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A L G E B R A

11

INEQUALITIES

The number line

THIS SIGN < means is less than..  This sign > means is greater than.  In each case, the sign opens towards the larger number.

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  > .

(If you are not viewing this page with Internet Explorer 6, then your browser may not be able to display the symbol , "is less than or equal to;" or , "is greater than or equal to.")

The number line

On the number line,  a < b  means:  a falls to the left of b.

 2 < 5 −1 < 2 −4 < −1 1 > −4

Problem 1.   Between each pair, place the correct sign of inequality.

Do the problem yourself first!

 3 < 7 8 > 5 −2 < 1 −6 < −2 −3 > −9 −10 > −12 0 > −9 −6 < 0

"Or" versus "and"

The following is called a compound inequality:

x > 1  and   x 5.

It is a compound sentence whose conjunction is "and."  It says that x takes on values that are greater than 1  and  less than or equal to 5.

It is within that interval that x takes its values.  The endpoint 1 is not included.  x is definitely greater than 1; we indicate that by placing a parenthesis "(".  The endpoint 5 is included; we indicate that by placing a bracket "]".

Now consider this compound inequaltiy:

x < 1  or  x > 5.

Here, the values of x are either less than 1  or  greater than 5.

It should be clear that x could not be a number that is less than 1 and greater than 5.  There is no such number.

So, when the conjunction is and,

x > 1  and   x 5.

the values of x fall inside a certain interval.  But when the conjunction is or,

x < 1  or  x > 5.

the values of x fall outside an interval.

Problem 2.   Graph these compound inequalities.

a)     x < −1   or   x 3.

b)     x −1   and   x < 3.

A continued inequality

a < x < b

That is called a continued inequality.  It means

a < x  and  x < b.

A continued inequality always implies the conjunction and.  The sense is always <  or  .

The continued inequality means:

x falls in the interval between a and b.

"a < x < b" illustrates that.

Note:  When neither end point is included, as in this example, we call that an open interval.  When both end points are included --

-- 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)   x > −3  and  x < 1.

−3 < x < 1.

b)  Graph that continued inequality.

Problem 4.   0 < x < 6.  Write that continued inequality as a compound inequality.

x > 0  and  x < 6.

Problem 5.   Write as a continued inequality:

x < 1  or  x > 5.

Not possible!  The conjunction must be and.

Problem 6.   Name four values that x might have.

a)   1 x < 3.

For example, 1,  1.2,  2.5,  2.999999999999999999999.

b)  x < −1,  or   x > 1.

For example, −2,  −3,456,987,  1.000005,  1023.

 * * *

Problem 7.   Write in symbols.

a)  x is a positive number.   x > 0

b)  x is a negative number.   x < 0

c)  x is a non-negative.   x 0.  "x is greater than or equal to 0."

Some theorems of inequalities

Algebraically, we say that  a > b  when  ab is positive.

a > b  if and only if  ab > 0.

On the basis of this definition of a > b, we can prove various theorems about inequalities.

Theorem 1.    We may add the same number to both sides of an inequality, and the sense will not change.

 If a > b, then a + c > b + c.

Note:  If c is a negative number, then the theorem implies that we may subtract the same number from both sides.  Just as with equalites, any theorem of addition is also true for subtraction.

We can prove the theorem as follows:

 a > b means a − b > 0. Therefore, upon adding and subtracting c : a − b + c − c > 0 (a + c) − (b + c) > 0  (Lesson 7) Therefore, a + c > b + c.

Which is what we wanted to prove.

 x + d > e implies x > e − d.

We could prove that by adding −d to both sides.

Note also that a theorem of inequality is true for an inequality of the opposite sense.  For, we could write:  b < a  implies   b + c < a + c.

Theorem 2.   We may multiply both sides of an inequality by the same positive number, and the sense will not change.

 If a > b,   and c > 0, then ca > cb.

For example:

 4 > −5. If we now multiply both sides by 3, for example, then 12 > −15.

The sense does not change.

The proof is similar to that of Theorem 1.  Simply apply the Rule of Signs.

 a > b implies a − b > 0. Therefore, if c is positive, then c(a − b) > 0 ca − cb > 0. This means ca > cb.

This theorem also allows us to divide both sides by the same positive number, because division is multiplication by the reciprocal.

Theorem 3.   If we multiply both sides of an inequality by the same negative number, the sense of the inequality changes.

 If a > b,   and c < 0, then ca < cb.

Here is an example:

 −2 < 5. If we now multiply both sides by −3, for example, then 6 > −15.

The sense changes.

Proof:

 a > b implies a − b > 0. Therefore, if c is negative, then c(a − b) < 0 ca − cb < 0. This means ca < cb.

Theorem 4.   If we change the signs on both sides of an inequality,
then the sense of the inequaltiy will change.

 If −a < −b, then a > b.

We can see that on the number line.

Here is an example:

 If −x < 2, then x > −2.

We may think of Theorem 3 as an instance of this theorem, because when we multiply or divide both sides by the same negative number, the signs on each side necessarily change.

Example.   If

 −5x < 30, then x > −6.

The signs on each side have changed.  Therefore the sense also changes.

We have divided each side, of course, by negative 5.

Theorem 5.   If a, b are both positive or both negative, then on taking reciprocals, the sense of the inequality changes.

 If a > b, then 1a < 1b .

In any event, since

 2 < 5, then 12 > 15

Problem 8.   Apply the theorems to complete the following with an inequality.

 a)  If x + 2 < 8 b)  If x − 2 < 8 then x < 8 − 2 then x < 8 + 2 x < 6 x < 10
 c)  If 3x < 12 d)  If −3x < 12 then x < 12 3 then x > 12−3 x < 4 x > −4

On dividing by a negative number, the sense changes.
Equivalenly, we can think of it as changing the signs on both sides.

Problem 9.    Use Theorem 2 to prove:  If a is a positive number less than 1, then a² is less than a.

For example, (½)² = ¼, which is less than ½.

 a < 1 implies a· a < 1· a That is, a² < a.

Problem 10.   Assuming that the literals all have positive values, complete the following with an inequality.

 If ab < cd then ba > dc ,  according to Theorem 5.

Problem 11.   −a > 0.  How is that possible?

a is not a number. It is a variable. It takes values which are numbers. In this case, a must have a negative value; for example, −5. Therefore, −a = −(−5)  is a positive number.

In fact, according to Theorem 4, we may change the signs on both sides if we change the sense. Therefore, on changing the signs (and since −0 = 0): a < 0.

Solving inequalities

A linear inequality has this standard form:

ax + b < c

 ax < c − b x < c − b   a

As with equations, the inequality is "solved" when positive x is isolated on the left.  (The above steps follow from Theorems 1 and 2.)

The only difference, then, between solving an inequality and solving an equation, is that when when we multiply or divide by a negative number, the sense changes.

For example,

 −2x + 5 < 11 −2x < 6 x > −3

(Alternatively, however, we could immediately make 2x positive -- by changing all the signs on both sides.  But then we must also change the sense; Theorem 4.   −2x + 5 < 11  implies  2x − 5 > −11,  and so on.)

Problem 12.   Solve each inequality for x.  Write a logical sequence of statements.

 a)  5x + 3 < 38 b)  5x − 3 < 12 5x < 35 5x < 15 x < 7 x < 3
 c)  −2x + 7 < 19 d)  −2x − 1 < −9 −2x < 12 −2x < −8 x > −6 x > 4
 e)   −3x < x − 8 f)   4x − 7 < 11x + 7 −3x − x < −8 4x − 11x < 7 + 7 −4x < −8 −7x < 14 x > 2 x > −2

Next Lesson:  Absolute Value

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