Which Is Bigger

Every problem in this lesson gives two quantities and asks which one is larger. The rules from 8.1 are the tools, and the work is choosing which one to reach for. Several of these pairs are built so computing both is slow, and one is out of reach by hand, so the route has to settle the comparison without producing either value.

Problem

Two sums are

A = \frac{5}{8} + \frac{7}{9}

and

B = \frac{5}{9} + \frac{7}{8}

. Do not compute either sum. Work out

A - B

instead and read its sign. Enter

A

B

, whichever is bigger.

Show a hint

Write $A - B$ as one string of four fractions, then group the two with denominator $8$ and the two with denominator $9$ before anything else.
Grouping the eighths gives $\frac{5}{8} - \frac{7}{8} = - \frac{1}{4}$ and grouping the ninths gives $\frac{7}{9} - \frac{5}{9} = \frac{2}{9}$ . Add those two results and read the sign.

Show the full solution

Group the eighths and the ninths.

A - B = (\frac{5}{8} - \frac{7}{8}) + (\frac{7}{9} - \frac{5}{9}) = - \frac{1}{4} + \frac{2}{9} = - \frac{9}{36} + \frac{8}{36} = - \frac{1}{36}

, which is negative, so the bigger sum is

B

. Both sums are built from the numerators

5

and

7

over the denominators

8

and

9

with the pairing swapped, so they differ by only

\frac{1}{36}

Problem

Let

A = 987654 \times 987656

and

B = 98765 5^{2}

. Both are twelve-digit numbers, so multiplying them out by hand is not the route. Enter the value of

B - A

Show a hint

There is no need to compute either number. Rewrite $987656$ as $987655 + 1$ and distribute, so that $A$ becomes $987654 \times 987655$ plus a leftover.
Split $B = 987655 \times 987655$ the same way, writing one factor as $987654 + 1$ . Both quantities then contain $987654 \times 987655$ , and only the leftovers differ.

Show the full solution

Since

987656 = 987655 + 1

, distributing gives

A = 987654 \times 987655 + 987654

, and writing

B = (987654 + 1) \times 987655

gives

B = 987654 \times 987655 + 987655

. The first piece is the same in both, so

B - A = 987655 - 987654 = 1

. Splitting each quantity so that a common piece appears turns a twelve-digit subtraction into a one-digit one.

Problem

Let

A = \frac{12}{49}

and

B = \frac{8}{33}

. The denominators share no factor, so a common denominator would be a large one. Enter A or B, whichever is bigger.

Show a hint

$12$ and $8$ both divide $24$ , so both fractions can be written with the same numerator.
Multiply $A$ top and bottom by $2$ and $B$ top and bottom by $3$ . Then only the denominators differ.

Show the full solution

Rewrite both with numerator

24

. Then

A = \frac{24}{98}

and

B = \frac{24}{99}

, and

98 < 99

, so the bigger one is

A

. Cutting the same

24

into

98

pieces gives bigger pieces than cutting it into

99

, so the larger denominator gives the smaller number.

Problem

Let

A = 9^{15}

and

B = 2 7^{10}

. Working out either power is a long job, and neither one has to be done. Enter A or B, whichever is bigger, or the word equal if neither is.

Show a hint

$9 = 3^{2}$ and $27 = 3^{3}$ , so both can be written as powers of $3$ .
Raising a power to a power multiplies the exponents, the rule from 3.1. Get both exponents, then compare them.

Show the full solution

Both bases are powers of

3

. Then

A = (3^{2})^{15} = 3^{30}

and

B = (3^{3})^{10} = 3^{30}

, so the answer is

equal

B

has the bigger base and

A

the bigger exponent, so neither one of those settles it. Written with base

3

, only the exponents are left to compare, and they match.

The two brackets are the only comparisons made. Each one is a single comparison between

m

and an outer point, and the two together are enough for

a < b

, with no direct comparison between

a

and

b

anywhere. Choosing

m

is the work, since it has to sit above one and below the other.

Neither number of a pair has to be measured against the other directly. Transitivity from 8.1 says that if $a < m$ and $m < b$ then $a < b$ , so choosing one number in the middle replaces a single hard comparison with two easy ones.

Problem

Let

A = \frac{138}{137}

and

B = \frac{265}{266}

. A common denominator here is

36442

. Enter A or B, whichever is bigger.

Show a hint

Building that common denominator is the long way round. Look instead for one simple number that separates the two.
A positive fraction sits above $1$ when its numerator is the larger of the two numbers, and below $1$ when its denominator is. Check $138$ against $137$ , then $265$ against $266$ .

Show the full solution

Compare each with

1

. The fraction

\frac{138}{137}

has the larger number on top, so it sits above

1

, and

\frac{265}{266}

has the larger number underneath, so it sits below

1

. That gives

A > 1 > B

, and the bigger one is

A

. The common denominator

36442

never had to be built.

Problem

Let

A = \frac{17}{40}

and

B = \frac{22}{51}

. Both sit below

\frac{1}{2}

and above

\frac{2}{5}

, so neither of those settles the pair. Enter A or B, whichever is bigger, or the word equal if neither is.

Show a hint

A middle number settles the pair only when the two land on opposite sides of it, so the one you want is a fraction strictly inside the gap the statement names.
Sevenths land in that gap, since $\frac{2}{5} < \frac{3}{7} < \frac{1}{2}$ . Compare each of the two with $\frac{3}{7}$ by cross-multiplying.

Show the full solution

Try sevenths. Since

17 \times 7 = 119

and

3 \times 40 = 120

A < \frac{3}{7}

, and since

22 \times 7 = 154

and

3 \times 51 = 153

B > \frac{3}{7}

. So

A < \frac{3}{7} < B

and the bigger one is

B

. Halves and fifths fail here because they leave both numbers on the same side.

Two numbers with awkward digits can have reciprocals that are easy to read. The bigger of two numbers of the same sign has the smaller reciprocal, so comparing the reciprocals settles the original pair, with the result read backwards.

Problem

Let

A = \frac{31}{218}

and

B = \frac{23}{162}

. Enter

A

B

, whichever is bigger.

Show a hint

Computing both decimals is the slow route here. Both numbers are positive, so both can be flipped, and of two positive numbers the bigger one has the smaller reciprocal.
Divide $218$ by $31$ , then $162$ by $23$ . Each division has remainder $1$ , so each reciprocal is $7$ plus a unit fraction.

Show the full solution

Both are positive, so flip both.

\frac{1}{A} = \frac{218}{31} = 7 + \frac{1}{31}

and

\frac{1}{B} = \frac{162}{23} = 7 + \frac{1}{23}

. Since

\frac{1}{31} < \frac{1}{23}

\frac{1}{A} < \frac{1}{B}

, so the bigger number is

A

. Flipping replaced a hard pair with an easy one, but the comparison reverses, since of two positive numbers the bigger one has the smaller reciprocal.

Problem

Two numbers

A

and

B

satisfy

\frac{1}{A} = \frac{4}{9} - \frac{7}{12}

and

\frac{1}{B} = \frac{5}{6} - \frac{3}{8}

. Enter the letter of the bigger number,

A

B

Show a hint

Work out $\frac{1}{A}$ and $\frac{1}{B}$ as single fractions first, then look at the sign of each.
One of the two reciprocals is negative, so the reciprocal rule does not apply here. A number and its reciprocal always have the same sign.

Show the full solution

\frac{1}{A} = \frac{16}{36} - \frac{21}{36} = - \frac{5}{36}

and

\frac{1}{B} = \frac{20}{24} - \frac{9}{24} = \frac{11}{24}

, so

A

is negative and

B

is positive, which makes the bigger number

B

. The reciprocal rule only applies when both numbers are on the same side of zero. Without that check,

\frac{1}{A} < \frac{1}{B}

would be reversed into the false claim

A > B

A second way to find the answer is to assume it. Suppose one of the two is bigger, then apply the same move to both sides until the statement is plainly true or plainly false. A false ending means the other one is bigger. That finds the answer, and whether it also proves it is a separate question.

Problem

Let

A = - (\sqrt{5} + \sqrt{11})

and

B = - (\sqrt{2} + \sqrt{14})

. Assume one is bigger, square both sides, and check whether that step can be run backwards. Enter A or B, whichever is bigger, or the word equal if neither is.

Show a hint

Both quantities are negative, so squaring does not preserve the direction of a comparison. Work out which square is larger first, then decide what that says.
$A^{2} = 16 + 2 \sqrt{55}$ and $B^{2} = 16 + 2 \sqrt{28}$ , so $A^{2} > B^{2}$ . For two negative numbers, the one with the larger square is the smaller number.

Show the full solution

Squaring gives

A^{2} = 16 + 2 \sqrt{55}

and

B^{2} = 16 + 2 \sqrt{28}

, and

55 > 28

, so

A^{2} > B^{2}

. Both

A

and

B

are negative, and for negative numbers a larger square means a smaller number, so

A < B

and the bigger one is

B

. Squaring is only reversible when both sides are known to be positive, and here running it backwards without that check returns

A

Problem

Let

A = (2 m + 7)^{2} + 3

and

B = 3 - (m - 1)^{2}

, where

m

is any real number. The same one of the two is bigger at every real

m

. Enter A or B, whichever it is.

Show a hint

Subtract and watch the two $3$ s cancel. What is left is built only out of squares.
From 8.1, a square of a real number is never negative. These two squares are zero at different values of $m$ , so they are never zero together.

Show the full solution

Subtract.

A - B = (2 m + 7)^{2} + 3 - 3 + (m - 1)^{2} = (2 m + 7)^{2} + (m - 1)^{2}

, which is zero only if

2 m + 7

and

m - 1

are both zero, and no single

m

is both

- \frac{7}{2}

and

1

. So the difference is positive at every real

m

and the bigger one is

A

. Testing one value of

m

would settle only that value, while the sign of the difference settles them all.

Four routes are available, subtracting, rewriting both into a common form, going through a third number, and flipping to reciprocals. Read the shape of the pair first. Nearly identical pieces suit subtracting, matching numerators or bases suit a rewrite, two numbers near a round value suit a middle number, and small fractions suit reciprocals.

Problem

Let

A = \frac{13}{30}

and

B = \frac{15}{34}

. Two routes from this lesson settle this pair. Enter A or B, whichever is bigger.

Show a hint

One route subtracts over the common denominator $1020$ . The other flips both and compares the reciprocals.
$\frac{30}{13}$ and $\frac{34}{15}$ are each $2$ plus a fraction, and those two fractions have the same numerator.

Show the full solution

Both are positive, so the bigger number is the one with the smaller reciprocal.

\frac{1}{A} = \frac{30}{13} = 2 + \frac{4}{13}

and

\frac{1}{B} = \frac{34}{15} = 2 + \frac{4}{15}

, and those leftovers share a numerator, so

\frac{4}{13} > \frac{4}{15}

and

\frac{1}{A} > \frac{1}{B}

. The smaller reciprocal is

\frac{1}{B}

, so the bigger number is

B

. Subtracting works too, since over the denominator

1020

they are

\frac{442}{1020}

and

\frac{450}{1020}

Problem

Let

A = 7^{18}

and

B = 1 8^{12}

. Enter A or B, whichever is bigger, or the word equal if neither is.

Show a hint

Neither base is a power of the other, so a common base is out. Look at the exponents $18$ and $12$ , which share the factor $6$ .
Write $7^{18}$ as $(7^{3})^{6}$ and $1 8^{12}$ as $(1 8^{2})^{6}$ . Both are sixth powers now, so only the bases $7^{3}$ and $1 8^{2}$ have to be compared.

Show the full solution

Both exponents are multiples of

6

, so write each as a sixth power. Then

A = (7^{3})^{6} = 34 3^{6}

and

B = (1 8^{2})^{6} = 32 4^{6}

, and

343 > 324

, so the bigger one is

A

B

has the bigger base and

A

has the bigger exponent, so neither of those on its own settles the pair.

Every comparison here was between two numbers already fixed. Lesson 8.3, Solving Linear Inequalities, puts a letter on one side and asks which values of that letter make the comparison true. The rules are these same ones, the reversal on multiplying by a negative included.

Practice these ideas

Practice

Let

A = \frac{11}{26}

and

B = \frac{11}{29}

. Enter A or B, whichever is bigger.

Show the solution

Both fractions have numerator

11

, and

26 < 29

, so

\frac{11}{26}

is the bigger number, which makes the answer

A

. Cutting

11

into more equal parts makes each part smaller, so with the numerator fixed the bigger denominator gives the smaller number.

Practice

Let

A = \frac{5}{9} + \frac{2}{11}

and

B = \frac{5}{9} + \frac{3}{14}

. Enter

A

B

, whichever is bigger.

Show the solution

Both carry

\frac{5}{9}

, so subtracting one from the other removes it.

A - B = \frac{2}{11} - \frac{3}{14} = \frac{28 - 33}{154} = - \frac{5}{154}

, which is negative, so the bigger one is

B

. Neither sum had to be computed.

Practice

Let

A = 3^{15}

and

B = 9^{7} \cdot 3

. Enter A or B, whichever is bigger, or the word equal if neither is.

Show the solution

Write both as powers of

3

B = 9^{7} \cdot 3 = (3^{2})^{7} \cdot 3 = 3^{14} \cdot 3 = 3^{15}

, the same power as

A

, so the answer is

equal

Practice

Let

A = \frac{23}{22}

and

B = \frac{21}{24}

. Enter A or B, whichever is bigger.

Show the solution

Compare each with

1

. Since

23 > 22

A > 1

, and since

21 < 24

B < 1

, so

A > 1 > B

and the bigger one is

A

Practice

Let

A = 46 \times 54

and

B = 5 0^{2}

. Enter A or B, whichever is bigger, or the word equal if neither is.

Show the solution

Splitting each product the same way gives

46 \times 54 = 46 \times 50 + 184

and

5 0^{2} = 46 \times 50 + 200

, so the bigger one is

B

. The shared

46 \times 50

never has to be worked out, only

184

against

200

Practice

Let

A = - \frac{1}{6}

and

B = - \frac{1}{13}

. Enter A or B, whichever is bigger, or the word unknown if it cannot be decided.

Show the solution

Both are negative, and

\frac{1}{6} > \frac{1}{13}

, so negating reverses that to

- \frac{1}{6} < - \frac{1}{13}

and the bigger one is

B

. Multiplying by

- 1

is the reversal from 8.1.

Practice

Let

v

be any real number, with

A = (v - 8)^{2} + 3

and

B = 2

. Enter A or B, whichever is bigger.

Show the solution

A square is never negative, so

(v - 8)^{2} + 3

is at least

3

, and

3

is more than

2

, which makes the bigger quantity

A

. Checking

A

at its smallest is enough here, since a quantity whose smallest value is above

B

is above

B

at every

v

Practice

Let

A = \frac{3}{50}

and

B = \frac{4}{67}

. Enter the letter

A

B

, whichever is bigger.

Show the solution

Both are positive, so the bigger number has the smaller reciprocal.

\frac{1}{A} = \frac{50}{3} = 16 + \frac{2}{3}

and

\frac{1}{B} = \frac{67}{4} = 16 + \frac{3}{4}

, and

\frac{2}{3} < \frac{3}{4}

, so

\frac{1}{A} < \frac{1}{B}

and the bigger one is

A

. Cross-multiplying settles it too, since

3 \cdot 67 = 201

is more than

4 \cdot 50 = 200

Practice

Let

A = \frac{337}{676}

and

B = \frac{449}{897}

. Enter A or B, whichever is bigger.

Show the solution

Compare each with

\frac{1}{2}

. Doubling the numerators gives

674 < 676

and

898 > 897

, so

A < \frac{1}{2} < B

and the bigger one is

B

Practice

Let

A = \sqrt{10} + \sqrt{14}

and

B = \sqrt{2} + \sqrt{22}

. Enter A or B, whichever is bigger, or the word equal if neither is.

Show the solution

Both are positive, so squaring preserves the comparison.

A^{2} = 10 + 14 + 2 \sqrt{140} = 24 + 2 \sqrt{140}

and

B^{2} = 2 + 22 + 2 \sqrt{44} = 24 + 2 \sqrt{44}

, and

140 > 44

, so

A^{2} > B^{2}

and the bigger one is

A

. The two sums have the same total under the radicals, so the pair with the closer parts is the larger sum.

Practice

Enter how many of these five statements are true.

(1) \frac{5}{12} > \frac{5}{13} (2) 2^{10} > 1 0^{3} (3) - \frac{3}{4} > - \frac{2}{3}

(4) \frac{1}{- 11} > \frac{1}{- 6} (5) \frac{7}{15} < \frac{1}{2}

Show the solution

Only statement 3 fails, since

\frac{3}{4} > \frac{2}{3}

and multiplying both sides by

- 1

reverses that to

- \frac{3}{4} < - \frac{2}{3}

. The other four hold, by the same numerator over the smaller denominator in 1, by

1024 > 1000

in 2, by

- \frac{1}{11} > - \frac{1}{6}

in 4, and by

\frac{14}{30} < \frac{15}{30}

in 5, so the count is

4

Practice

Let

u = - \frac{5}{6}

and

v = \frac{4}{3}

, and let

A = \frac{1}{u}

and

B = \frac{1}{v}

. Enter A or B, whichever is bigger, or the word unknown if it cannot be decided.

Show the solution

Flip each one.

A = \frac{1}{u} = - \frac{6}{5}

and

B = \frac{1}{v} = \frac{3}{4}

, and a negative value is less than a positive one, so the bigger one is

B

. The reciprocal rule needs

u

and

v

on the same side of zero, and they are not, so comparing the two values directly is the reliable route.

Practice

Let

n

be a nonzero real number, with nothing else known about it, and let

A = \frac{9}{n}

and

B = \frac{5}{n}

. Enter A or B, whichever is bigger, or the word unknown if it cannot be decided.

Show the solution

n = 2

the values are

\frac{9}{2}

and

\frac{5}{2}

, so

A

is bigger, while at

n = - 2

they are

- \frac{9}{2}

and

- \frac{5}{2}

, so

B

is bigger, leaving the answer

unknown

. Subtracting settles it the same way, since the sign of

A - B = \frac{4}{n}

is the sign of

n

, and no sign is given.

Practice

Let

A = \frac{27}{40}

and

B = \frac{19}{28}

. Multiplying both by

280

clears both denominators. Enter A or B, whichever is bigger.

Show the solution

Multiplying by

280

gives

7 \times 27 = 189

for

A

and

10 \times 19 = 190

for

B

, so the bigger one is

B

. Since

280

is positive, the order is the same before and after the scaling, so

189 < 190

settles the original pair.

Practice

Enter the smallest integer

n

for which

2^{n} > 5^{7}

Show the solution

5^{7} = 78125

, and

2^{16} = 65536

is less than that while

2^{17} = 131072

is greater, so the smallest such integer is

17

. A middle number like

1 0^{5}

settles the upper end at once, since

2^{17} > 1 0^{5} > 5^{7}

, but it cannot settle the lower end, because

2^{16}

and

5^{7}

are both below it.

Practice

Let

A = \frac{8}{17}

and

B = \frac{9}{19}

. Enter A or B, whichever is bigger.

Show the solution

Over the common denominator

17 \times 19 = 323

\frac{9}{19} - \frac{8}{17} = \frac{9 \times 17 - 8 \times 19}{323} = \frac{153 - 152}{323} = \frac{1}{323}

, which is positive, so the bigger one is

B

. Comparing each to

\frac{1}{2}

is no help, since doubling the numerators gives

16 < 17

and

18 < 19

, so both fractions are below

\frac{1}{2}

Practice

Let

A = \frac{1}{2^{45}}

and

B = \frac{1}{5^{18}}

. Enter A or B, whichever is bigger.

Show the solution

Both have numerator

1

, so the smaller denominator gives the bigger fraction. Power of a power rewrites

2^{45} = (2^{5})^{9} = 3 2^{9}

and

5^{18} = (5^{2})^{9} = 2 5^{9}

, so

5^{18} < 2^{45}

and the bigger fraction is

B

. A common exponent works whenever the two exponents share a factor, since it leaves only the bases to compare.