Lesson 4 of 42 · Numbers and sets

Order properties of the real numbers

Know which operations keep an inequality and which reverse it, and compare fractions, roots and powers by hand without testing a single convenient number.

About 20 minutes

In this lesson

  • State which operations preserve an inequality and which reverse it, and when squaring or taking reciprocals is safe.
  • Order xx, x2x^2, x\sqrt{x} and 1x\frac{1}{x} for xx in different ranges.
  • Compare fractions, roots and sums of roots without a calculator.

The idea

The rules that keep or flip an inequality

For real numbers with a<ba < b:

Operation Result Condition
add or subtract any cc a+c<b+ca + c < b + c always
multiply or divide by cc ac<bcac < bc c>0c > 0
multiply or divide by cc ac>bcac > bc (reversed) c<0c < 0
take reciprocals 1a>1b\frac{1}{a} > \frac{1}{b} (reversed) aa and bb have the same sign
square a2<b2a^2 < b^2 0≤a<b0 \leq a < b
take an odd power or odd root a3<b3a^3 < b^3, a3<b3\sqrt[3]{a} < \sqrt[3]{b} always

Odd powers and odd roots are increasing on all reals, so they always preserve order. Even powers preserve order only between non-negative numbers: −3<2-3 < 2 but 9>49 > 4. Reciprocals reverse order within one sign, but if a<0<ba < 0 < b then 1a<0<1b\frac{1}{a} < 0 < \frac{1}{b} and the order stays as it was.

Two inequalities in the same direction can be added: from a<ba < b and c<dc < d, a+c<b+da + c < b + d. They cannot in general be subtracted or divided, and they can be multiplied only when every term is positive.

Powers and roots of numbers between 0 and 1

For x>1x > 1, raising to a higher power makes the number larger: x<x<x2<x3\sqrt{x} < x < x^2 < x^3. For 0<x<10 < x < 1 the order turns round, because multiplying by a number less than 1 makes things smaller:

0<x3<x2<x<x<1<1x.0 < x^3 < x^2 < x < \sqrt{x} < 1 < \frac{1}{x}.

Test with x=14x = \frac{1}{4}: x2=116x^2 = \frac{1}{16}, x=12\sqrt{x} = \frac{1}{2}, 1x=4\frac{1}{x} = 4. For −1<x<0-1 < x < 0 the signs matter too: odd powers stay negative, even powers become positive, and 1x<−1\frac{1}{x} < -1.

Comparing fractions

To compare ab\frac{a}{b} and cd\frac{c}{d} with positive denominators, cross-multiply: ab<cd\frac{a}{b} < \frac{c}{d} exactly when ad<bcad < bc. This is just multiplying both sides by the positive number bdbd. If a denominator could be negative, cross-multiplying may need a reversal, so check signs first.

Other quick comparisons: compare each fraction with a benchmark such as 12\frac{1}{2} or 1, or compare how far each falls short of 1. For instance 1213\frac{12}{13} is 113\frac{1}{13} below 1 and 1415\frac{14}{15} is 115\frac{1}{15} below 1, so 1415\frac{14}{15} is larger.

Comparing roots

Two positive numbers compare in the same way as their squares, cubes or any positive power. So to compare roots, raise both to a power that clears every root:

  • 33\sqrt[3]{3} against 2\sqrt{2}: raise to the 6th power, giving 99 against 88, so 33>2\sqrt[3]{3} > \sqrt{2}.
  • 2+3\sqrt{2} + \sqrt{3} against 10\sqrt{10}: both are positive; square them. (2+3)2=5+26(\sqrt{2} + \sqrt{3})^2 = 5 + 2\sqrt{6} and (10)2=10(\sqrt{10})^2 = 10, so compare 262\sqrt{6} with 55. Square again: 24<2524 < 25. Hence 2+3<10\sqrt{2} + \sqrt{3} < \sqrt{10}.

Squaring is safe here only because both sides are positive. Before squaring, always confirm the sign of each side.

"Must be true" questions

A typical question gives an inequality between letters and asks which statement must follow. A statement that must be true has to hold for every allowed value, including negatives, zero (if allowed) and fractions. The fastest test is to look for a counterexample: try a negative number, a number between 0 and 1, and numbers of opposite signs. One failure removes an option; the survivor should then be justified by a rule from the table, not by the examples.

Intervals

Solution sets are written as intervals: (a,b)(a, b) excludes both ends, [a,b][a, b] includes them, and ∞\infty always takes a round bracket. Between any two different real numbers there is another (their average), so an open interval like (0,1)(0, 1) has no smallest or largest element.

Worked examples

Example 1: Ordering powers of a fraction

If 0<x<10 < x < 1, put xx, x2x^2, x\sqrt{x} and 1x\frac{1}{x} in increasing order.

  1. Multiplying by x<1x < 1 makes a positive number smaller, so x2=x⋅x<xx^2 = x \cdot x < x.

  2. x\sqrt{x} is the number whose square is xx; since squaring makes numbers in (0,1)(0, 1) smaller, x\sqrt{x} must be larger than xx, but still less than 1.

  3. 1x>1\frac{1}{x} > 1 because x<1x < 1, so it is the largest.

  4. Check with x=14x = \frac{1}{4}: 116<14<12<4\frac{1}{16} < \frac{1}{4} < \frac{1}{2} < 4.

Answer

x2<x<x<1xx^2 < x < \sqrt{x} < \frac{1}{x}. Assuming squares are always larger, x<x2x < x^2, is the habit from numbers greater than 1 that this question targets.

Example 2: Two close fractions

Which is larger, 79\frac{7}{9} or 1114\frac{11}{14}?

  1. The denominators are positive, so cross-multiply: compare 7⋅14=987 \cdot 14 = 98 with 11⋅9=9911 \cdot 9 = 99.

  2. Since 98<9998 < 99, 79<1114\frac{7}{9} < \frac{11}{14}.

  3. Second route: 79\frac{7}{9} is 29\frac{2}{9} below 1 and 1114\frac{11}{14} is 314\frac{3}{14} below 1. Compare 29\frac{2}{9} with 314\frac{3}{14}: 28>2728 > 27, so 79\frac{7}{9} falls further short of 1 and is the smaller.

Answer

1114\frac{11}{14} is larger, by 1126\frac{1}{126}. Comparing numerators or denominators alone does not settle it, since both are larger in 1114\frac{11}{14}.

Example 3: What follows from a<b<0a < b < 0?

If a<b<0a < b < 0, decide whether each is true: a2>b2a^2 > b^2; 1a>1b\frac{1}{a} > \frac{1}{b}; ab>0ab > 0; ab>1\frac{a}{b} > 1.

  1. Both numbers are negative and aa is further from 0, so ∣a∣>∣b∣\lvert a \rvert > \lvert b \rvert and a2>b2a^2 > b^2. True.

  2. Same sign, so reciprocals reverse the order: 1a>1b\frac{1}{a} > \frac{1}{b}. True. With a=−3a = -3, b=−2b = -2: −13>−12-\frac{1}{3} > -\frac{1}{2}.

  3. A product of two negatives is positive: ab>0ab > 0. True.

  4. Divide a<ba < b by the negative number bb, reversing the sign: ab>1\frac{a}{b} > 1. True; with the same numbers, −3−2=1.5\frac{-3}{-2} = 1.5.

Answer

All four are true. The usual slip is to keep the direction when dividing by bb and conclude ab<1\frac{a}{b} < 1.

Common traps

  • Multiplying by a negative without reversing

    From −2x<6-2x < 6, dividing by −2-2 gives x>−3x > -3, not x<−3x < -3. The same applies when you multiply by a letter whose sign you do not know: then you cannot multiply at all until you split into cases.

  • Squaring an inequality with negative numbers

    −5<3-5 < 3 does not give 25<925 < 9. Square only when both sides are non-negative; otherwise compare absolute values.

  • Flipping reciprocals across zero

    −2<3-2 < 3 gives −12<13-\frac{1}{2} < \frac{1}{3}: no reversal, because the two numbers have opposite signs. Reciprocals reverse the order only within one sign.

  • Proving "must be true" with one friendly number

    Checking a=1a = 1, b=2b = 2 shows a statement can be true, not that it must be. Try a negative value and a fraction before accepting an option.

Check yourself

Answer each question before you open its answer.

Question 1: Ordering when xx is between −1-1 and 00

Numbers

If −1<x<0-1 < x < 0, which ordering is correct?

  1. A.x3<x<x2<1xx^3 < x < x^2 < \frac{1}{x}
  2. B.1x<x3<x<x2\frac{1}{x} < x^3 < x < x^2
  3. C.1x<x<x3<x2\frac{1}{x} < x < x^3 < x^2
  4. D.x<1x<x3<x2x < \frac{1}{x} < x^3 < x^2
Show the answer

Answer: C. 1x<x<x3<x2\frac{1}{x} < x < x^3 < x^2

Working

Take x=−12x = -\frac{1}{2}: 1x=−2\frac{1}{x} = -2, x=−12x = -\frac{1}{2}, x3=−18x^3 = -\frac{1}{8}, x2=14x^2 = \frac{1}{4}. In general, ∣x3∣<∣x∣<1<∣1x∣\lvert x^3 \rvert < \lvert x \rvert < 1 < \lvert \frac{1}{x} \rvert with all three negative, and x2>0x^2 > 0. So 1x<x<x3<x2\frac{1}{x} < x < x^3 < x^2.

Why the other options are wrong

  • A. This uses the habits of positive numbers: for negative xx, 1x\frac{1}{x} is negative and x3x^3 is closer to 0 than xx.
  • B. Cubing a number between −1-1 and 00 moves it towards 0, so x3>xx^3 > x, not x3<xx^3 < x.
  • D. The reciprocal of a number between −1-1 and 00 is less than −1-1, so 1x<x\frac{1}{x} < x.

What carries over

For negative numbers, order by distance from zero and then reverse: the one furthest from zero is the smallest.

Question 2: Which statement must hold?

Numbers

If aa and bb are real numbers with a<ba < b, which statement must be true?

  1. A.a2<b2a^2 < b^2
  2. B.−3a>−3b-3a > -3b
  3. C.1a>1b\frac{1}{a} > \frac{1}{b}
  4. D.ab<1\frac{a}{b} < 1
Show the answer

Answer: B. −3a>−3b-3a > -3b

Working

Multiplying a<ba < b by the negative number −3-3 reverses the sign: −3a>−3b-3a > -3b. This holds for every pair with a<ba < b. Each other statement has a counterexample.

Why the other options are wrong

  • A. Fails for a=−3a = -3, b=1b = 1: 9>19 > 1. Squaring keeps the order only for non-negative numbers.
  • C. Fails for a=−1a = -1, b=2b = 2: −1<12-1 < \frac{1}{2}. Reciprocals reverse the order only when both numbers have the same sign, and they do not exist at 0.
  • D. Fails for a=−4a = -4, b=−2b = -2: ab=2\frac{a}{b} = 2. Dividing by a negative bb reverses the inequality, and bb could also be 0.

What carries over

A statement that must be true survives negative values, fractions and mixed signs; multiplying by a known negative constant always reverses the order.

Question 3: Ordering sums of roots

Numbers

Which ordering of 5+3\sqrt{5} + \sqrt{3}, 7+1\sqrt{7} + 1 and 44 is correct?

  1. A.7+1<5+3<4\sqrt{7} + 1 < \sqrt{5} + \sqrt{3} < 4
  2. B.5+3<7+1<4\sqrt{5} + \sqrt{3} < \sqrt{7} + 1 < 4
  3. C.7+1<4<5+3\sqrt{7} + 1 < 4 < \sqrt{5} + \sqrt{3}
  4. D.4<7+1<5+34 < \sqrt{7} + 1 < \sqrt{5} + \sqrt{3}
Show the answer

Answer: A. 7+1<5+3<4\sqrt{7} + 1 < \sqrt{5} + \sqrt{3} < 4

Working

All three are positive, so compare squares. (5+3)2=8+215(\sqrt{5} + \sqrt{3})^2 = 8 + 2\sqrt{15}, (7+1)2=8+27(\sqrt{7} + 1)^2 = 8 + 2\sqrt{7}, 42=16=8+84^2 = 16 = 8 + 8. Now 27<2152\sqrt{7} < 2\sqrt{15}, and 215<82\sqrt{15} < 8 because 60<6460 < 64. So 7+1<5+3<4\sqrt{7} + 1 < \sqrt{5} + \sqrt{3} < 4.

Why the other options are wrong

  • B. Both sums square to 8 plus a term: 2152\sqrt{15} against 272\sqrt{7}. Since 15>715 > 7, 5+3\sqrt{5} + \sqrt{3} is the larger.
  • C. Rough estimates such as 5≈2.3\sqrt{5} \approx 2.3 and 3≈1.8\sqrt{3} \approx 1.8 overshoot. Squaring shows 8+215<168 + 2\sqrt{15} < 16, because 60<6460 < 64.
  • D. 7<3\sqrt{7} < 3, so 7+1<4\sqrt{7} + 1 < 4.

What carries over

To compare positive expressions with roots, square them, cancel the common part, and square again if needed.

Key points

  • Multiplying or dividing by a negative reverses an inequality; adding never does.

  • Squares keep the order only for non-negative numbers; reciprocals reverse it only within one sign.

  • For 0<x<10 < x < 1: x2<x<x<1<1xx^2 < x < \sqrt{x} < 1 < \frac{1}{x}.

  • Compare positive roots by raising both sides to a power that clears the roots.

All questions here are original and written for UniExamer; none comes from the test maker.

Written by the UniExamer team with AI assistance. Every worked example and check question was re-solved by an independent checker before publication. No subject expert has reviewed this lesson yet. If you find a mistake, write to support@uniexamer.com.

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