Order on the Real Numbers

Addition and multiplication tell us how to combine real numbers; order tells us how to compare them. The multiplicative inverse results from the previous tutorial will also be useful: taking reciprocals changes the size comparison between positive numbers. Here we introduce the usual order on $\mathbb{R}$ and establish rules for using it.

The usual order is part of the structure of the real numbers, alongside the field operations. One way to describe it is to specify which real numbers are positive. Let $P$ denote the set of positive real numbers. The usual positive reals have these properties: $1\in P$; the sum and product of two members of $P$ are in $P$; and every nonzero real number is either in $P$ or the negative of a member of $P$, but not both. These properties let us define comparisons and prove the basic order rules.

Definition (Order on the real numbers). For $x,y\in\mathbb{R}$, define $x<y$ if and only if $y-x\in P$. We say $x$ is less than $y$, or equivalently $y$ is greater than $x$. Define $x\leq y$ to mean $x<y$ or $x=y$. The symbols $>$ and $\geq$ are defined by reversing the two numbers.

In this definition, $y-x$ is positive precisely when $y$ lies to the right of $x$ on the number line. In particular, $0<x$ means $x\in P$. The special properties of $P$ ensure that exactly one of $x<y$, $x=y$, or $y<x$ holds. This is the trichotomy property: any two real numbers are comparable in exactly one of these three ways.

Comparing Numbers and Translating Inequalities

To compare two numbers, it is often enough to subtract the first from the second. A positive difference means the second number is larger; a negative difference means it is smaller. This approach also makes clear why adding the same real number to both sides of an inequality does not change its direction.

Theorem (Trichotomy and translation of order). For any $x,y\in\mathbb{R}$, exactly one of $x<y$, $x=y$, and $y<x$ holds. Moreover, for any $a,x,y\in\mathbb{R}$, $$ x<y \quad\Longleftrightarrow\quad x+a<y+a. $$

Proof. Consider $y-x$. If $y-x=0$, then $y=x$. If $y-x\ne0$, the defining properties of $P$ say that exactly one of $y-x$ and $-(y-x)$ is positive. If $y-x\in P$, then $x<y$. If $-(y-x)=x-y\in P$, then $y<x$. The three alternatives are mutually exclusive, so exactly one holds.

For translation, the definition of order gives $x<y$ if and only if $y-x\in P$. But

$$ (y+a)-(x+a)=y+a-x-a=y-x. $$

Thus $(y+a)-(x+a)\in P$ if and only if $y-x\in P$, which proves $x+a<y+a$ if and only if $x<y$. $\square$

Worked Example: Comparing Two Negative Numbers

Compare $-\frac{11}{6}$ and $-\frac{7}{5}$. Subtract the first number from the second:

-\frac{7}{5}-\left(-\frac{11}{6}\right) =-\frac{42}{30}+\frac{55}{30} =\frac{13}{30}.

Since $\frac{13}{30}>0$, the difference is positive. By the definition of order,

-\frac{11}{6}<-\frac{7}{5}.

The calculation illustrates why comparing negative numbers by their signs alone is not enough: both are negative, so their difference determines which is greater.

The translation theorem works in both directions. If the same number is added to each side, the inequality is preserved; if the same number is subtracted from each side, it is preserved as well, since subtraction is addition of an additive inverse. For example, $x+4<9$ is equivalent to $x<5$, because subtracting $4$ from both sides leaves the inequality direction unchanged.

Multiplying Inequalities and Determining Signs

Multiplication requires more care than addition. Multiplying an inequality by a positive number preserves its direction, but multiplying by a negative number reverses it. The sign of the multiplier is essential; multiplying by zero does not preserve a strict inequality.

Theorem (Multiplication and order). Suppose $x<y$.

If $c>0$, then $cx<cy$.
If $c<0$, then $cy<cx$.

Proof. From $x<y$, the difference $y-x$ is positive. If $c>0$, closure of $P$ under multiplication gives $c(y-x)>0$. Since $c(y-x)=cy-cx$, this means $cx<cy$.

If $c<0$, then $-c>0$. The product $(-c)(y-x)$ is positive, and

$$ (-c)(y-x)=cx-cy. $$

Therefore $cx-cy>0$, which means $cy<cx$. In this case multiplication reverses the inequality. $\square$

The positive-multiplier case also explains why division by a positive number preserves an inequality: dividing by $c>0$ means multiplying by $c^{-1}$, and $c^{-1}>0$. Division by a negative number reverses the inequality for the same reason as multiplication by a negative number. The reciprocal sign fact is proved below.

Worked Example: Solving an Inequality with a Negative Coefficient

Solve $-3x+4<10$. Subtract $4$ from both sides. Translation preserves the direction:

$$ -3x<6. $$

Now multiply both sides by $-1$. Because the multiplier is negative, the inequality reverses:

$$ 3x>-6. $$

Since $3>0$, division by $3$ preserves the direction, giving $x>-2$. To check the boundary and direction, $x=-2$ gives $-3(-2)+4=10$, which is equality rather than a solution to the strict inequality. Any $x>-2$, such as $x=0$, gives $-3(0)+4=4<10$.

Order rules also determine the signs of products. A product of two positive numbers is positive by the defining properties of $P$. A product of two negative numbers is positive because if $x<0$ and $y<0$, then $-x>0$ and $-y>0$, so $(-x)(-y)>0$; the field identities give $(-x)(-y)=xy$. A product of a positive and a negative number is negative, since negating the negative factor makes the product positive, so the original product is its negative.

Theorem (Squares are nonnegative). For every $x\in\mathbb{R}$, $x^2\geq0$. If $x\ne0$, then $x^2>0$.

Proof. If $x=0$, then $x^2=0$. If $x\ne0$, trichotomy gives either $x>0$ or $x<0$. In the first case, $x\cdot x>0$ because it is a product of positive numbers. In the second case, $-x>0$, so $(-x)(-x)>0$. The field identity $(-x)(-x)=x^2$ then gives $x^2>0$ in this case as well. Therefore every square is nonnegative, and every nonzero square is positive. $\square$

Worked Example: Determining the Sign of a Product

Suppose $-\frac{5}{8}<0$ and $\frac{7}{3}>0$. Their product is negative. Indeed, $\frac{5}{8}>0$ and $\frac{7}{3}>0$, so

\frac{5}{8}\cdot\frac{7}{3}=\frac{35}{24}>0.

Using $-\frac{5}{8}= -\left(\frac{5}{8}\right)$, the product in question is

\left(-\frac{5}{8}\right)\left(\frac{7}{3}\right) =-\left(\frac{5}{8}\cdot\frac{7}{3}\right) =-\frac{35}{24}<0.

The sign conclusion comes from the order rules; the fraction calculation confirms the particular value.

Positive Reciprocals and Reversed Order

The previous tutorial established that every nonzero real has a multiplicative inverse. Order adds information about the sign of that inverse. A positive number has a positive reciprocal, and comparing two positive numbers shows why their reciprocals appear in the opposite order.

Theorem (Reciprocals reverse positive order). If $0<x<y$, then $$ 0<y^{-1}<x^{-1}. $$

Proof. First, $y>0$ and $yy^{-1}=1>0$. If $y^{-1}<0$, then the product of $y>0$ and $y^{-1}<0$ would be negative, contradicting $yy^{-1}=1>0$. Also $y^{-1}\ne0$, since $yy^{-1}=1$. Trichotomy therefore gives $y^{-1}>0$.

Next, $y-x>0$, and $x>0$, $y^{-1}>0$. Their product with the positive reciprocal is positive:

(y-x)y^{-1}>0.

We can rewrite this difference using $yy^{-1}=1$:

(y-x)y^{-1}=1-xy^{-1}.

Thus $xy^{-1}<1$. Since $x>0$, the reciprocal $x^{-1}$ is positive by the same sign argument used for $y^{-1}$. Multiplying $xy^{-1}<1$ by $x^{-1}>0$ preserves the inequality, so

y^{-1}<x^{-1}.

Together with $y^{-1}>0$, this proves $0<y^{-1}<x^{-1}$. $\square$

Worked Example: Comparing Positive Reciprocals

Since $0<\frac{3}{8}<\frac{5}{6}$, the reciprocal-order theorem gives

\left(\frac{5}{6}\right)^{-1} < \left(\frac{3}{8}\right)^{-1}, \qquad\text{so}\qquad \frac{6}{5}<\frac{8}{3}.

The original comparison can be checked by subtraction:

\frac{5}{6}-\frac{3}{8} =\frac{20}{24}-\frac{9}{24} =\frac{11}{24}>0.

The reciprocal comparison can also be checked directly:

\frac{8}{3}-\frac{6}{5} =\frac{40}{15}-\frac{18}{15} =\frac{22}{15}>0.

The reciprocals are still positive, but the larger original number has the smaller reciprocal.

Using Order Carefully

The most common error when manipulating inequalities is to change their direction incorrectly. Adding or subtracting the same real number never reverses an inequality. Multiplying or dividing by a positive number preserves its direction, while multiplying or dividing by a negative number reverses it. If a quantity's sign is unknown, one cannot decide which rule applies until that sign is established.

A second useful distinction is between $<$ and $\leq$. The inequality $x<y$ excludes equality, whereas $x\leq y$ permits it. For example, $x^2\geq0$ holds for every real $x$, but $x^2>0$ requires $x\ne0$. Keeping track of whether equality is possible prevents an otherwise correct sign argument from becoming too strong.

Key takeaway. Real numbers are comparable by trichotomy. Adding the same number preserves an inequality; multiplying by a positive number preserves it, while multiplying by a negative number reverses it. Positive reciprocals reverse the order of positive numbers, and every real square is nonnegative.

Check Your Understanding

Use the definitions and order rules from this tutorial to answer the following questions.

How can the sign of $y-x$ be used to compare $x$ and $y$?
Why does adding the same real number to both sides preserve a strict inequality?
If $x<y$ and $c<0$, which of $cx<cy$ or $cy<cx$ holds? Explain why.
Why is $x^2>0$ guaranteed when $x\ne0$, but not when $x=0$?
If $0<a<b$, which is larger, $a^{-1}$ or $b^{-1}$?
Solve $5-2x\geq1$, stating where the inequality direction changes or stays the same.

Order on the Real Numbers

What You'll Learn