The Basics of Inequality

An inequality says one quantity is larger than another rather than equal to it. Nearly every move chapter 4 used on equations, adding the same number to both sides or scaling both sides, carries over unchanged. Exactly one does not, and finding which one, and why it fails, is the work of this lesson.

Problem

Three warehouses each store a whole number of pallets. Warehouse A stores more than warehouse B, and warehouse B stores more than warehouse C. Warehouse C stores

47

pallets. Enter the smallest number of pallets warehouse A could store.

Show a hint

Comparing $A$ with $C$ in one step gives a true statement, but a weaker one than the two given comparisons allow. Work out what $B$ must be first.
$B$ is a whole number above $47$ , so the smallest $B$ can be is $48$ . Now take the same step one more time, from $B$ to $A$ .

Show the full solution

Since

B > 47

and

B

is a whole number,

B

is at least

48

, and since

A > B

A

is at least

49

. The triple

49, 48, 47

satisfies every condition, so the smallest count is

49

. Comparing

A

with

C

directly gives the weaker

A > 47

, which is true but allows

A = 48

, a value the step through

B

rules out.

Problem

At the start of a shift the first bin holds

415

bolts and the second holds

392

. A machine then drops

168

bolts into each bin. Enter how many more bolts the first bin holds than the second after the drop.

Show a hint

Ask what happens to the gap between the two bins, not to either count on its own.
The two counts become $415 + 168$ and $392 + 168$ . Subtract one from the other.

Show the full solution

After the drop the counts are

415 + 168 = 583

and

392 + 168 = 560

, and

583 - 560 = 23

, the same gap as

415 - 392

before it. Both counts go up by the same

168

, so the gap is untouched. Adding

168

to the gap instead gives

191

, which would be right only if the bolts all went into the first bin.

Problem

A hiking club logged two trips. The first covered more than

34

kilometers and the second covered more than

57

kilometers, and neither distance has to be a whole number. Enter the largest number of kilometers that the two trips together are guaranteed to exceed.

Show a hint

Each trip has its own number to beat. The two comparisons point the same way, so they can be combined into one comparison about the total.
Write each trip as a comparison, then add the two left sides together and the two right sides together. Test the result on a sample pair such as $34.2$ and $57.3$ .

Show the full solution

The first distance is more than

34

and the second is more than

57

, and two comparisons pointing the same way may be added, so the total is more than

34 + 57 = 91

. Writing the two distances as

f

and

s

, both

f - 34

and

s - 57

are positive, so

(f + s) - 91

is positive as well. Distances just above

34

and

57

give a total just above

91

, so no larger number works.

Problem

A swim club beat a tennis club in wins two years running,

41

12

in the first year and

44

39

in the second. Subtracting the first comparison from the second, the way two equations are subtracted, gives the conclusion that the swim club's gain in wins was the larger one. Enter how many more wins the tennis club gained from the first year to the second than the swim club gained.

Show a hint

Both stated comparisons are true, so if subtracting one from the other were a legal move, its conclusion would have to be true as well. Work out each club's gain on its own.
The swim club went from $41$ to $44$ and the tennis club went from $12$ to $39$ . Subtract within each club, then compare the two gains.

Show the full solution

The swim club gained

44 - 41 = 3

wins and the tennis club gained

39 - 12 = 27

, so the tennis club's gain is larger by

27 - 3 = 24

. The subtraction gives

3 > 27

, which is false even though both starting comparisons are true, so one inequality may not be subtracted from another.

Two numbers sit to the right of zero with

a

nearer to it, so

a < b

. Multiplying both by the same negative number sends each to the other side of zero, and they cross on the way, so the image of

a

finishes to the RIGHT of the image of

b

. The crossing is the whole rule: the direction of the comparison reverses.

Adding shifts both sides by the same amount and leaves the difference alone. Multiplying does not, since $c x - c y = c (x - y)$ scales that difference by $c$ . Scaling a positive number by a positive $c$ keeps it positive, and scaling by a negative $c$ makes it negative, so multiplying splits into two cases. The next two problems take one of each.

Problem

A press stamped

t

parts on Monday, with

t > 185

. On Tuesday it stamped

3

times Monday's number and then

40

more. Apply those same two operations to both sides of

t > 185

. The result says Tuesday's count is greater than a certain number. Enter that number.

Show a hint

Tuesday's count is $3 t + 40$ . Build it up from $t > 185$ one operation at a time.
Multiplying both sides by $3$ is multiplying by a positive number, so the direction holds. Now add $40$ to both sides.

Show the full solution

Multiplying

t > 185

by the positive number

3

keeps the direction, giving

3 t > 555

, and adding

40

to both sides keeps it as well, so Tuesday's count satisfies

3 t + 40 > 595

. The direction holds through both steps because

3

is positive and adding

40

changes both sides by the same amount.

Problem

A student divides both sides of

- 7 x > 91

- 7

, leaves the inequality symbol pointing the same way, and concludes that the smallest integer

x

can be is

- 12

. Enter the largest integer

x

can actually be.

Show a hint

Test the student's answer against the original statement before doing anything else. Put $x = - 12$ into $- 7 x$ and compare the result with $91$ .
Dividing both sides by $- 7$ reverses the direction, so the allowed values sit below $- 13$ rather than above it.

Show the full solution

Dividing

- 7 x > 91

- 7

reverses the direction to

x < - 13

, so the largest integer available is

- 14

. Checking,

- 7 (- 14) = 98 > 91

while

- 7 (- 13) = 91

is not above

91

. The student's

- 12

fails outright, since

- 7 (- 12) = 84

sits below

91

Choosing between those two cases takes the sign of the multiplier. With a number written out the sign is there to read. With a letter it is not, and there are three cases rather than two, since multiplying by zero gives $0$ on both sides. The next problem asks what that leaves.

Problem

It is given that

a > b

, and

k

is a real number whose sign is not stated. Enter one word, greater, less, equal or unknown, for how

k a

compares with

k b

Show a hint

Write $k a - k b = k (a - b)$ . Since $a - b$ is positive, the sign of $k a - k b$ matches the sign of $k$ .
Fix $a = 9$ and $b = 2$ , then try $k = 4$ , $k = - 4$ and $k = 0$ in turn, comparing $k a$ with $k b$ each time.

Show the full solution

Here

k a - k b = k (a - b)

and

a - b

is positive, so the sign of that difference is the sign of

k

. With

a = 9

and

b = 2

k = 4

gives

36 > 8

k = - 4

gives

- 36 < - 8

, and

k = 0

gives

0 = 0

, so the comparison is

unknown

. All three cases are possible, so an inequality cannot be multiplied by a letter whose sign is not known.

Problem

It is given that

m > n

, with nothing said about the signs of

m

and

n

, and

t

is a real number. Statement 1 is

m - 9 > n - 9

. Statement 2 is

- 2 m > - 2 n

. Statement 3 is

\frac{m}{5} > \frac{n}{5}

. Statement 4 is

m t > n t

. Statement 5 is

m + n > 2 n

. Statement 6 is

m - n > 0

. Enter how many of the six must be true.

Show a hint

Each statement is one move applied to both sides of $m > n$ . Name the move first, then ask whether that move preserves the direction.
Statement 5 is $m > n$ with $n$ added to both sides, and statement 6 is what $m > n$ means written out. Check the other four against what each operation does to the direction.

Show the full solution

Statement 1 subtracts

9

from both sides, statement 3 divides by the positive number

5

, statement 5 adds

n

to both sides, and statement 6 restates

m > n

, so those four hold. Statement 2 multiplies by

- 2

without reversing the direction and statement 4 multiplies by a letter whose sign is not stated, so the count is

4

. The signs of

m

and

n

never matter here. Only the sign of each multiplier does.

A square is never negative, whatever real number is put into it. Squaring means multiplying a number by itself, so the two factors are the same number and have the same sign. A positive times a positive is positive, a negative times a negative is positive, and $0 \times 0$ is $0$ . So $x^{2}$ is never negative, whatever real $x$ is.

Problem

For every real number

x

, the quantity

x^{2} + 82 - 18 x

is equal to

(x - 9)^{2} + 1

. How does

x^{2} + 82

compare with

18 x

at every real

x

? Enter one word, greater, less, equal or unknown.

Show a hint

Deciding which is larger is the same as deciding the sign of the difference $x^{2} + 82 - 18 x$ , and the problem already gives that difference in a second form.
$(x - 9)^{2}$ is the square of a real number, so it is never negative. Add $1$ and ask how small the whole thing can get.

Show the full solution

The difference

x^{2} + 82 - 18 x

equals

(x - 9)^{2} + 1

, and the square of a real number is never negative, so that difference is at least

1

at every real

x

. Since the difference is always positive,

x^{2} + 82

greater

. Rearranging the two quantities is not enough on its own. You need

x^{2} \geq 0

, applied to

x - 9

Problem

Each of these four pairs has

x > y

. Pair 1 is

x = 9

y = 4

. Pair 2 is

x = - 2

y = - 5

. Pair 3 is

x = 7

y = - 1

. Pair 4 is

x = - 4

y = - 8

. Enter how many of the four pairs satisfy

\frac{1}{x} < \frac{1}{y}

Show a hint

Work out both reciprocals in each pair and compare them. A negative number has a negative reciprocal, and of two negative numbers the one further from zero is the smaller.
Watch for the pair where $x$ is positive and $y$ is negative. There $\frac{1}{x}$ is positive while $\frac{1}{y}$ is negative, and no positive number is less than a negative one.

Show the full solution

Pair 1 gives

\frac{1}{9} < \frac{1}{4}

, pair 2 gives

- \frac{1}{2} < - \frac{1}{5}

and pair 4 gives

- \frac{1}{4} < - \frac{1}{8}

, while pair 3 gives

\frac{1}{7} > - 1

, so the count is

3

. The three that flip have

x

and

y

on the same side of zero, and pair 3 does not.

Every rule in this lesson is one question about a sign. Adding or subtracting the same number never changes a direction. Multiplying or dividing changes it exactly when the multiplier is negative, and leaves it undetermined when the multiplier is a letter of unstated sign. The last two problems run several of these moves in one chain.

Problem

Two numbers satisfy

p > 11

and

q > - 3

, and nothing else is known about either one. Enter the smallest number that

4 - 6 (p + q)

must stay below.

Show a hint

Two inequalities pointing the same way may be added, so pin down what $p + q$ must exceed before touching anything else.
That gives $p + q > 8$ . Multiplying both sides by $- 6$ is multiplying by a negative, so settle the direction before adding $4$ .

Show the full solution

Adding

p > 11

and

q > - 3

gives

p + q > 8

, multiplying by

- 6

reverses the direction to

- 6 (p + q) < - 48

, and adding

4

to both sides gives

4 - 6 (p + q) < - 44

. Skipping the reversal would give

4 - 6 (p + q) > - 44

, which is false at

p = 12

and

q = - 2

, since those give

- 56

Problem

Two numbers satisfy

15 - 8 s > 15 - 8 t

, and their product

s t

is positive. Neither one is known to be positive on its own. Enter one word, greater, less, equal or unknown, for how

\frac{1}{t}

compares with

\frac{1}{s}

Show a hint

Two moves. First turn $15 - 8 s > 15 - 8 t$ into a comparison between $s$ and $t$ alone, watching the step that reverses the direction.
That gives $s < t$ . The reciprocal rule needs the two numbers to share a sign, and $s t > 0$ says exactly that even though neither sign is known on its own.

Show the full solution

Subtracting

15

from both sides gives

- 8 s > - 8 t

, and dividing by

- 8

reverses the direction to

s < t

. Since

s t > 0

the two share a sign, so the reciprocal rule applies and

\frac{1}{t}

compared with

\frac{1}{s}

less

. Both sign cases check out,

s = 1, t = 2

giving

\frac{1}{2} < 1

and

s = - 2, t = - 1

giving

- 1 < - \frac{1}{2}

. Drop

s t > 0

and the answer is unknown, since

s = - 1, t = 2

gives

\frac{1}{t} > \frac{1}{s}

Every rule here says what one move does to a comparison already in hand. Lesson 8.2, Which Is Bigger, uses them in the other direction. It starts with two specific quantities too awkward to compute outright and looks for the move that settles which one is larger.

Practice these ideas

Practice

w

is an integer and

w - 18 > 45

. Enter the smallest value

w

can be.

Show the solution

Adding

18

to both sides leaves the direction unchanged, giving

w > 63

, so the smallest integer available is

64

. Checking,

64 - 18 = 46 > 45

while

63 - 18 = 45

is not above

45

Practice

Integers

a

b

, and

c

satisfy

a > b

and

b > c

, with

c = 61

. Enter the smallest possible value of

a

Show the solution

Since

b > 61

and

b

is an integer,

b

is at least

62

, and

a > b

makes

a

at least

63

. The triple

63, 62, 61

meets every condition, so the answer is

63

. Chaining to

a > 61

first is true but weaker, since

a = 62

would need an integer

b

with

61 < b < 62

, and there is none.

Practice

A number

g

satisfies

g > 16

. Enter the largest number that

7 g

must exceed.

Show the solution

Multiplying

g > 16

by the positive number

7

leaves the direction unchanged, so

7 g > 112

. Nothing larger works, since

g

can sit just above

16

, which puts

7 g

just above

112

Practice

Real numbers

m

and

n

satisfy

m > 30

and

n > 17

. Enter the largest number that

m + n

must exceed.

Show the solution

Two inequalities pointing the same way may be added, so

m + n > 30 + 17 = 47

. Nothing larger works, since taking

m

and

n

just above

30

and

17

makes

m + n

as close to

47

as we like.

Practice

q

is an integer and

\frac{q}{8} > 13

. Enter the smallest value

q

can be.

Show the solution

Multiplying both sides by the positive number

8

keeps the direction, giving

q > 104

, so the smallest integer is

105

. Checking,

\frac{105}{8} = 13.125

is above

13

while

\frac{104}{8} = 13

is not.

Practice

A number

k

satisfies

k > 3

, and nothing else is known about it. Enter the smallest number that

- 8 k

must stay below.

Show the solution

Multiplying

k > 3

- 8

reverses the direction, so

- 8 k < - 24

. Keeping

>

would give the false claim

- 8 k > - 24

, since

k = 4

gives

- 32

, and

- 32

is not above

- 24

Practice

n

is an integer and

- 5 n > 37

. Enter the largest value

n

can be.

Show the solution

Dividing

- 5 n > 37

- 5

reverses the direction to

n < - 7.4

, so the largest integer is

- 8

. Checking,

- 5 (- 8) = 40 > 37

while

- 5 (- 7) = 35

is not above

37

Practice

Both

20 > 3

and

b > 1

are true, and

b

is an integer. Subtracting the second comparison from the first would claim

20 - b > 3 - 1

. Enter the smallest integer

b

that makes that claim false.

Show the solution

The claim reads

20 - b > 2

, which holds only while

b < 18

. At

b = 18

it reads

2 > 2

, which is false, so the answer is

18

. Both starting comparisons still hold at

b = 18

, so subtracting one comparison from another is not a legal move.

Practice

It is given that

a > b

, and

t = 0

. Enter one word, greater, less, equal or unknown, for how

t a

compares with

t b

Show the solution

With

t = 0

, both

t a

and

t b

are

0

, so the two are

equal

. Zero is neither positive nor negative, so neither multiplication rule applies, and the given

a > b

makes no difference to the comparison.

Practice

It is given that

u > v

. Statement 1 is

u + 3 > v + 3

. Statement 2 is

4 u > 4 v

. Statement 3 is

\frac{u}{- 2} > \frac{v}{- 2}

. Statement 4 is

u - v > 0

. Statement 5 is

- u < - v

. Statement 6 is

u + v > 2 v

. Enter how many of the six must be true.

Show the solution

Dividing both sides by

- 2

reverses the direction, so statement 3 is the only one that fails and the count is

5

. Statements 1, 4 and 6 come from adding or subtracting the same quantity on both sides, statement 2 from multiplying by the positive

4

, and statement 5 from multiplying by

- 1

with the direction reversed.

Practice

It is given that

x > y

. For how many of

k = - 3

k = 0

and

k = 5

k x > k y

guaranteed? Enter that count.

Show the solution

The sign of

k x - k y = k (x - y)

is the sign of

k

, since

x - y

is positive. So

k = 5

gives

k x > k y

k = - 3

gives

k x < k y

, and

k = 0

gives

k x = k y

, leaving

1

. Three values of

k

and three different outcomes, so you may only multiply both sides by a letter when you know its sign.

Practice

A real number

c

satisfies

c^{2} \leq 0

. Enter the value of

(c + 5) (c + 8)

Show the solution

Since

c^{2} \geq 0

for every real

c

, the condition

c^{2} \leq 0

can hold only when

c^{2} = 0

, so

c = 0

and

(0 + 5) (0 + 8) = 40

. The trivial inequality pins

c

to a single value, so nothing else about

c

has to be found.

Practice

For every real number

x

, the quantity

x^{2} + 121 - 22 x

is equal to

(x - 11)^{2}

. Enter the value of

x^{2} + 121

at the one real number

x

where it equals

22 x

Show the solution

x^{2} + 121

equals

22 x

exactly when

x^{2} + 121 - 22 x = 0

, which is

(x - 11)^{2} = 0

, so

x = 11

. Then

x^{2} + 121 = 121 + 121 = 242

, and

22 (11) = 242

agrees. A square is never negative, so

(x - 11)^{2}

0

at one value of

x

and positive everywhere else, which is why

x^{2} + 121

is never below

22 x

Practice

The numbers

x = 15

and

y = 6

are both positive, and

x > y

. Enter the value of

\frac{1}{y} - \frac{1}{x}

as a fraction in lowest terms.

Show the solution

With a common denominator of

30

\frac{1}{6} - \frac{1}{15} = \frac{5}{30} - \frac{2}{30} = \frac{3}{30} = \frac{1}{10}

. The reciprocal rule says that when

x > y

and both are positive,

\frac{1}{x} < \frac{1}{y}

, so this difference had to come out positive.

Practice

The numbers

x = 10

and

y = - 4

satisfy

x > y

, so the reciprocal rule would claim

\frac{1}{x} < \frac{1}{y}

. Enter the value of

\frac{1}{x} - \frac{1}{y}

as a fraction in lowest terms.

Show the solution

The reciprocals are

\frac{1}{10}

and

- \frac{1}{4}

, so

\frac{1}{10} - (- \frac{1}{4}) = \frac{2}{20} + \frac{5}{20} = \frac{7}{20}

. The difference is positive, which says

\frac{1}{x} > \frac{1}{y}

, the opposite of the claim. These two numbers sit on opposite sides of zero, and the rule needs them on the same side.

Practice

A number

z

satisfies

z < 6

, and

y = 9 - 4 z

. Enter the largest number that

y

must exceed.

Show the solution

Multiplying

z < 6

- 4

reverses the direction to

- 4 z > - 24

, and adding

9

to both sides gives

9 - 4 z > - 15

. Skipping the reversal gives

y < - 15

instead, which is false at

z = 0

, where

y = 9

Practice

An integer

m

satisfies

\frac{1}{m} < \frac{1}{7}

and

m > 0

. Enter the smallest value

m

can be.

Show the solution

Both numbers are positive, so taking reciprocals reverses the comparison and

\frac{1}{m} < \frac{1}{7}

becomes

m > 7

, making the smallest integer

8

. Checking,

\frac{1}{8} < \frac{1}{7}

. The condition

m > 0

matters, since

m = - 3

also satisfies

\frac{1}{m} < \frac{1}{7}

but is not greater than

7