 For many lectures now, we've been dealing with equations. But what if the left-hand side is not equal to the right-hand side? In this video lecture, I want to talk to you about inequality. So we start to talk about inequalities. We've always had a left-hand side equal to the right-hand side or the right-hand side equal to the left-hand side, but now we're gonna have an inequality. The left and the right are not equal to each other. Before we get there though, I really want to talk about intervals. So let's start talking about intervals. Intervals, if we think of the real number line, so let's have a real number line right here that goes off to positive infinity on the side, on the side it goes off to negative infinity, we're talking about some interval of values that we would find x, these are all the values x, our horizontal axis, that we have some part of this line in which we find the solution x. Now that can take various forms. We can have a lower bound and an upper bound, and we would write it with square brackets, denoting that a is included in that possible set of solutions and so is b. So if we have something like the solution is between three and four, the solution can absolutely be three and it can absolutely be four and anything in between. So these we're gonna call closed, that's a closed interval, closed interval. We can also have this scenario a comma b, but we put parentheses there. So if that is any solution between three and four, that is an open interval and what we do now, we exclude three and four. We can have 3.0000001 and we're gonna go 3.999999, but three and four are excluded, anything in between three and four is possible, but certainly not three and not four. Then we also get this idea of half open intervals, so we can have something like from a to b, and now say for instance that would be three to four, it is still possible for four to be a solution, but not three, something just bigger than three, but four is definitely included. So we can call this semi open, semi closed depending on the textbook that you read. So let's say it's semi closed or semi open and we can also get this situation a comma b. Now three is definitely possible and then anything bigger than three up to four, but four is definitely not included. You can call that semi open, semi closed, et cetera. We can have other ways that we can write this. We can say a is less than or equal to x, which is less than or equal to b. Yeah, we'd have a is less than x is less than b, so a, so this x at least is between a and b, but there's no equality there. A and b are not included there, there isn't equality there, so a and b are definitely included. Here we would have a is less than x is less than or equal to b, it is possible to be b, and here we have a is less than or equal to x, which is less than b, b now is not included, up to b, but not with b, and then on the left hand side, a and a can be included in that solution. So that's just something that we have to consider when we look at inequalities. We're thinking about these possible solutions that we can have. So now there are a couple of rules that we really have to pay attention to if we know these, then these inequalities are no problem. So in order to do this, let's create some constants, a, b, c, and d. We're gonna say they can be any real numbers, so they are members of the set of real numbers, but we're gonna subtract from that real number one of the elements of the real numbers, and that's zero. Let's just remove zero because we wanna divide, and we don't want to divide by zero. So let's have these rules. Now the first one that I want to do, I'm gonna say let's do this first one. If a is less than b, so we're gonna have a is less than b, now c can be anything there, we're gonna say then a plus c is still less than b plus c. So I've got a is less than b, I'm adding c to both sides, it's still going to be, a plus c is still going to be less than equal to b plus c. So let's say for instance three is definitely less than four, then we could say, say for instance three plus five, that is still going to be less than four plus five, because what do we have here? We have eight is less than nine. So that one checks out for us. Let's do the following. If a is larger, if a is larger than b, then if I take a plus some value c, that is still going to be larger than b plus c. So an instance here, let's say four is larger than three, and if I set something like four plus four, it's still going to be larger than three plus four, because eight is larger than seven. So that one checks out as well. Now these needn't be positive numbers, let's just have a look at this, let's still have four is larger than three, we would all agree with that. But let's add to four, let's add to that a negative four. We are saying that's still going to be larger, there's our three on the right hand side, plus we add a negative four, that's still going to hold, whether that's a positive number or a negative number, because four minus four, that's going to be zero, and that's still going to be bigger than three minus four, which is negative one. So it doesn't matter what you add on both sides, nothing is going to happen to this inequality. If it was less than, it's going to stay less than, if it was greater than, it's going to stay greater than. So let's have the following rule, if A is larger than B, but now we're going to say, and C is larger than zero. So we're going to constrain C now, then we're going to have the following, we're going to say that A plus B, let's have it there then, is to A times C is still going to be bigger than B times C. So those were additions, now we're going to do multiplication. There we had addition, addition, even though that addition can be a negative number, now we're constraining C to be a positive number. So let's have four is larger than three, and then let's multiply both sides by anything that's larger than zero, let's make it five, let's then, it's four times five, it's still going to be greater than three times five. We have that 20 is larger than 15, nothing happens there. Let's have the following rule, if let's do A, let's stick with it, larger than B, and now we're going to make C less than zero, it's going to be negative. What's going to happen then? Then, A times C, for the first time we're going to swap around that inequality, it's now going to be less than B times C. So in this instance, A was larger than B, now suddenly becomes less than, if we multiply both sides by a negative number. So let's have four is larger than three, and let's say then we multiply both sides by negative five. So now we're going to have four times negative five, and on this side we're going to have three times negative five. And so what do we have here? We have negative 20, and here we have negative 15, and now suddenly this is smaller than that, it is smaller than, it is smaller than, so we went from a greater than to a less than. Now let's have the next rule, let's say if A is less than B, that was greater than, now we're going to make it less than, and we have C is greater than zero, then nothing's going to change. Then we're going to have the following, A times C is going to remain less than B times C. Let's have an example, we're going to say three is definitely less than four, we're going to multiply both sides by a positive number, let's make it five again, three times five, and on this side four times five, it's still going to be less than. We have that 15 is less than 20, so that works out. But now if for instance A is less than B, and C is now negative it's less than zero, then we're going to have a swap around again. Then AC is going to be bigger than BC. So let's have three is definitely less than four, let's multiply both sides by a negative five, and we have four on this side and a negative five, so we have negative 15, and we have negative 20, now negative 15 is larger than, this is larger than, and we can see this sign, this inequality has changed. Let's have the following, if, now let's stick with this, A is going to be larger than B, and we're going to make C is also as positive, then if I take A divided by C, now remember we said these can up is zero, what happens now if we do B divided by C given, A was larger than B, and C was positive, this remains exactly the same. So let's have some example, let's say for instance eight is larger than four, and I'm going to divide both of them by two, two is certainly larger than zero, then we have eight divided by two, and we have four divided by two, and that gives us four, and that gives us two, and it's still larger than, it's still larger than, it's still larger than. Now though, if A is larger than B, and C is now suddenly less than zero, then if I have A divided by C, and I have B divided by C, then we're going to swap around. Now A divided by C is going to be less than, whereas before A was larger than B. So let's have the following, we have eight is larger than four, and let's divide both of them by negative two, negative two is certainly smaller than zero, so I'm going to have eight divided by a negative two, I'm going to have a four divided by a negative two, on this side I'm going to have a negative four, on this side I'm going to have a negative two, now this one is smaller than, it's smaller than, and now that sign has swapped. So let's have now if A was less than B, and we have that C is positive, then we're going to have the following, if I take A divided by C, and I take B divided by C, this is positive, that was less than, that's going to stay less than. So let's say four is less than eight, we agree with that, let's divide both of them by some positive value, let's make it two, we take eight divided by two, four divided by two is two, eight divided by two is four, it's still less than, it's still less than, it's still less than. And then if I have that A is less than B, and C is now negative, well then we're going to have, if we take A divided by C, and we take B divided by C, we're going to have that sign now swaps, and it's going to be like that. So let's have four is less than eight, and let's divide both sides by a negative two, once again negative two is less than zero, I have eight and I'm dividing that by negative two, four divided by negative two, well there's negative two, eight divided by negative two is negative four, well now the sign swaps because negative two is larger than negative four, it's larger than, and that sign has swapped. Let's have our very last rule, we're going to say if A divided by C is less than, let's make it B divided by D, if I take the reciprocal of that, instead of A divided by C, I have C divided by A, let's do C divided by A, and on this side we have D divided by B, if I swap those around, then the sign is also going to swap around, so there's the sign. So A divided by C less than B divided by B means C divided by A is going to be greater than the D divided by B, so let's have a little example of that, let's have two divided by seven, that is less than for instance five divided by eight, why do I know that? Well I can do common denominators seven and eight, seven times eight is 56, so let's make both sides 56 as the denominator, let's have the two numerators, so I divided this denominator, multiplied it by eight, I've also got to multiply the numerator by eight, this would be the same as multiplying by eight over eight, which is this multiplying by one, nothing changes, eight times two is 16, this side I'm going to multiply by seven over seven, taking any number, multiplying by one, nothing changes, seven times five is 35, and clearly 35 is larger than 16, so that's it, but let's swap that around, let's make it seven over two, and let's make this eight over five, now common denominators now, let's make that 10, so I've got 10 as a common number, 10 to get from two to 10, that's multiplied by five, so I'm going to take seven times five, that's 35, five to get to 10, that's multiplied by two, eight times two is 16, now clearly 35 over 10 is larger than 16 over 10, so seven over two is larger than eight over five, so that sign has swapped, so let's solve some inequalities now that we know these rules, let's start with an example, so our example is going to be the following, let's do three x minus four, and let's have that being less than five, so what interval of values for x would solve this, it's not a quality anymore, so I'm not going to get a specific value for x, I'm going to get some interval, so what I can do of course, I can add four to both sides, and if I add some c that's larger than zero, nothing's going to happen to the sign, so I could say three x minus four plus four is less than five plus four, so I've done the plus four to both sides, so I haven't changed anything, of course negative four, positive four, that's just zero, so I've got three x is less than nine, now what I'm going to do, I'm going to divide both sides by three, and remember three is larger than zero, so nothing is going to happen to this sign, I can cancel out the threes, and nine divided by three, well that's just three, so x is less than three, any value less than three, so three is not included, if I plug any value in less than three, I should get this, and let's do that, let's choose the value two, let's have three times two minus four, well that should be less than five, three times two is six, six minus four is less than five, and d two is less than five, so that's absolutely correct, now there's other ways that we can write this, we can for instance say that x is on this interval, we can say from negative infinity, all the way to three, but three is not included, so it's open interval, negative infinity, remember infinity and negative infinity, they are not numbers, they're not some exact numbers, so you can never have that that's included, so we'll always have this open interval as far as negative and on the other side positive infinity is concerned, so we can say that x is in this interval, maybe it's an element of that interval, or it's equal to that interval, we can also write negative infinity, that's going to be less than x, and x is going to be less than three, three is on that interval, so we think of a real number line, and perhaps that is zero, and that is one, and that is two, and that is three, and it goes off to positive infinity, the side goes off to negative infinity, we're going to say that this interval is something like this, three is not included, and then all the values here, you can plug in any value there, and you're going to get that inequality holds. Let's have another example, let's have the fact that negative two, let's make that x minus four, let's say that that is less than, let's make it five, let's make it five. Now what I want to do is I want to get rid of this negative two, so let's divide both sides by negative two, so I'm going to have negative two x minus four, and on the side I have five, I want to divide both sides by a negative two, negative two and negative two, now I have divided by a negative number, a value that we see is less than zero, so that sign has got to swap around. Now those two would cancel, and I'm left with the fact that x minus four, let's have that x minus four, let's have that x minus four, well that's going to be, let's try and rescue this, it's going to be larger than five over negative two, so let's make it negative five over two, I can add four to both sides, so x minus four, let's add four to the side, that's going to do nothing to that sign, I have negative five over two, and I'm going to add another four to that, so on this side negative four plus four, nothing remains there, so that's going to be larger than, and let's have negative five over two, I'm going to add to that, and I'll have eight over two, eight divided by two is four, I have a common denominator there, so I can say x is larger than negative five plus eight, negative five plus eight over a common denominator two, that is x is larger than three over two, and again we can write something like, if x is indeed larger than that number, it means that we have three over two is not included, and it goes up to infinity, or I could say three over two is smaller than x, if I read it from this side, it's x is larger than three over two, x is larger than three over two, and it's going to be less than positive infinity. Now let's do this very complicated example, let's have x squared minus one, let's divide this by x plus three, and I want this to be larger than or equal to both possibilities zero, so let's deal with the two cases, it can be equal or it can be less than, so let's deal with case number one, in case number one we're going to have x squared minus one, we're going to divide that by x plus three, and we set that equal to zero, it's certainly possible to be equal to zero, what can we do, well we can multiply both sides by the denominator, so we'll have x squared minus one divided by x plus three, let's multiply that by x plus three over one, and so we just have to do that on the right hand side as well, x plus three divided by one, now of course zero times anything is to zero on the right hand side, I'm just left to the zero, on the left hand side remember this is one single value, that's one single value, there's a multiplication, which means I can cancel those two, on the left hand side I'm left with x squared minus one, I can bring that across to the other side or add one to both sides, so on the left hand side negative one plus one, nothing there, on the right hand side I have equal to one, so in other words I take the square root of both sides and I have that x is equal to plus or minus one, so x can be positive one, x can be negative one, by the way let's have a look at this denominator, remember one thing that x definitely cannot be, it cannot be negative three, negative three plus three is zero, I can't divide by zero, so at least I know that, and now we have x can be one, or x can be negative one, and if we plug those in there, I get either one of those two, I get zero in the numerator, some number in the denominator, and that is definitely equal to zero, this is the case that we have here, so let's have the second case, I now want x squared minus one, divided by x plus three, I want that now strictly to be less than zero, so I've dealt with the equal, and I wanna deal with the less than, because I have less than or equal to, so how can I have a numerator divided by denominator and it's negative, well I can have a negative divided by a positive or I can have a positive divided by negative, so I have to look at either of these, and remember between this set of, this case and this case, it's less than or equal to, so there's definitely an all between these two, there's gonna be an all between these two, but definitely I have to have this and this, or I have to have this and that, so let's start with this, I've got to have the numerator, it's less than zero, it's negative, and both of these things must be true, I have to have that x plus three has gotta be positive, so both of these, yeah I would have x squared is less than one, I just added one to both sides, and I can have, I have to think this through, what values can I square that it's still less than one? Well it turns out if I go negative one is less than x is less than one, if I have this value on this open interval from negative one to positive one, if I square anything there, I'm gonna get a value less than one, as soon as I get beyond that, say negative two or one plus 1.1, I will no longer have the case that the value is less than that, on this side, and I've gotta have that x is larger than negative three, now how can I have the fact that x is in this interval and x is larger than negative three? So what can I have so that both these things are true? So certainly I can't have negative two or negative two and a half or negative one and a half, and I can't go beyond one anyway, so if I look at x being, now both of these things have to be true, then certainly this is the part that remains, because if this is true, then by necessity this is also true, but if I have a value say negative 2.5, which is certainly there, it's not in there, but all the values in there are definitely within that, so this one has gotta win out, because remember, I have to have that both of these things are true. Now let's have the other case, I'm gonna have that x squared minus one that is larger than zero, or I have x plus three is less than, so and remember, is less than zero, so it's a positive enumerator negative in the denominator, so I'm in this case, and between these cases remember, there's an or between, so I really have to keep track of the ands and ors, so here we are gonna have that x squared is larger than one, and the only way that that's going to happen is if I have that x is less than one, or x is larger than, so I've gotta put that actually negative one, or x is larger than positive one, if I have again this scenario, I'm gonna land up with a value that's less than one, I've gotta be beyond these values, if I square them to get something that's larger than that, and on this side, on the denominator side, I'm gonna have that x is less than negative three, now again I've gotta think that or that and that, and certainly what is going to win out is gonna be this one, because this x is less than negative one, so let's make that negative two, well that's certainly not here, and definitely if it's beyond one, it's certainly not there, so certainly this is the one that's going to win out here, so now have a look at this, it can definitely not be negative three, it can be plus or minus one, so let's have that, those must exist as well, and I've got this, and this, so it's this, or, this, or, this, and now if I have one and negative one there, that's less than, but it can be equal, it's less than but it can be equal, so that's going to change to negative one is less than or equal to x is less than or equal to one, so that's the one part, or I have to have the fact that x is less than negative three, so what I'm gonna leave for you is to plug in values there into there, and see that it's less than or equal to zero, or plug in a value less than negative three, plug that into x there, and you'll see that's also less than negative three, if you plug in any value outside of these, so any value that's larger than one, or value that's here between negative one and negative three, that's not going to hold, now the last thing I want to deal with is absolute values, absolute values, so let's deal with that, we're gonna say that the absolute value of x, now yeah, let me just be sure that x is a member of the set of real numbers, the absolute value of x is the following, now this is a piecewise defined function, I'm gonna say the absolute value of x is x, if x is larger than or equal to zero, and it's gonna be negative x, if x is less than zero, now that might seem a bit strange, so let's look at a couple of numbers, let's look at anything that's equal to a larger than zero, so let's say x is equal to positive three, well the absolute value of three, while we say that is so, then x is just what it is, it's just itself, so in this instance, well that's just gonna equal to three, let's have a value where x is less than zero, let's say x is negative two, while we say that the absolute value of negative two, that's gonna be negative x, that's negative, negative two, and that's also positive two, so we're just saying that we take any number and we turn it into something that's non-negative, because if we plug in zero, it's just gonna be zero, but any negative, negative times a negative gives me a positive, so we're turning anything into a non-negative number, so let's look at an example problem, let's have that the absolute value of x minus three is equal to five, how can we solve for that? Well, let's think of this definition, so we're definitely going to have a couple of things, we're gonna have that x minus three equals five, that is if x is larger than equal to zero, and we're gonna have minus x minus three is equal to five if x is less than zero, so I'm just looking at the definition, I'm taking this, now it's not just x, it's x minus three, I'm leaving in that as is, when x is positive, I'm just putting a negative in front when x is negative, so if we carry on with that, we're gonna have that x equals eight, if x is larger than equal to zero, well, now we have a specific value, it is eight, while we're going to have on this side, we're gonna have negative x plus three equals five, and if we just carry that on, we'll have that negative x is equal to, and we're gonna take negative three on both sides, that leaves me with two, or we're going to have the fact that x equals negative two, so we have two solutions, x equals eight, or x equals negative two for this absolute value problem to hold. Now let's do one more, let's have that x minus three must now be less than five, how are we gonna do that? Well, these are actually quite a bit of fun to do, so we're gonna say x minus three is less than five, that is if x is non-negative, and we're gonna have negative x minus three is less than five, if x is less than zero, so we've got to have both of those cases, cases number one and case number three. This one is easy, I'm just gonna take this across to the other side, so I'm gonna have that x is less than positive three on both sides, that's eight, if x is larger than or equal to zero, so now I just gotta be careful, it's less than eight, but it's gotta be there, so I'm gonna say zero is less than or equal to x is less than eight, that's my one solution. Here, let's multiply both sides by negative one, but do remember that means that sine has to swap, so I'm gonna have x minus three is larger than negative five, and that is if x is less than zero. So what we're gonna have here, we're gonna add three on both sides, so that's gonna be x is larger than negative two, but remember x is less than zero, so we're gonna have this, negative two, we're gonna have x, and it's gonna go up to zero, now x is less than zero, but it's larger than negative two, so there are my two solutions, so it's between zero, so let's have it, let's put it down here, it's gonna be between zero, let's put it out here, zero to eight, but eight is not included, and we've gotta have this union here, and that goes from negative two to zero, and these are open intervals on both sides, there's no equality there, and I want the union of those two, remember with the union of those two, I can also say that that must be all, which is the symbol instead of the symbol, so if I have on this open interval, the semi-open, semi-closed interval, either of those two, this is an or a union, then this inequality is going to hold.