11 ## INEQUALITIESTHIS SIGN < means For example, 2 < 5 ("2 is less than 5"). Equivalently, 5 > 2 ("5 is greater than 2"). These are the two senses of an inequality: < and > . The number line On the number line,
Problem 1. Between each pair, place the correct sign of inequality. To see the answer, pass your mouse over the colored area.
"Or" versus "and" The following is called a compound inequality:
It is a compound sentence whose conjunction is "and." It says that It is within that interval that Now consider this compound inequaltiy:
Here, the values of It should be clear that So, when the conjunction is
the values of
the values of Problem 2. Graph these compound inequalities. a)
b)
A continued inequality
That is called a continued inequality. It means
A continued inequality always implies the conjunction The continued inequality means:
"
-- we call that a closed interval. (Otherwise, like the proverbial glass, the interval is half-open, or half-closed.) Problem 3. Write as a continued inequality. a)
−3 < b) Graph that continued inequality.
Problem 4. 0 <
Problem 5. Write as a continued inequality:
Not possible! The conjunction must be
Problem 6. Name four values that a) 1 ≤ For example, 1, 1.2, 2.5, 2.999999999999999999999. b)
For example, −2, −3,456,987, 1.000005, 10
Problem 7. Write in symbols. a) b) c) Some theorems of inequalities To prove a statment whose predicate is "is greater than," we must have a definition of "is greater than." We shall adopt the following. We shall define "
On the basis of this definition, we can prove various theorems about inequalities.
Theorem 1.
It follows, from this Theorem, that we may transpose.
We could prove that by adding − We can prove Theorem 1 as follows:
Which is what we wanted to prove. Note that any theorem of inequality is true for an inequality of the opposite sense. For, we could write:
Theorem 2.
For example:
The sense does not change. The proof is similar to that of Theorem 1. Simply apply the Rule of Signs.
This theorem also allows us to
Theorem 3.
Here is an example:
The sense changes. (As in Theorem 2, this one also implies that when we Proof:
Theorem 4.
We can We could prove it simply by transposing − For example, since
We may think of Theorem 3 as an instance of Theorem 4, because when we multiply or divide both sides by the same negative number, the Example 1. If
The signs on each side have changed. Therefore the sense also must change. The signs changed because we divided each side by negative 5.
Theorem 5.
We can prove that by dividing both sides by In any event, since
Problem 8. Apply the theorems to complete the following with an inequality.
On dividing by a
Problem 9. Use Theorem 2 to prove: If For example, (½)
Problem 10. Assuming that the literals all have positive values, complete the following with an inequality.
Problem 11. −
That is, since −0 = 0,
Solving inequalities A linear inequality has this standard form:
When
As with equations, the inequality is "solved" when positive The only difference between solving an inequality and solving an equation, is the following: When when we multiply or divide by a Equivalently, when the signs on both sides change, then the sense also must change (Theorem 4), as in Example 1. Example 2.
On going to the last line, the signs on both sides changed. Therefore, the sense also changes. The signs changed, of course, because we divided both sides by Alternatively, we could immediately make 2 −2 and so on.
Problem 12. Solve each inequality for
Problem 13. In each of the following, what can you conclude about the a)
Lesson 4: The Rule of Signs. b)
c)
Either
Same as a).
Same as b).
The numerator must be 0.
Example 3. Solve this inequality for
1) or 2) Now, 1) implies
Which numbers are these that are both greater than 3 Clearly, any number greater than 3 will also be greater than −5. Therefore, 1) has the solution
Next, 2) implies
Which numbers are these that are both less than 3 Clearly, any number less than −5 will also be less than 3. Therefore, 2) has the solution
The solution, therefore, is
Please make a donation to keep TheMathPage online. Copyright © 2014 Lawrence Spector Questions or comments? E-mail: themathpage@nyc.rr.com |