For examples, consider the possible inequalities relating the expressions 2x − 4 and 3x − 9. Let's evaluate these five inequalities, each at x = 4 and each at x = 5. As before, when x = 4, then 2x − 4 = 4 and 3x − 9 = 3; when x = 5, then 2x − 4 = 6 and 3x − 9 = 6. So here is a table of the results:
|Inequality:||When x = 4:||Result:||When x = 5:||Result:|
|2x − 4 < 3x − 9;||4 < 3;||False;||6 < 6;||False.|
|2x − 4 > 3x − 9;||4 > 3;||True;||6 > 6;||False.|
|2x − 4 ≤ 3x − 9;||4 ≤ 3;||False;||6 ≤ 6;||True.|
|2x − 4 ≥ 3x − 9;||4 ≥ 3;||True;||6 ≥ 6;||True.|
|2x − 4 ≠ 3x − 9;||4 ≠ 3;||True;||6 ≠ 6;||False.|
t = 1 or t = −1.
A compound statement consists of simpler statements (like equations or inequalities) joined by either the word ‘and’ or the word ‘or’. (If you study Logic, you'll see that there are other words that might be used here, like ‘if’ or ‘but not’. However, we won't need such connectives in this course.) For example:
Particularly common are compound inequalities, each of which consists of two inequalities joined with ‘and’ and sharing one expression. An example is
x > 2 and x ≤ 3above. This example can be summarised simply as
2 < x ≤ 3;that is its form as a compound inequality. (Notice that x > 2 is equivalent to 2 < x, and remember that a bunch of statments run together always have implicit ‘and’s between them.)
Here are some more examples:
|Compound inequality:||Expanded meaning:|
|2 < x ≤ 3;||2 < x and x ≤ 3;|
|−1 < y < 6;||−1 < y and y < 6;|
|5 > a ≥ 3;||5 > a and a ≥ 3;|
|2 < 5x − 9 ≤ 3;||2 < 5x − 9 and 5x − 9 ≤ 3.|
The words ‘and’ and ‘or’ are pretty fundamental English words, but for purposes of Mathematics I should tell you exactly what they mean. A compound statement with ‘and’ is true only if both statements are true, and false if either statement (or both!) is false. Conversely, a compound statement with ‘or’ is true if either statement (or both!) is true, and false only if both statements are false. Finally, if any statement within a compound statement is meaningless, then so is the compound statement.
In particular, mathematicians always use ‘or’ in an inclusive sense, so that if both statements are true, then the compound statement is true. For example, consider the compond statement x ≤ 2 or x < 3. This is really just equivalent to the simple statement x < 3, because both parts are true when x ≤ 2, so the first part (x ≤ 2) is superflouous.
To begin with, consider the equation
x = 4.It's easy to draw a picture of this on a number line:
This is so easy that there may not seem much point to it. However, there are a few points worth mentioning:
Now consider a more complicated solution, the compound statement
x = −2 or x = 4 or x = 6.Now the graph looks a little more interesting:
You can see here what I mean when I say that the parts of a compound statement should be listed in increasing order;
x = 4 or x = 6 or x = −2,means the same thing, but it doesn't match the graph as nicely.
Now consider the inequality
y < 3.Here, y could be any real number less than 3, so the graph is spread out along that entire part of the line:
Since y = 3 is not itself a solution, I indicate this with a round parenthesis. The parenthesis also faces towards the solutions, which helps if it's hard to see the shading.
Compare this with
y ≥ 3:
Now I have a square bracket instead of a round parenthesis, to indicate that now y = 3 is a solution. Also, the bracket faces the other way, because now the solutions include values larger than 3.
There is another way to graph inequalities, which you may have already learned before this course. I think that it's less clear than the method above, but you can use it if you like. In this method, you use a solid dot (instead of a square bracket) to indicate that a boundary point is a solution, and you use a hollow circle (instead of a round parenthesis) to indicate that a boundary point is not a solution. Then you get these graphs:
If you use the circle/dot method, then be sure to shade in well the region where the solutions are; this isn't so important when you use the parenthesis/bracket method.
Now consider the compound inequality
−3 < x ≤ 2.This means that −3 < x and x ≤ 2. In other words, x is between −3 and 2, and it can't be exactly −3, but it might be exactly 2. So I shade in the region between −3 and 2, put a round parenentheses (or hollow circle) at −3, and put a square bracket (or solid dot) at 2:
You could just as well write this compound inequality as 2 ≥ x > −3, but again we prefer to write things in increasing order.
Finally, consider the compound statement
x ≤ −3 or x > 2.This comes in two pieces, one where x ≤ −3, and another where x > 2, so I simply mark these two pieces separately on the same graph:
Again, you could just as well write x > 2 or x ≤ −3, but once more we prefer to write things in increasing order.
Notice that these two pieces have no overlap. Compare this with a statement like
x ≤ 3 or x > −2.This statement is always true, one way or another, so I should really simplify it further to the simple statement
True.That has a very simple graph; everything is filled in:
The flip side of this idea is a compound inequality like
2 < x ≤ −3.Because 2 > −3, no real number can possibly be both greater than 2 and less than −3, so this statement is simply
False.Its graph is the simplest of all, completely empty:
These continuous ranges (as I've been calling them) are intervals in the real line; writing solutions sets this way is called interval notation. (The first couple of examples aren't really interval notation.)
Compare the interval notation to the graphs; you'll see that (except for −∞ and ∞) the round parentheses and square brackets match up perfectly; they're used in the same places, facing in the same directions. This is no coincidence, of course; the notation is designed to work this way!
To solve an order inequality, in contrast, is a little trickier. You can basically use the same techniques as for solving equations, but there are complications. The first technique —replacing one expression by an equivalent expression— is exactly the same. The last technique —swapping the sides— is almost the same, but you also have to switch the direction on the inequality. (For example, 2 < x is equivalent to x > 2.) But the main technique —doing the same invertible operation to each side— is more complicated; the operation must also preserve order.
An order-preserving operation is any operation on a real number that always takes smaller numbers to smaller numbers and larger numbers to larger numbers. Notice that this is relative; adding a million may take small numbers to large numbers, but as long as one number is smaller than another, then it will still be smaller after you add a million to both of them.
Addition and subtraction always preserve order (as long as you are subtracting from the two sides of the inequality). Multiplication or division by a positive quantity is also order-preserving. But multiplication or division by a negative quantity does not preserve order! For example, start with 2 and 3; notice that 2 < 3. After you multiply these by the negative number −4, you get (−4)2 = −8 and (−4)3 = −12, but −8 > −12.
In fact, multiplication or division by a negative quantity reverses order. An order-reversing operation is any operation on a real number that always takes smaller numbers to larger numbers and larger numbers to smaller numbers. You can still use such an operation to solve order inequalities, but you must switch the direction of the inequality when you do so!
So here's a summary of techniques for solving order inequalities:
Here's an example where I have to change the direction of the inequality:
To solve compound statements, you can just treat each statement separately. However, if you have a compound inequality, like
2 < x + 1 ≤ 5,then instead of breaking this up into 2 < x + 1 and x + 1 ≤ 5, then solving each of these by doing the same order-preserving operation to both sides, you simply do that operation to all sides of the original compound inequality. So if I subtract 1 from all sides, then I get
1 < x ≤ 4;that's the answer. To check when x = 1 (which should be barely false), 1 + 1 = 2, 2 = 2, and 2 < 5, so the original inequality is just barely false. To check when x = 4 (which should be barely true), 4 + 1 = 5, 2 < 5, and 5 = 5, so the original inequality is just barely true. Finally, I'll try x = 2 (which should be quite true); 2 + 1 = 3, 2 < 3, and 3 < 5, so the original inequality is quite true. So this checks; the solution set for x is (1, 4].
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