Integer (positive or negative whole numbers or zero) arithmetic

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Lionel Lindsey
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1 Integer (positive or negative whole numbers or zero) arithmetic The number line helps to visualize the process. The exercises below include the answers but see if you agree with them and if not try to work out why! 15 8 = = = = = (-) = +1 Rules - in subtraction two minuses (next to each other) make a plus (-5) = -9 In multiplication and division two minuses make a plus. -7 x 2 = x (-) = -6-4 x (-4) = = -1-2 (-2) = = -4 If brackets appear around more than one number it means that the quantity contained within the brackets must be worked out first (BODMAS brackets, other, division, multiplication, addition, subtraction). 2 x ( + 4) = (7-4) = (5-8) = +7 (7 + 9) - (8-6) = +14 (9-15) - (10-15) = -1 1

2 2 x ( + 4) = 14 4 x (5-7) = -8 8 (7-9) = -4 - (10-5) = -/5-9 (10-19) = +1 (4 + 5) x (9 + ) = -108 (7 + 8) (6 + 9) = +1 (28 + ) (0-61) = -1 Arithmetic using fractions (rational numbers) Simplification - e.g. 2/4 = 1/2, /9 = 1/ etc Simplify: 12/24 = 1/2 4/48 = 1/12 4/6 = 2/ Adding and subtracting fractions: /8 + 2/8 = 5/8 ½ +¼ = /4 Rule: denominators (bottoms) must be made equal 2/ + /4 = 17/12 5/9 - /9 = 2/9 2/5 - /10 = 1/10 1/2 - /8 = 1/8 Multiplying Fractions easy! Simply multiply numerators (top) and denominators (bottom) 1/ x 2/5 = 2/15 1/4 x /9 = 1/12 2

3 1/15 x 4/18 = 2/15 Dividing fractions (Rule: turn the second fraction upside down and then multiply) 1/2 1/4 = 2 /9 5/18 = 6/5 15/5 /7 = 1 Decimal (base 10) arithmetic Decimal points divide the integer part from the fractional part of a decimal number. The fractional part uses the following convention: first number = number of tenths second number = number of hundredths third number = number of thousandths etc. Any number can be written this way although some become unfeasibly long! e.g. 1/9 = forever! If the decimal expansion repeats the convention is to put a dot over the repeating number(s). Convert the following fractions into decimals, e.g. ¼ = 0.25 Rule: just do the division. ½ = 0.5 2/8 = 0.25 /9 = /10 = 0.6 If a decimal number stops or repeats it can always be written as a fraction or rational number. If a decimal number never stops or repeats it is not rational an irrational number. The most famous of which is π however there are an infinite number of them! All square roots of numbers are irrational e.g. 2, etc. excepting for (fairly obviously) the square numbers. Convert the following decimals into fractions, e.g = 24/100 = 6/ = 1/ = 1/ = 1/ = 7/8

4 Indices (powers or exponents) A short hand notation consisting of a small number or other quantity as a superscript. Positive indices e.g. 2 x 2 = 4 = squared = 4 the area of a 2 x 2 square numbers of this type are referred to as square numbers e.g. 2 x 2 x 2 = 8 = 2 2 cubed = 8 the volume of a 2 x 2 x 2 cube Generally, 2 x 2 x 2 x...x 2, n times = 2 n e.g. 2 = 9, = 27, 4 = 81. etc. e.g = = 1000 for powers of ten, the index = number of zeros. Multiplication In multiplication we add indices. e.g. 2 2 x 2 = 2 5 (base number has to be the same) Generally a p x a q = a p+q the first law of indices Division e.g. 2 5 /2 = /2.2.2 = 2.2 = 2 2 using indices: 2 5 / 2 = 2 2 = 2 (5-) Generally a p a q = a p-q the second law of indices Power of a power In the first law if p=q then a p x a p = a p+p = a 2p or (a p ) 2 a p times itself q times = a p xa p xa p..q times = (a p ) q = a pq the third law of indices e.g. (2 2 ) = 2 2 x2 2 x2 2 = = 2 (2+2+2) = 2 (2x) = 2 6 4

5 Negative indices In the second law of indices concerning division what if p is less than q? e.g = 1/2 2 by the second law this would be = 2 (-5) = 2 2 hence we can write 1/2 2 = 1/4 = 2-2 Generally: 1/a p = a -p e.g. 1/ 4 = 1/81 = -4 and 1/10 2 = 1/100 = 10-2 = 0.01 For powers of ten the negative index = number of places point moves to the left from that in 1.0 Since we know that = 1 (since 4 4 = 1) then using the second law of indices we see that = 2 (2-2) = 2 0 = 1 We could have chosen any number to begin with so any number to the power zero = 1 Dividing is just the same as multiplying by the inverse so we see that this procedure works for multiplying with a negative index. e.g. 2 x -2 = 0 = 1 etc. To summarise: a 0 = 1 Fractional indices Square root - the number which when multiplied by itself gives the original number. 2 = can be written using index notation as 2 1/2 Generally, n = n 1/2 since 2 1/2 x 2 1/2 = 2 (1/2 + 1/2) = 2 1 = 2 Similarly, cube roots 2-1/2 = 1/ 2 etc. 2 = 2 1/, 8 = 8 1/ = 2 5

6 Generally: a 1/p = p a the pth root of a. Finally, by the third law (a 1/q ) p = a p/q = (a p ) 1/q since 1/q x p = p x 1/q Exercise: Simplify the following expressions using the index notation answers in brackets. a. 2 x 2 (2 2 ) b. 2 x 2 x 2 x x (2 2 ) c. (10 x 10) (10 x 10 x 10) (10-1 ) Simplify the following leaving answers in the form of one number with one index. a. 2 2 x 2 x (2 9 ) b. 2 x 8 ( - ) c. -2 x -1 ( 2 ) d x (10-1 ) e. 8 x (8 0 = 1) f. 8 1/2 8 /2 (8-1 ) g. 10 1/2 x 10 1/4 (10 /4 ) h. (2 x2 7 )/4 (2 4 ) i. 16 5/4 8 4/ (=2, hint, write 16 and 8 as powers of two) Evaluate (work out) the following. a. (1/4) 5/2 (1/2) b. 1/4-2 (16) c. (-1/5) -1 (-5) 6

7 Algebra - the part of mathematics which studies sets of symbols and the rules of operations on them. Used in the solution of equations where symbols are used for the unknown quantities = algebraic representation e.g. If x = and y = 4 then x + y = 7, 2x = 6, x 2 = 9, x 2 + y 2 = 25 etc. Note that the multiplication symbol never appears in algebra. If two quantities are next to each other it is assumed that they are multiplied e.g. 4y = 4 times y. An equation contains two expressions which are equal and are joined by the = symbol. If the whole expression is true for all values of the symbols then it is an identity. If not then the rules of algebra can be used to find the value of the unknown (if only 1 unknown). Essentially the equation is manipulated until the unknown quantity is left on its own on one side of the = sign. Depending on the type of equation there may be more than one solution (or root) of an equation. General rule: whatever you do to one side of an equation you must do to the other in order to preserve the equality. Coefficients of like terms (having the same power of the variable) can be added. e.g. 5x + x - x = 7x, xy + xy - 8xy = -4xy Quantities outside brackets multiply all terms inside. e.g. solve for x the equation, (x + 4) - 8x = 10 multiplying out the bracket gives x x = 10 gathering like terms in x gives -5x +12 = 10 subtracting 12 from both sides gives, -5x = -2 dividing both sides by -5 gives, x = 2/5 Always check the answer by substituting back into the original equation and seeing if it works! e.g. ( x + ) (4x + 7) 9 = 1 multiply both sides by 9, (x+) - (4x + 7) = 9 multiply out brackets, x x - 21 = 9-11x - 18 = 9 The three dots is the therefore symbol. 7

8 -11x = 27 x = -27/11 e.g. (4 + x) 9 = x Multiply both sides by (4 + x) gives 9(4 + x) = x Multiply both sides by x gives x = 9(4 + x) The previous two stages are usually carried out together - an operation called cross multiplication Finally, x = 6 + 9x -6x = 6 x = -6 Exercise in substitution If x =, y = 2 and z = 1 find the values of the following expressions answers given x + y + z = 6 2x + y + 2z = 14 x - y - z = -2 x 2 + y 2 + z 2 = 14 Solving elementary equations Find the value of x in the following expressions x + = 5 (x=2) 7 - x = 4 (x=) 2x = 6 (x=18) x/2 = 6 (x=12) 8

9 x/ + 4 = 9 (x=15) 2x - 6 = x 2 (x=4) Find the unknown quantity in the following expressions (a + 2) = 9 (a=1) 4(b - 5) = 40 (b=15) 2(c + 6) = 50 (c=19/) 2(d - 7) + 4d - 5 = 19 (d=19/) Simplify the following expressions. a. x + 5x -7x (x) b. 4xy -10xy - 18xy (-24xy) c. 2x 2 + 4y 2 - x + 4x - 10y 2-5y (2x 2-6y 2 +x-5y) d. 4p + 5p - 15p + 28p (22p) 9

10 Geometry Triangles A triangle is a polygon having three sides. Equilateral - all sides equal Isosceles - two sides equal Scalene - no sides equal. Angles - in a whole circle - 60 degrees - 60 o in a half circle or on one side of a line = 180 o quarter circle = right angle = 90 o Obtuse angles are between 90 and 180 degrees Acute angles are between 0 and 90 degrees Reflex angles are between 180 and 60 degrees The three angles in a triangle always add up to 180 o Circles segment chord arc - this is the minor arc - the other part is the major arc radius twice radius = diameter sector Perimeter - The total length of the edges of a polygon The perimeter of a circle = circumference, C By definition π = C/D where, D = diameter of circle So C = π D = 2 π r where, r = radius of circle 10

11 Quadrilaterals Polygons with 4 sides Special cases i Trapezium - one pair of opposite sides is parallel. ii Parallelogram - both pairs of opposite sides are parallel and hence of equal length. Opposite angles are equal. iii Rhombus - parallelogram with all four sides of equal length. iv Rectangle - parallelogram with all four angles right angles. v Square - rectangle with all four sides equal. Other Polygons 5 sides - Pentagon 6 sides - Hexagon 7 sides - Heptagon 8 sides - Octagon 9 sides - Nonagon (Enneagon more correct all Greek) 10 sides - Decagon Area The measure of the amount of space inside a finite region obtained by approximating to a sum of areas of small rectangles whose areas are defined by length x breadth. Area of a triangle = 1/2 base x height (perpendicular to the side defining the base) = ½ bh a Area of a square = side x side = a x a = a 2 Area of a rectangle = length x breadth = a x b a a b h b 11

12 Area of a parallelogram = base x height (perpendicular to side defining the base) = b x h h r Area of a circle = π r 2 b Volume The measure of the amount of space inside a solid obtained by approximating to the sum of volumes of cuboids which are defined to have volume = length x breadth x height. Volume of a cube, all sides length a = a x a x a = a a Volume of sphere radius r = 4 πr l Volume of a prism = area of end section x length = Al A r Volume of a cylinder = πr 2 l l 12

13 Write down the value of angle x in the following figures N.B. tick marks on edges indicate equal length. x x 150 x x x 40 Answers to above, clockwise from top left, 60 degrees, 0, 40, 100, 50. Work out the perimeters (distance around whole edge) in cm and areas in cm 2 of the following figures. Assume π = all angles right angles Answers: from top left clockwise: 20cm, 21cm 2 ; 16cm, 16cm 2 ; circle - 24cm, 48cm 2 ; triangle -12cm, 6cm 2 ; 16cm, 12cm

Preliminary chapter: Review of previous coursework. Objectives

Preliminary chapter: Review of previous coursework Objectives By the end of this chapter the student should be able to recall, from Books 1 and 2 of New General Mathematics, the facts and methods that