College Algebra & Trig I: Real Numbers

thanks for downloading this lecture today we’re going to talk about the topic of real numbers to begin with we’re gonna start by talking about working with sets a set is a collection of elements which satisfy some given rule if a set has no elements is called the empty set or the null set so if you know what rule defines your set you can generate the set of numbers that belong to it there are two general methods for describing a set the first method is called the roster method in the roster method you simply list every element of the set so for example we have d equals the set of numbers 0 1 2 3 4 5 6 7 8 & 9 so we’ve listed each number that’s a part of the set the other method for listing a set is called set builder notation or the set builder method so for this method what you’re going to do is you’re gonna write the rule for your set so we have in in brackets the brackets denote this is a set the set of numbers X the vertical line means such that so X such that X is a digit set builder notation is really useful if you have a set of very large numbers where it would take you a long time to list every single element so let’s do an example where we’re gonna write out some set the phone number for the math department at NC A&T is two eight five two zero three three given that information we want to write the set of numbers that are represented in the math departments phone number but we’re gonna do this both using set-builder notation and using roster notation so first we’re going to write the set and set builder notation remember a set builder notation just basically lists the rule that you’re using to generate your set so for this example we could write P is equal to the set of X such that X is a digit in a math departments phone number and the second method is the roster method so remember the roster method we just list out every element of the set so if we look at the phone number we see that the numbers that are there are two eight five two zero three and three but we only need to write each number once so if I write them in order my set would be P equals the set zero two three five and eight those are the numbers that are represented in the phone number next we’re gonna talk about subsets the definition of a subset is as follows if every element of a is also present in B then a is a subset of B denoted as follows a or the sideways you want a line under it B now if two sets have the exact same elements then they’re called they’re said to be equal and so then we would say set a equals set B now we’ll talk about intersections and unions the definition of the intersection of two sets a and B written a with an upside-down U and a B is the set of elements that appear in both a and B so to find the intersection you’ll look at your two sets a and B and you’ll pick out all of the numbers or elements that are common between the two and then the union of two sets a and B written a with a cup or u B is the set of elements that appear in either A or B so to find the union of two sets you’re gonna look at set a and you’re gonna look at set B and you’re gonna write down every single element that appears in either one so let’s do an example we’re going to work with intersections and unions given the sets a equals 1 3 4 5 and 9 b equals 2 4 6 7 & 8 & c equals 1 3 4 & 6 find the following a union B a intersection C and a union B intersected with C we’ll start by finding the union of a and B so remember Union means all of the elements that are represented in either of the sets so let’s look at set a and set B and we want to look at the numbers that are in either set so if I look is 1 in a or b is 2 and a or b write down all of the numbers that appear in either a or b and so if we write down all of the numbers that appear in either a or b will see that the union of a and B is the set of numbers 1 2 3 4 5 6 7 8 and 9 so all of those numbers are in both or are in either A or B next we’ll look at

the intersection of a and C so remember intersection means the numbers that appear in both a and C so look at set a and look at set C and figure out which numbers appear in both of them you should have seen that this set that’s the set of numbers that appear in both a and C are the numbers 1 3 & 4 so since 1 3 & 4 appear in both set a and set C those are the numbers that make up the intersection of a and C the last part of this example is probably the trickiest one part because it’s a compound statement we want to find out the intersection of C with the union of a and B so in order to do this we want to take it one step at a time first we want to find out what the union of a and B are is since that’s in parentheses we want to find that first now fortunately that was the first part of the example so we know that the union of a and B are the set of numbers 1 2 3 4 5 6 7 8 and 9 so we can look at that along with C to find the intersection remember intersection is going to be the set of numbers that appear in both of them so which numbers appear in both a union B and the set C and you should have come up with a set of numbers 1 3 4 and 6 these are the numbers that appear in both a union B and in C the last thing that we’re going to talk about when dealing with sets is the universal set and complements the universal set denoted capital u is the set of all possible elements and the complement of a set a denoted as a with a bar on top of it is the set of all elements that are in the universal set but are not in the set a so for example if we’re given that the universal set u is equal to 1 2 3 4 5 6 7 8 9 and the set a is equal to 1 2 3 5 7 9 we should be able to find the complement of a or a bar so remember the complement of a where a bar is going to be the set of numbers that are possible so they’re part of you but they’re not actually in a so look at the set a and figure out which numbers are not there and if we look for the numbers that are not in a we find that those are 4 6 & 8 those are the only numbers that are possible the part of U but are not an A so that would be the complement of a or a bar now we’re going to move on and talk about classifying numbers the first set of numbers that we’re going to talk about is the largest set and those are the real numbers those are all of the numbers that you can plot on the real number line note that the set of real numbers is denoted by an R with a double-leg the real numbers can be split into two subsets the first subset is the set of irrational numbers irrational numbers are numbers that cannot be expressed expressed as a ratio or a fraction examples of irrational numbers include the number pi the square root of 2 the square root of 3 or any decimal that doesn’t terminate and doesn’t repeat if a number isn’t irrational then it’s part of the other subset of real numbers and those are the rational numbers rational numbers are numbers that can be expressed as a rate ratio or fraction examples of rational numbers are to one-third 0.25 an in decimal that terminates or any decimal that repeats the set of rational numbers can be split up into further subsets one subset of the rational numbers which we’ll talk about a lot in this course are the integers integers are numbers that do not have a decimal or a fraction so for example negative 3 negative 2 negative 1 0 1 2 3 these numbers don’t have fractions or decimals and so these are the integers a subset of the integers are the whole numbers the whole numbers are basically the non-negative integers so 0 1 2 3 4 5 so on and so forth and finally a subset of the whole numbers which is also a subset of the integers and a subset of the rational numbers are the natural or counting numbers the natural or counting numbers are just the positive integers the ones that you would think about counting with 1 2 3 4 5 so on and so forth so let’s do an example where we’re going

to classify numbers within a set if we’re given the set a equals negative 6 1/2 negative 1 point 3 3 3 repeating 0 PI 2 & 5 we want to identify which of those numbers fit into the following categories the natural numbers the integers the rational numbers the irrational numbers were the real numbers for the natural numbers remember the natural numbers are the numbers that we count with those are the positive integers so the only numbers in our set that are positive integers are 2 & 5 the integers are all of the numbers that don’t have decimals or fractions they can be positive or negative so the numbers within our set which are integers are the numbers negative 6 0 2 & 5 the rational numbers are numbers that can be written as a fraction so remember integers are rational numbers because you can always write them as the number divided by one so for example negative six we could write as negative six over one so it can be written as a fraction that makes it a rational number also any decimal that is terminating or repeating is a rational number and anything that can be written as a ratio or fraction is a rational number so from our set a the rational numbers are negative 6 1/2 negative 1 point 3 3 3 3 repeating 0 2 & 5 the irrational numbers are those that cannot be written as a fraction or ratio so the only one that we have in this set is the number pi remember pi represents a decimal that goes on forever and ever for lots and lots of digits so it can’t be written it doesn’t terminate so it can’t be written as a ratio so pi is our only irrational number and finally for the real numbers all of the numbers that are in our set a are real numbers so the set of real numbers or the numbers that are real numbers would be negative 6 1/2 negative 1 point 3 3 3 repeating 0 pi 2 & 5 the entire set a next we’re going to talk about finding the approximation z’ of numbers we’re going to talk about two types of approximations the first is truncation in truncation what you do is you drop all of the digits that follow the specified final digit so if the problem asks you to truncate two four digits or four decimal places you would count the four decimal places out and then drop all the numbers after that the other method of approximation that we’ll talk about is rounding so for rounding again we’ll find the specified digit and then we’ll look at the next digit over so if we’re asked to round to three decimal places we would count the three decimal places and then look at the fourth decimal place if the next digit is bigger than or equal to five then we’re going to add one to our specified digit if the next digit is less than five then we’re going to truncate and just leave it the way that it is all right so let’s do an example where we’re going to approximate some numbers approximate the following numbers to three decimal places by first truncation and second rounding the numbers that we want to approximate are twenty-eight point six five three one nine and then zero point zero six two nine one and again we want to approximate each of these to three decimal places by truncation and then by rounding so first will truncate twenty-eight point six five three one nine two three decimal places if we count over three decimal places that will put us at the three that’s highlighted in red and remember for truncation you just drop everything beyond that so the one in the nine are just going to go away so the truncated approximation will be twenty eight point six five three next we’ll round 28 point six five three one nine all right so again we count to the three decimal places that’s the three that’s in red and then we’ll look at the next place which is a one now since one is less than five will truncate we’ll just drop the one in the nine out and leave the three the way that it is now we’ll move to the other number we’ll start by truncating 0.06 – 9 1 again we count 3 decimal places that puts us at the 2 that’s highlighted in red and since we’re truncating we will just drop the rest of the numbers so the 9 and the 1 will go away so the truncated approximation is 0.06 – and finally we will approximate 0.06 – 9 1 by rounding

again we go to 3 decimal places that’s the 2 in red and when we round we look at the next number so the next number is a 9 9 is greater than 5 so we will round the 2 up to a 3 so the rounded approximation is 0.06 3 next we’re going to talk about using operations to simplify and combine numbers together so there there’s some basic operations that you probably are already familiar with but there’s a table here that expresses what the operation is how we would write that out in symbols and some sample words if you were dealing with word problems so the first operation we’ll deal with is addition addition can be written a plus B and the sample words that make you think that you’re dealing with addition are some a plus B or B more than a the next operation is subtraction we could write subtraction as a minus B and some sample words that tell us we’re dealing with subtraction our difference a minus B a less B or B less than a note that those last two examples a less B and be less than a are very similar it just depends on how things are worded if you have five less two that means five subtract two but if you have five less than two that means you subtract 5 from two the next operation is multiplication which we can write a B meaning a times B or we could do a B with parentheses around them the sample words that tell us we’re dealing with multiplication our product or a times B next we have division which we would write as a divided by B or a over B and the sample words that tell us that we’re dealing with division are quotient or a divided by B and then the final operation is equality or equals and that is written with the equal sign and the sample words that tell us that we’re dealing with that are equals or is so let’s do a couple of examples where we’re gonna take some words and write them in terms of symbols to express the operations given the first that we want to do is the difference two less y equals six now remember the order two less Y is very important and so written in sin this would be 2 minus y equals 6 next we have the sum of 3 and Z is the sum of 2 & 2 so remember sum means addition and is means equality so the sum of 3 and Z means 3 plus Z is would say right an equal sign and the sum of 2 & 2 gives us 2 plus 2 so we could write this as 3 plus Z equals 2 plus 2 and we could simplify that that III plus Z is equal to 4 but we just want to get the hang of writing symbols out from words now we want to talk about evaluating numerical expressions specifically using the order of operations so so when you’re given an expression to evaluate an expression that contains many different operations the order in which we do those operations are very important one way to remember the order that we do them is using the acronym PEMDAS you might have learned this is please excuse my dear Aunt Sally or something to that effect where each of the letters PEMDAS in PEMDAS represent an operation that you would do in order the first letter in PEMDAS is P this tells us that we want to do anything that’s in parentheses first so the very first step to simplifying an expression is to simplify whatever’s inside the parenthesis that may be present next we have the letter E which stands for exponents so after we clear our parentheses we want to evaluate any exponents that we may have in our expression next we have M and D which stand for multiplication and division and multiplication and division can be done at the same time but it’s important that you go from left to right so you’ll do all the multiplication and division going from left to right and then finally we have a and S which stand for addition and subtraction and again we can do addition and subtraction at the same time keeping in mind that it’s important to go from left to right so go from the left of the expression to the right of the expression so let’s do an example evaluate the following expression 2 times bracket 8 minus 3 parentheses 4 plus 2 close parenthesis

close bracket minus 3 so if we want to evaluate this again we want to look at the order of operations the order of operations says we need to clear our parentheses first so we’re going to start by simplifying what’s inside the square brackets so specifically we’re going to start with the parentheses inside the square brackets if we add the 4 and the 2 together inside those parentheses that’s going to give us so inside our square brackets we now have 8 minus 3 times 6 so next we’ll do multiplication within the square brackets we need to simplify 3 times 6 and 3 times 6 is 18 so now inside the square brackets we have 8 minus 18 so the final step to simplifying what’s inside the square brackets is to do the subtraction there so 8 minus 18 gives us negative 10 and we’ve completely evaluated everything inside the square brackets so now our expression is 2 times negative 10 minus 3 so we can continue going down our order of operations we’ve cleared the parentheses so the next thing we’re going to do is multiplication 2 times negative 10 is negative 20 so we have negative 20 minus 3 we can combine those together using the rules of addition and subtraction to give us our final answer of negative 23 so this whole expression simplifies to be negative 23 so let’s do another example where we want to evaluate the expression using the order of operations we want to simplify to minus 4/5 minus 3 ok in order to simplify this we need to treat the numerator and denominator as if they are individual expressions so we’re going to put those in parentheses so that gives us parenthesis 2 minus 4 divided by parenthesis 5 minus 3 so going through our order of operations we first need to evaluate what’s inside the parentheses now we can do both of those at the same time if we evaluate 2 minus 4 in the numerator that gives us negative 2 and if we evaluate 5 minus 3 in the denominator that gives us positive 2 so we’re left with negative 2 divided by 2 we’ve cleared our parentheses so the next thing we want to do is go on to multiplication and division so what do we get if we divide negative 2 by 2 dividing negative 2 by 2 gives us negative 1 and so that’s this that’s the expression simplified the simplification of this expression is negative 1 next we want to talk about the least common multiple and least common denominator to find the least common multiple of two or more numbers what we want to do is list the multiples of the numbers that were given and find the smallest one that they have in common for example let’s find the least common multiple of six and eight we’ll start by writing down the two numbers whose multiples we want to find six and eight next we’ll list out the multiples of six remember multiples are just what you would get if you multiply 6 by numbers one two three four so the multiples of six are 6 12 18 24 and 30 and then we’ll list out the multiples of eight those are 8 16 24 32 so the least common multiple will be the smallest number that they have in common and if we look at our list that number will be 24 so 24 is the least common multiple of 6 and 8 the least common multiple is important in finding the least common denominator the least common denominator is needed to add or subtract fractions and can be found by finding the least common multiple of the denominators present so let’s look at an example we want to add together the fractions 1/6 plus 5/8 now remember in order to add fractions together you need them to have the same denominator so we need to find the least common denominator and the least common denominator is simply going to be the least common multiple of the denominators that you have so we have the denominators of 6 and of 8 which we found out in the previous example the least common multiple would be 24 so we need to transform our fractions so that they have that common multiple as a denominator so if I have the fraction one-sixth and I want to have a 24 in the denominator I need to multiply the denominator by 4 but I can’t just multiply the denominator by 4 because that’ll change my number so I need to multiply the numerator by 4 as well so if I multiply 1/6 by 4 over 4 which is the same thing as 1 so things don’t change that will give me a fraction that has a denominator of 24 similarly if I multiply five eighths by

3 over 3 that will give me another fraction that has a denominator of 24 so if I multiply those together and simplify 1/6 times 4 over 4 gives me 4 over 24 and 5/8 times 3 over 3 gives me 15 over 24 so now I have a common denominator all I need to do is add the numerators together and I’ll find that 1/6 plus 5/8 is equal to 19 over 24 finally we’re going to end this section on real numbers by talking about the properties of real numbers these are properties that we’re going to use throughout this course to solve problems and to evaluate real expressions we’ll start with the properties of equality the first property of equality is the reflexive property which can be written as a equals a this basically just says that any number is equal to itself the second property of equality is the symmetric property that says if a equals B then B equals a so essentially the order that you write them with the equal sign doesn’t matter the third property of equality is the transitive property that says if a equals B and B equals C then a has to equal C and finally the fourth property of equality is the substitution property that says if you know that a equals B then you can substitute B into any expression for a next we’ll talk about the commutative property of addition and multiplication the commutative property of addition says that a plus B equals B plus a so essentially the commutative property says that for addition the order doesn’t matter so an example that shows this if we look at 2 plus 3 that’s equal to 5 but if we switch the order around and do 3 plus 2 that’s also equal to 5 so this shows the commutative property of addition the commutative property of multiplication basically says the same thing if I have a times B that should equal B times a so the order doesn’t really matter for multiplication either and an example that illustrates the commutative property of multiplication if I take 2 times 4 that equals 8 but if I switch the order around and do 4 times 2 that also equals 8 next we’ll talk about the associative property of addition and multiplication the associative property of addition says that if you have the quantity a plus B plus C that equals a plus the quantity B plus C so we’re dealing with addition the placement of the parentheses doesn’t really matter so an example that illustrates the associative property of addition if we take the quantity 4 plus 6 plus 2 well 4 plus 6 is equal to 10 and 10 plus 2 is equal to 12 but if we move the grouping from the 4 plus 6 to the 6 plus 2 and we look at 4 plus the quantity 6 plus 2 well that would give us 4 plus 8 which is also equal to 12 the associative property of multiplication is very similar saying that the order of our grouping doesn’t really matter if we take the quantity a times B and multiply it by C that’ll be the same thing as a time’s the quantity B times C and an example of this if I look at 3 times the quantity 2 times 4 well 2 times 4 is 8 so 3 times 8 is 24 but if I move the grouping to be the quantity 3 times 2 times 4 well 3 times 2 is 6 and 6 times 4 is 24 so we see it’s the same thing next we have the distributive property which states that if you have a time’s the quantity B plus C you can distribute the a throughout to get a times B plus a times C and note the distributive property will go both ways so if you have a B plus AC you can pull the a out in front to rewrite it as a times B plus C so here’s an example of the distributive property if you have 2 times X plus 3 you can distribute the 2 throughout the parenthesis to give you 2 times X plus 2 times 3 and then you simplify that to give you 2x plus 6 and an example that goes the other way if you’re given 3x plus 5x well we have an X in both terms that we can pull out and so we would have 3 plus 5 times X which gives us 8x next we have the identity properties the first is the additive identity property that says that any number plus zero is equal to itself so zero plus a will give you a and we also have the multiplicative identity property that

says that any number times one will give you give it give you itself so one times a results in a next we have the additive inverse which states that negative a plus a is equal to zero so a and negative a are opposites and if you add them together they cancel out and we have the multiplicative inverse which says that if you multiply a times one over a or one over a times a they will cancel each other out and give you one assuming of course that a is not equal to zero because we can’t have a zero in the denominator the quantity 1 over a is called the reciprocal so if you multiply a number by its reciprocal they will cancel each other out and give you 1 next we have the difference of two numbers which can be rewritten using addition as follows a minus B is the same thing as a plus a negative B similarly for a quotient we can rewrite the quotient a divided by B as a times 1 over B again assuming that B is not equal to 0 with the property of multiplication by zero which states that any number multiplied by zero will give you zero and we have a couple of division properties the first says that zero divided by any number is going to give you zero and any number divided by itself so a divided by a will give you one again assuming that a does not equal zero we have some rules of signs that we’ll be using the first says that number times the opposite of another number will equal the opposite of the product of the two numbers so a times minus B will equal minus a B similarly minus a times B will give you minus a be third we have minus a times minus B equals a B so the two minus signs will cancel each other out fourth we have minus minus a equals a so again if you have two negative signs they will cancel each other out and make a positive fifth a divided by negative B or negative a divided by B will give you negative a over B so a one positive and one negative will results in a quotient that is negative and lastly the quotient of two negative numbers negative a over negative B is the same as the quotient of their positive counterparts a divided by B so again the negatives will cancel each other out we have a couple of cancellation properties which we will use fairly regularly throughout this course the first says that if you have a product AC that equals a product BC the C’s can cancel out from both sides leaving you with a equals B assuming the C does not equal zero and the second is similar says that if you have a times C divided by B times C again the C’s can cancel each other out leaving you with a divided by B assuming that B doesn’t equal zero and C doesn’t equal zero so let’s look at a couple of examples that use these cancellation properties first we have 3x equals 12 in order to use the cancellation property we want to rewrite 12 so that it has a factor that’s common with 3 times X so we can rewrite 12 as 3 times 4 so we have 3 x equals 3 times 4 the 3 that is common on both sides of the equation will cancel out leaving us with x equals 4 for a second example we want to examine the fraction 4 divided by 12 which we can rewrite as 4 times 1 in the numerator and 4 times 3 in the denominator the four being a common factor in both the numerator and denominator will cancel out leaving us with 1/3 or 1/3 next we have the zero product property this property is going to be extremely important when we start talking about factoring and solving equations and the zero product property states that if you have the product of two numbers a and B and that product equals zero then you know that either a equals zero or B equals zero or they both equal zero and the final property of real numbers that we’re going to talk about in this section is the arithmetic of quotients first if you have the addition of two quotients or fractions a divided by B plus C divided by D remember in order to add them together you need a common denominator so you could multiply the a and the B by D and the C and the D by B

to give you a do over BD plus BC over B D and once you have that common denominator you can add the numerator together to give you a d-plus BC over B D assuming that B and D are both not equal to 0 multiplication of quotients is a little bit more straightforward maybe a little bit easier if you have a divided by B times C divided by D all you’re going to do is multiply straight across so multiply your numerators together to give you AC and multiply your denominators together to give you a denominator of BD and again B and D cannot equal 0 lastly we want to look at the division of quotients or the quotient of quotients so if you have a divided by B divided by C divided by D the way that we deal with this is we multiply by the reciprocal of the denominator so since C over D is in our denominator we’re going to multiply by its reciprocal which would be D over C so a divided by B divided by C over D is the same thing as a divided by B times D over C and again for multiplication we’re just going to multiply straight across so this will give us ad over BC assuming that B C and D are all not zero so let’s do some examples where we’re gonna deal with the arithmetic of quotients first we want to add together 4 divided by 3 plus 1 divided by 2 so 4 thirds plus 1/2 remember in order to add these we first need to find a common denominator so we’ll multiply 4/3 by 2 over 2 and 1/2 by 3 over 3 to create a common denominator of 6 on both terms now we can just add the quotient the the numerators together so 4 times 2 plus 1 times 3 all divided by 3 times 2 and then we’ll simplify to get our answer so 4 times 2 is 8 and 1 times 3 is 3 so that gives us 8 plus 3 in the numerator and 3 times 2 is 6 so that gives us a 6 in the denominator if we add these together we’ll get 11 over 6 as our final answer next we want to do an example where we deal with multiplications so we want to multiply 5 divided by 9 times 3 divided by 10 remember with multiplication you’re just gonna go small to fly straight across multiply the numerators and then multiply the denominators so if we multiply our numerators together we get 5 times 3 is 15 and if we multiply the denominators 9 times 10 is 90 so this is going to give us 15 over 90 but we can simplify that we can divide a 15 out of both the numerator and denominator to give us a reduced fraction of 1 over 6 so five ninths times 3/10 gives us 1 over 6