Inequalities and their properties

An is taller than Bình. A year later both have grown by $5$ cm. Is An still taller than Bình? Certainly. But if we know $a>b$ , which is larger, $-2a$ or $-2b$ ? The answer surprises many students.

This lesson helps us answer questions like these with certainty, without guessing.

Inequalities

A statement of the form $a<b$ (or $a>b$ , $a\le b$ , $a\ge b$ ) is called an inequality; $a$ is the left-hand side and $b$ is the right-hand side.

$a\le b$ is read " $a$ is less than or equal to $b$ ", or " $a$ is not greater than $b$ ".
$a\ge b$ is read " $a$ is greater than or equal to $b$ ", or " $a$ is not less than $b$ ".

On a number line, $a<b$ means that the point $a$ is to the left of the point $b$ . If $a$ is to the left of $b$ and $b$ is to the left of $c$ , then $a$ is to the left of $c$ .

Order and addition

Explore: sliding along the number line

We have $2<3$ . Add $4$ to both sides: $6<7$ . Add $-5$ : $-3<-2$ . Adding a number to both sides slides both points the same distance along the number line, so the order does not change.

The rule also holds for $>$ , $\le$ , $\ge$ , and for subtraction (since subtracting $c$ is adding $-c$ ).

Order and multiplication

Explore: multiplying by a negative number

We have $2<3$ . Multiplying both sides by $4$ gives $8<12$ : the direction stays the same. Multiplying both sides by $-1$ gives $-2$ and $-3$ , and $-2>-3$ : the direction is reversed. On a number line, multiplying by $-1$ reflects every point in $0$ , so the point on the left ends up on the right.

Multiplying by

-1

reverses the order

Dividing by $c\neq0$ is multiplying by $\dfrac1c$ , so the same rule applies.

Putting the properties together, we can prove many inequalities.

To prove an inequality with letters, we often use a simple fact: the square of any number is non-negative.

Key terms

English	Tiếng Việt
inequality	bất đẳng thức
left-hand side / right-hand side	vế trái / vế phải
less than or equal to	nhỏ hơn hoặc bằng
greater than or equal to	lớn hơn hoặc bằng
transitive property	tính chất bắc cầu
keep the direction (of the inequality)	giữ nguyên chiều
reverse the inequality sign	đổi chiều
equality holds	dấu bằng xảy ra

Summary

Inequalities: $a<b$ , $a>b$ , $a\le b$ , $a\ge b$ .
Transitive: $a<b$ , $b<c\Rightarrow a<c$ .
Adding (subtracting) the same number on both sides: the direction stays the same.
Multiplying (dividing) by a positive number: keep the direction; by a negative number: reverse it.
Proofs: reduce to $(\dots)^2\ge0$ .

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Exercise 1

Write as inequalities: a) the number of students $n$ in a class is at most $45$ ; b) the ticket price $p$ (thousand đồng) is at least $50$ ; c) the temperature $T$ ( $^\circ$ C) is below $0$ .

What you need and how to start

What you need: "at most" is $\le$ , "at least" is $\ge$ , "below" is $<$ .

Steps:

Decide the sign for each phrase.
Write the inequality.

Show solution

a) $n\le45$ ; b) $p\ge50$ ; c) $T<0$ .

Answer. $n\le45$ ; $p\ge50$ ; $T<0$ .

Exercise 2

Is each statement true or false? a) $-5<-3$ ; b) $2\cdot(-3)\ge-6$ ; c) $(-2)^2<3$ ; d) $7-10\le-4$ .

What you need and how to start

What you need: work out each side, then compare; $\ge$ is also true when the two sides are equal.

Steps:

Work out each side.
Compare them on a number line.

Show solution

a) True. b) $-6\ge-6$ : true. c) $4<3$ : false. d) $-3\le-4$ : false.

Answer. True, True, False, False.

Exercise 3

Suppose $a<b$ . Compare: a) $a+5$ and $b+5$ ; b) $-3a$ and $-3b$ ; c) $2a-1$ and $2b-1$ .

What you need and how to start

What you need: adding keeps the direction; multiplying by a positive number keeps it, by a negative number reverses it.

Steps:

In b) multiply by $-3$ .
In c) multiply by $2$ , then add $-1$ .

Show solution

a) $a+5<b+5$ . b) $-3a>-3b$ . c) $2a<2b\Rightarrow 2a-1<2b-1$ .

Answer. a) $<$ ; b) $>$ ; c) $<$ .

Exercise 4

Given that $-2a+3<-2b+3$ , compare $a$ and $b$ .

What you need and how to start

What you need: add $-3$ to both sides, then divide by $-2$ (reverse the sign).

Steps:

Remove the $3$ from both sides.
Divide by $-2$ .

Show solution

$-2a+3<-2b+3\Rightarrow -2a<-2b\Rightarrow a>b$ (dividing by $-2<0$ reverses the sign).

Answer. $a>b$ .

Exercise 5

Prove that if $a>2$ , then $3a-1>5$ .

What you need and how to start

What you need: multiply by a positive number, then add the same number to both sides.

Steps:

Multiply both sides by $3$ .
Add $-1$ to both sides.

Show solution

$a>2\Rightarrow 3a>6\Rightarrow 3a-1>5$ .

Answer. Proved.

Exercise 6

A lift can carry at most $600$ kg. Mr Tư, who weighs $70$ kg, wants to take some bags of cement, each of $50$ kg. Write an inequality for the load with $10$ bags, and say whether he can take $10$ bags or $11$ bags.

What you need and how to start

What you need: "at most $600$ " means the total mass is $\le600$ .

Steps:

Work out the total mass with $10$ bags and with $11$ bags.
Compare with $600$ .

Show solution

$10$ bags: $70+10\cdot50=570\le600$ , allowed. $11$ bags: $70+11\cdot50=620>600$ , not allowed.

Answer. He can take $10$ bags but not $11$ .

Exercise 7 Review

What are the solutions of the equation $(x-1)(x+2)=0$ ? Which of them satisfies $x>0$ ?

What you need and how to start

What you need: equations in factorised form (previous lesson).

Steps:

Solve $x-1=0$ and $x+2=0$ .
Compare each solution with $0$ .

Show solution

The solutions are $x=1$ and $x=-2$ . Since $1>0$ and $-2<0$ , only $x=1$ works.

Answer. $x=1$ .

Exercise 8

Suppose $a>b>0$ .

a) Prove that $a^2>ab$ .

b) Prove that $ab>b^2$ .

c) Deduce that $a^2>b^2$ .

What you need and how to start

What you need: multiplying both sides by a positive number keeps the direction; the transitive property.

Steps:

a) Multiply $a>b$ by $a>0$ .
b) Multiply $a>b$ by $b>0$ .
c) Use the transitive property.

Show solution

a) $a>0$ , so $a\cdot a>b\cdot a$ , that is $a^2>ab$ . b) $b>0$ , so $ab>b\cdot b=b^2$ . c) $a^2>ab$ and $ab>b^2$ , so $a^2>b^2$ .

Answer. Proved; so for two positive numbers, the larger one has the larger square.

Exercise 9 Challenge

Prove that $a^2+b^2+2\ge2(a+b)$ for all numbers $a$ , $b$ . When does equality hold?

What you need and how to start

What you need: the square of any number is non-negative.

Steps:

Move everything to the left-hand side.
Write the left-hand side as $(a-1)^2+(b-1)^2$ .

Show solution

$a^2+b^2+2-2a-2b=(a^2-2a+1)+(b^2-2b+1)=(a-1)^2+(b-1)^2\ge0$ . Equality holds when $a-1=0$ and $b-1=0$ .

Answer. The inequality holds; equality when $a=b=1$ .

Inequalities

Order and addition

Explore: sliding along the number line

Order and multiplication

Explore: multiplying by a negative number

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

a) $a+2>b+2$ .	True	False
b) $-5a>-5b$ .	True	False
c) $a-b>0$ .	True	False
d) $a^2>b^2$ .	True	False