Transitivity of Inequalities

From Three Terms to a Finite Chain

Transitivity lets us pass from comparisons involving three quantities to a comparison between the first and last. In “Properties of Inequalities,” the Theorem (Chaining strict and non-strict inequalities) established this for two successive comparisons, including the cases in which one comparison is strict. We now use that result to organize longer chains. This is useful when several estimates have been proved separately: if the right side of one comparison is the left side of the next, the comparisons can be linked.

Throughout this tutorial, $F$ denotes an ordered field. The results therefore apply to the real numbers. A chain is useful only when its comparisons connect: from $a\leq b$ and $b\leq c$, the shared middle quantity $b$ links the two statements. By contrast, knowing only $a\leq b$ and $c\leq d$ does not, by itself, compare $a$ with $d$. The endpoints of successive comparisons must match, or an additional comparison must connect them.

Theorem (Finite chaining of inequalities). Let $n\geq2$, and let $a_1,\ldots,a_n\in F$. Suppose $$ a_1\leq a_2\leq\cdots\leq a_n. $$ Then $a_1\leq a_n$. If at least one of the comparisons $a_i\leq a_{i+1}$ is strict, then $a_1<a_n$.

Proof. We use induction on the number of comparisons in the chain. If there is one comparison, then $n=2$, and the conclusion $a_1\leq a_2$ is one of the assumptions. If that comparison is strict, then $a_1<a_2$, as required.

Now suppose the result holds for chains with $k$ comparisons, where $k\geq1$, and consider a chain with $k+1$ comparisons: $a_1\leq a_2\leq\cdots\leq a_{k+1}\leq a_{k+2}$. The first $k$ comparisons give $a_1\leq a_{k+1}$ by the induction hypothesis. Combining this with $a_{k+1}\leq a_{k+2}$, the Theorem (Chaining strict and non-strict inequalities) gives $a_1\leq a_{k+2}$.

If one of the first $k$ comparisons is strict, the induction hypothesis gives $a_1<a_{k+1}$; chaining this with $a_{k+1}\leq a_{k+2}$ gives $a_1<a_{k+2}$. If instead the last comparison is strict, we have $a_1\leq a_{k+1}<a_{k+2}$, which also gives $a_1<a_{k+2}$. Thus the conclusion holds whenever at least one link is strict. This completes the induction. $\square$

The strictness condition is about the links in the chain, not just the endpoints as originally written. A single strict link cannot be undone by later non-strict links: the chain still forces the final quantity to be larger than the first. If every link is non-strict, the theorem guarantees only a non-strict comparison of the endpoints.

Worked Example: Tracking a Strict Link

Suppose $r\leq s$, $s<t$, and $t\leq u$. These comparisons form one connected chain. The middle comparison is strict, so the finite chaining theorem gives a strict comparison from the first quantity to the last:

r\leq s<t\leq u \qquad\Longrightarrow\qquad r<u.

The locations of the strict and non-strict links matter. For instance, $r=2$, $s=2$, $t=3$, and $u=3$ satisfy all the assumptions: $2\leq2$, $2<3$, and $3\leq3$. The endpoint comparison is $2<3$, as claimed. The first or last link may be equality; the strict middle link still ensures strictness between the endpoints.

Equality at Both Ends

A nondecreasing chain cannot leave a value and later return to it. This gives a useful converse perspective: if the endpoints of a chain are equal, every intermediate quantity must equal them. The statement can help check a proposed equality case in a sequence of estimates, or show that equality in an overall bound requires equality at every step.

Theorem (Equality throughout a nondecreasing chain). Let $a_1,\ldots,a_n\in F$, where $n\geq2$, and suppose $$ a_1\leq a_2\leq\cdots\leq a_n. $$ If $a_1=a_n$, then $a_i=a_1$ for every $i$ with $1\leq i\leq n$.

Proof. Fix any index $i$ with $1\leq i\leq n$. The finite chaining theorem applied to the beginning of the chain gives $a_1\leq a_i$ when $i>1$; for $i=1$, this comparison is equality. Applied to the part from $a_i$ to $a_n$, it gives $a_i\leq a_n$ when $i<n$; for $i=n$, this comparison is equality. Since $a_n=a_1$, in every case we have $a_1\leq a_i\leq a_1$. By trichotomy in the ordered field $F$, an element cannot be both greater than $a_1$ and less than $a_1$; hence $a_i=a_1$. Since $i$ was arbitrary, every term equals $a_1$. $\square$

This theorem also explains why a strict link rules out equal endpoints. If one link in a nondecreasing chain is strict, the finite chaining theorem gives $a_1<a_n$, so $a_1=a_n$ is impossible. Conversely, if the endpoints are equal, the equality-throughout theorem shows that no link can be strict.

Worked Example: Equality in a Chain of Bounds

Suppose $2\leq p\leq q\leq r\leq2$. The endpoints agree, so the equality-throughout theorem forces every term to equal $2$. In particular,

$$ p=q=r=2. $$

This conclusion can also be checked link by link: $2\leq p\leq2$ forces $p=2$, $2\leq q\leq2$ forces $q=2$, and $2\leq r\leq2$ forces $r=2$. If even one link had been strict, the endpoint comparison would have been strict, contradicting the stated equality of the endpoints.

Using Chains to Combine Bounds

A common use of transitivity is to connect two bounds that share a middle expression. The outer comparisons then follow without calculating the quantity in the middle. This is especially helpful when the middle expression is complicated, or when it is the result of a previous estimate.

Worked Example: Combining Lower and Upper Bounds

Suppose $x\geq-3$, $x\leq y$, and $y<4$. First, the assumptions $x\geq-3$ and $x\leq y$ give the connected chain $-3\leq x\leq y$. Appending $y<4$ gives

-3\leq x\leq y<4.

Finite chaining yields $-3\leq y$ from the first part of the chain, and the full chain yields $-3<4$. More informatively, the links from $x$ to $4$ include the strict comparison $y<4$, so $x<4$. Together with the original lower bound, this gives

-3\leq x<4.

For a direct check with values, $x=-3$, $y=0$ satisfy the assumptions and give $-3\leq x<4$. The strict upper bound does not make the lower bound strict: $x$ is allowed to equal $-3$.

Expressions need not be single variables for this reasoning to work. Once the comparisons have been established, transitivity applies to their values in $F$. For example, an expression such as $2t+1$ can serve as the middle quantity, just as a number or a variable can.

Worked Example: Chaining Bounds on an Expression

Suppose $-5\leq 2t+1$ and $2t+1<7$. The expression $2t+1$ is the shared middle quantity, so the comparisons can be written as a chain:

-5\leq 2t+1<7.

Now subtract $1$ from all three parts. Translation of order preserves the comparisons, giving $-6\leq2t<6$. Multiplying each part by the positive number $1/2$ preserves order as well, so

-3\leq t<3.

The endpoint directions are preserved: the original lower bound was non-strict and the upper bound strict. As a check, $t=-3$ gives $2t+1=-5$, satisfying the lower bound with equality, while $t=2$ gives $2t+1=5<7$. Transitivity connected the two bounds; the order rules for translation and positive multiplication then let us express them in terms of $t$.

What a Chain Does Not Allow

The middle terms have to connect. From $a\leq b$ and $c\leq d$ alone, one cannot conclude $a\leq d$, because these comparisons contain no information linking $b$ to $c$. For example, $10\leq11$ and $0\leq1$ are both true, but the proposed conclusion $10\leq1$ is false. To join separate comparisons, first establish a link between their relevant endpoints.

Direction matters too. Transitivity applies to a sequence that moves consistently upward, such as $a\leq b\leq c$. A statement such as $a\leq b$ together with $c\leq b$ does not form that chain: both comparisons end at $b$. The two quantities $a$ and $c$ may be ordered either way or may be equal. For example, $1\leq3$ and $2\leq3$ do not imply $1\leq2$ by transitivity alone, even though that particular conclusion happens to be true; changing the first two values to $2\leq3$ and $1\leq3$ shows that the opposite ordering is also possible.

When a chain combines strict and non-strict comparisons, preserve the symbol at each link before drawing a conclusion. The endpoint comparison is strict if at least one link is strict; it is not necessary for every link to be strict. If all links are non-strict, do not claim strictness unless some additional information supplies it.

Write the links in order.
Arrange the comparisons so that the right-hand quantity of one link is the left-hand quantity of the next.

Check the direction.
Confirm that the chain proceeds consistently from smaller to larger quantities.

Track strictness.
If any link is strict, the endpoint comparison is strict; if every link is non-strict, the endpoint comparison is non-strict.

Inspect equality cases.
If a nondecreasing chain has equal endpoints, every term in the chain must equal those endpoints.

Key takeaway. A finite connected chain $a_1\leq a_2\leq\cdots\leq a_n$ gives $a_1\leq a_n$, and any strict link makes the endpoint comparison strict. If the endpoints are equal, every link must be equality. Always verify that the comparisons connect before applying transitivity.

Check Your Understanding

Use the finite chaining theorem and the equality result to answer the following questions.

If $m\leq n<q\leq s$, what comparison follows between $m$ and $s$, and why?
If $a\leq b\leq c\leq d$ and $a=d$, what must be true of $b$ and $c$?
Do $x\leq y$ and $z\leq w$ by themselves form a chain? What additional kind of comparison could connect them?
If $-2\leq v<3$ and $3\leq u$, what comparison follows between $v$ and $u$?
Can a chain with only non-strict links have strictly ordered endpoints? Explain what the finite chaining theorem does and does not guarantee.

What You'll Learn