Week 1 · Day 3 Lesson 1.7 About 8 minutes Free

Inequality properties: test with numbers

By the end: Know which inequality rules always hold, which fail for negative numbers or zero, and use a small set of test numbers to find the one statement that must be true.

This is the “must be true” family: Questions 22–24 in all five sittings we studied. You almost never need to prove anything. You need to find the options that fail.

Understand it

The question is always some version of: “If $a>b$ , which of the following must be true?” Three of the four options fail for some numbers. One holds for all of them.

What always holds (for any real numbers):

Adding or subtracting the same number: $a>b\Rightarrow a+c>b+c$ and $a-c>b-c$ .
Multiplying by a positive number keeps the direction: $c>0$ gives $ac>bc$ .
Odd powers keep the direction: $a>b\Rightarrow a^3>b^3$ and $a^5>b^5$ .
Adding two inequalities that point the same way: $a>b$ and $c>d$ give $a+c>b+d$ .

What can fail:

Multiplying by a negative number reverses the direction: $c<0$ gives $ac<bc$ . If $c=0$ both sides are equal, so “ $ac>bc$ ” is not always true.
Even powers and absolute values do not keep the direction: $1>-2$ but $1^2<(-2)^2$ .
Reciprocals reverse the direction only when both numbers have the same sign: $a>b>0\Rightarrow\frac1a<\frac1b$ . When $a>0>b$ , then $\frac1a>0>\frac1b$ , so the order is kept instead of reversed.
You cannot subtract or divide two inequalities. You can multiply them only when all four numbers are positive.

The test-with-numbers method. Positive numbers hide the traps, so do not test with $a=3$ , $b=1$ . Use numbers that expose them:

$a=1,\ b=-2$ (mixed signs, so squares and absolute values fail),
$a=-1,\ b=-2$ (both negative, so reciprocals and products change),
$c=-1$ and $c=0$ when the option has a third letter.

Substitute each pair into every option and cross out any option that fails even once. A single test can kill an option but it can never prove one, so run a second pair on the options that survive. In a single-answer question the last option standing is the answer.

Essential formulas

a>b\ \Rightarrow\ a+c>b+c,\qquad a^3>b^3,\qquad ac>bc\ \ (c>0)

These hold for all real numbers

a,b

. Adding the same number, taking an odd power, and multiplying by a positive number never change the direction.

a>b,\ c<0\ \Rightarrow\ ac<bc\qquad a>b>0\ \Rightarrow\ \frac1a<\frac1b

A negative multiplier reverses the inequality. The reciprocal rule needs

a

and

b

to be positive (or both negative).

a>b,\ c>d\ \Rightarrow\ a+c>b+d

Adding inequalities with the same direction is always safe. Subtracting them is not: from

a>b

and

c>d

you cannot conclude

a-c>b-d

See it

Given 1 < x < 3 and 2 < y < 5: −5 < −y < −2, so adding gives −4 < x − y < 1. Subtracting the bounds directly (1 − 2 and 3 − 5) would give the wrong interval.

Worked examples

Worked example 1

If $a>b$ , which of these must be true? A. $a^2>b^2$ B. $a^3>b^3$ C. $\dfrac1a<\dfrac1b$ D. $ac>bc$

Test $a=1$ , $b=-2$ (so $a>b$ ). A: $1>4$ is false. C: $1<-\tfrac12$ is false. D with $c=-1$ : $-1>2$ is false.
Only B survives: $1>-8$ is true.
Confirm with a second pair, $a=-1$ , $b=-2$ : $-1>-8$ is true again. This agrees with the rule that odd powers keep the direction.

Answer: B

Worked example 2

If $a>b$ and $c>d$ , which must be true? A. $a-c>b-d$ B. $a+c>b+d$ C. $ac>bd$ D. $\dfrac ac>\dfrac bd$

Test $a=3$ , $b=1$ , $c=2$ , $d=-5$ . A: $1>6$ is false, so A is out.
C and D also survive that pair ( $6>-5$ and $\tfrac32>-\tfrac15$ ), so try a pair with negatives: $a=-1$ , $b=-2$ , $c=-3$ , $d=-4$ . C: $3>8$ is false, and D: $\tfrac13>\tfrac12$ is false. Both are out.
B holds in both tests ( $5>-4$ and $-4>-6$ ), and it is the rule “add inequalities that point the same way”.

Answer: B

Worked example 3

Given $1<x<3$ and $2<y<5$ , find the range of $x-y$ .

Never subtract inequalities. Rewrite $x-y=x+(-y)$ .
Multiplying $2<y<5$ by $-1$ reverses it: $-5<-y<-2$ .
Now add the two inequalities that point the same way: $1+(-5)<x+(-y)<3+(-2)$ , which is $-4<x-y<1$ .

Answer: $(-4,1)$

Try it yourself

Work it on paper first. Open a hint only when you are stuck, then compare with the solution.

Given $2<a<4$ and $-3<b<-1$ , find the range of $2a-b$ . Then decide which of these must be true if $x>y$ and $z<0$ : $xz>yz$ , $x^2>y^2$ , $x+z>y+z$ .

Hint 1 of 2

Handle each term separately: bound $2a$ , then bound $-b$ (remember that $-1$ reverses an inequality).

Hint 2 of 2

For the second part, test $x=1$ , $y=-2$ , $z=-1$ in all three.

Show the step-by-step solution

From $2<a<4$ : $4<2a<8$ . From $-3<b<-1$ , multiplying by $-1$ gives $1<-b<3$ .
Add: $4+1<2a-b<8+3$ , so $5<2a-b<11$ .
Test $x=1$ , $y=-2$ , $z=-1$ : $xz=-1>yz=2$ is false; $x^2=1>y^2=4$ is false; $x+z=0>y+z=-3$ is true. Only $x+z>y+z$ must be true.

Answer: $(5,11)$ ; only $x+z>y+z$ .

Common mistakes

Testing with positive numbers only. $a=3$ , $b=1$ makes squares, reciprocals and absolute values look fine when they are not.
Stopping after one test and treating a survivor as proven. Run a second pair on every option that survived the first.
Forgetting that $c$ can be negative or zero: “ $ac>bc$ ” is not always true.
Subtracting or dividing inequalities. To find the range of $x-y$ , write $x+(-y)$ and add.
Taking reciprocals without checking signs. $\frac1a<\frac1b$ needs $a$ and $b$ to have the same sign.

Exam tips

Start with $a=1$ , $b=-2$ . It breaks even powers, absolute values and reciprocals in one test.
If two options survive, test $a=-1$ , $b=-2$ , and a case with $c=-1$ or $c=0$ if the option has a $c$ .
Adding or subtracting the same number to both sides always survives, and so do odd powers.
For “find the range” questions, bound each piece, flip the ones with a negative sign, then add.

In short

Always true: add or subtract the same number, multiply by a positive number, odd powers, adding same-direction inequalities.
Can fail: negative or zero multipliers, even powers, absolute values, reciprocals across zero, subtracting inequalities.
Test with $a=1,b=-2$ then $a=-1,b=-2$ , and $c=-1$ or $0$ ; cross out every option that fails once.

Knowledge check

4 questions on this lesson, marked as you go. Take them in ACE CSCA with a free account: passing them completes the lesson, and your progress is saved on every device.

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