2. Solve for x: x 2 < 2x
Solving inequalities continued Find solutions to the following inequalities and write the solutions in interval notation. 1. 4 3x 2 < 13 2. x 2 5x + 6 0 3. x 3 + 3x 2 > 4x (Don t divide by x!) 4. x2 x 6 x 2 + x 6 0 5. x2 6 x 1 (Don t multiply by x!)
Absolute value The absolute value of a number a, denoted by a is the distance from a to 0 along the real line. A distance is always positive or equal to 0 so we have Examples a 0 for all a R. 3 = 3 3 = 3 0 = 0 2 3 = 2 3 3 π = π 3
In general we have a = a if a 0 a = a if a < 0 We can write the absolute value function as a piecewise defined function. { x if x 0 x = x if x < 0 We can also replace the x above with something more complicated.
Sketching y = x :
If f(x) = x, calculate: f( 5) = ( 5) = 5 f(4) = 4 f(0) = 0 Note: for any a R, a = a 2
Example Write 3x 2 without using the absolute value symbol. 3x 2 = Hence 3x 2 = { 3x 2 if 3x 2 0 (3x 2) if 3x 2 < 0 { 3x 2 if x 2 3 2 3x if x < 2 3
Below is a sketch of y = 3x 2. Note how the function changes at x = 2 3.
Properties of Absolute Values Suppose a, b R and n Z. Then 1. ab = a b 2. a b = a b (b 0) 3. a n = a n Let a > 0. Then 4. x = a if and only if x = a or x = a. 5. x < a if and only if a < x < a. 6. x > a if and only if x > a or x < a. Example: Solve 2x 5 = 3.
Solving absolute value inequalities: Solve: x 5 < 2. Solve: 3x + 2 4. To solve the first inequality above we must solve 2 < x 5 < 2 For the second inequality we have 3x + 2 4 OR 3x + 2 4
Solving x 5 < 2 gives x (3, 7).
Solving 3x + 2 4 gives x (, 2] [ 2 3, )
Next time
The triangle inequality If a, b R, then a + b a + b. Applications: Example: If x 4 < 0.1 and y 7 < 0.2, use the Triangle Inequality to estimate (x + y) 11. Exercise: show that if x + 3 < 1 2, then 4x + 13 < 3.