The Rational Root Theorem

Learning goals

State that $p$ divides the constant term and $q$ the leading coefficient
Build the full candidate list $\pm\tfrac{p}{q}$ , reduced and deduplicated
Test each candidate and factor out every root that is found
Explain why passing the divisibility test does not make a number a root
Prove a number irrational from a fully failed candidate list

A pattern hiding in the factored form

Take a factored quadratic whose factors have integer coefficients and expand it:

(2x - 1)(3x + 2) = 6x^2 + x - 2.

Its roots are $x = \tfrac{1}{2}$ and $x = -\tfrac{2}{3}$ . Now look at where each part of those fractions came from. The constant term $-2$ of the expanded polynomial is the product $(-1)\times 2$ of the two constants, so each root’s numerator ( $1$ and $2$ ) divides the constant term. The leading coefficient $6$ is the product $2 \times 3$ of the two $x$ -coefficients, so each root’s denominator ( $2$ and $3$ ) divides the leading coefficient.

That is not a coincidence of this example. When a polynomial with integer coefficients has a rational root, the fraction’s numerator is always locked to the constant term and its denominator to the leading coefficient. The theorem below says exactly that, and the proof shows why the two end coefficients, and only they, control where rational roots can live.

The theorem

The Rational Root Theorem. Let

P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0

be a polynomial whose coefficients $a_n, a_{n-1}, \ldots, a_0$ are all integers, with $a_n \ne 0$ . If the rational number $\tfrac{p}{q}$ , written in lowest terms (so $p$ and $q$ are integers, $q \ge 1$ , and $\gcd(p, q) = 1$ ), is a root of $P(x)$ , then

p \mid a_0 \qquad \text{and} \qquad q \mid a_n,

that is, $p$ divides the constant term and $q$ divides the leading coefficient.

Two things to keep in mind before you use it. First, the coefficients must be integers; if yours are fractions, multiply through by a common denominator, which changes no roots and hands you a polynomial the theorem can read. Second, and this is the one to remember: the theorem is a one-way implication. It says every rational root must pass the divisibility test; it does not say that every number passing the test is a root. Most passers fail when you actually substitute them in.

Why $p$ divides $a_0$ and $q$ divides $a_n$ #

The pattern is easiest to see on a cubic, and the same moves work at any degree. Take $P(x) = ax^3 + bx^2 + cx + d$ with integer coefficients and $a \ne 0$ , and suppose $\tfrac{p}{q}$ , in lowest terms ( $\gcd(p, q) = 1$ , $q \ge 1$ ), is a root. Substituting $x = \tfrac{p}{q}$ into $P(x) = 0$ and multiplying every term by $q^3$ to clear the fractions gives

a p^3 + b p^2 q + c p q^2 + d q^3 = 0.

Every term here is an integer, since the coefficients, $p$ , and $q$ all are.

One value of $p$ needs no argument at all: $p = 0$ . Lowest terms forces $\gcd(0, q) = 1$ , and the only $q \ge 1$ satisfying that is $q = 1$ , so $q \mid a$ holds because $1$ divides every integer. And $p = 0$ makes $x = 0$ the root, so $d = P(0) = 0$ , which makes $p \mid d$ read $0 \mid 0$ , true by definition. Both divisibilities hold with no prime factor in sight. What follows handles the remaining case, $p \ne 0$ .

Why $p \mid d$ . Move $dq^3$ to the other side. Every remaining term still has a factor of $p$ in it, so $p$ factors out:

d q^3 = -\left(ap^3 + bp^2q + cpq^2\right) = -p\left(ap^2 + bpq + cq^2\right).

The bracket is an integer, so $p$ divides the product $dq^3$ . Here is where lowest terms earns its keep: since $\gcd(p, q) = 1$ , no prime factor of $p$ is hiding inside $q$ , not even a repeated one, so none is hiding inside $q^3$ either. So when $p$ divides $dq^3$ , none of $p$ ‘s prime factors can be coming from the $q^3$ side of that product; every one of them, however many times it repeats in $p$ , has to be a factor of $d$ . That is exactly what $p \mid d$ means.

Why $q \mid a$ . Run the same move from the other end. Move $ap^3$ across instead, and every remaining term has a factor of $q$ :

a p^3 = -\left(bp^2q + cpq^2 + dq^3\right) = -q\left(bp^2 + cpq + dq^2\right).

So $q$ divides $ap^3$ . By the same coprimality reasoning, none of $q$ ‘s prime factors are hiding inside $p$ , so none are hiding inside $p^3$ either, and they must all divide $a$ instead. So $q \mid a$ .

Nothing here used the fact that $P$ had exactly four terms, one for each of $a$ , $b$ , $c$ , $d$ . Every term of the cleared equation except the very last one, $dq^3$ , carries a factor of $p$ , which is why isolating that last term caught $p$ ; every term except the very first one, $ap^3$ , carries a factor of $q$ , which is why isolating that first term caught $q$ . Clearing the fractions on a polynomial of any degree always leaves that same pattern, the last term is the only one missing a $p$ , and the first term is the only one missing a $q$ , so the same two moves catch both divisibilities regardless of degree.

Building the candidate list

The theorem turns root-hunting into a bookkeeping exercise. For a polynomial with integer coefficients and nonzero constant term:

List the positive divisors of the constant term $a_0$ . These are the possible numerators $p$ .
List the positive divisors of the leading coefficient $a_n$ . These are the possible denominators $q$ .
Form every quotient $\tfrac{p}{q}$ , reduce each to lowest terms, discard duplicates, and attach a $\pm$ sign to each survivor.

(If $a_0 = 0$ , skip straight to factoring out the largest power of $x$ instead: $x = 0$ is already a root, and the cofactor left behind has a nonzero constant term the theorem can read.)

Take $P(x) = 2x^3 + 3x^2 - 8x + 3$ . The constant term is $3$ , with divisors $1$ and $3$ ; the leading coefficient is $2$ , with divisors $1$ and $2$ . The quotients are $\tfrac{1}{1}, \tfrac{3}{1}, \tfrac{1}{2}, \tfrac{3}{2}$ , so the complete candidate list is

\pm 1, \quad \pm 3, \quad \pm \tfrac{1}{2}, \quad \pm \tfrac{3}{2}.

Eight numbers. Any rational root of $P$ must be one of these eight; every other rational number in existence is ruled out before you compute a single value.

Two small housekeeping points. First, the signs: roots can be negative, so the list always carries both signs even when every coefficient is positive. Second, duplicates: reducing to lowest terms in step 3 can send two different raw quotients to the same fraction. For $2x^3 - x^2 - 8x + 4$ , $a_0 = 4$ gives $p \in \{1, 2, 4\}$ and $a_n = 2$ gives $q \in \{1, 2\}$ :

Raw quotients $\tfrac{p}{q}$	Reduced
$\tfrac{1}{1}$ , $\tfrac{2}{2}$	$1$
$\tfrac{2}{1}$ , $\tfrac{4}{2}$	$2$
$\tfrac{4}{1}$	$4$
$\tfrac{1}{2}$	$\tfrac{1}{2}$

Six raw quotients collapse to four reduced values, so the candidate list is $\pm 1, \pm 2, \pm 4, \pm\tfrac{1}{2}$ : eight candidates, not twelve.

Steps 1 and 2 are one question asked twice: does this whole number divide that coefficient? The figure below is that question with its answer drawn: the Dots control sets a count from $2$ up to $12$ , the Rows of control sets a row width from $2$ up to $6$ , and the rows either come out even (a divisor) or leave dots hollow (not a divisor). Neither control reaches every divisor by itself: a row width of $1$ is never available, and testing the count against itself only works when the count is $6$ or smaller, since the row width stops at $6$ . So $1$ and the number itself belong on the list regardless of what the controls show, since every whole number is divisible by both. Whenever a constant term or leading coefficient you are working with is $12$ or smaller, set the count to it, walk the row width up through what the control reaches, note every width that leaves nothing hollow, and finish the list with $1$ and the number itself; for a coefficient larger than $12$ , or a negative one (test its absolute value), apply the same divisor test by hand.

Testing whether a row width divides the count

Dots Rows of

12 dots in rows of 6 make 2 full rows with none left over. So 12 = 2 times 6. So 6 is a factor of 12.

A number of dots tested against a row width, with any dot short of completing a row drawn hollow. A width that leaves nothing hollow is a divisor of the count, and that single test is what both halves of the candidate list are built from.

Check your understanding

According to the Rational Root Theorem, which of these numbers CANNOT be a root of $4x^3 - x + 6 = 0$ ?

-\tfrac{3}{2}

\tfrac{2}{3}

\tfrac{1}{4}

-6

The theorem is a filter, not an oracle. For 2x³ + 3x² - 8x + 3 it cuts the infinitely many rationals down to eight candidates; testing each one shows only three are roots.

When the leading coefficient is 1

A polynomial is monic when its leading coefficient is $1$ . Then the theorem’s second condition $q \mid 1$ forces $q = 1$ , and something pleasantly strong follows. Every rational root of a monic integer-coefficient polynomial is an integer, and that integer must divide the constant term. No fractions can occur at all. This special case is sometimes called the integer root theorem.

It bites quickly. Consider $P(x) = x^3 - 4x + 2$ . Any rational root must be an integer dividing $2$ , so the whole candidate list is $\pm 1, \pm 2$ . Substituting gives $P(1) = -1$ , $P(-1) = 5$ , $P(2) = 2$ , and $P(-2) = 2$ . None is zero, so this cubic has no rational roots whatsoever: whatever its roots are, every one of them is irrational or nonreal. Four quick evaluations settled a question about all infinitely many rational numbers.

From candidates to roots

The theorem supplies the shortlist; the tools from earlier in this chapter do the checking. Here is the workflow that grows out of it, and step 5 is worth reading before you trust it: a polynomial of degree $3$ or higher is not guaranteed to have any rational root at all, so the candidate list can fail completely without the factorization finishing.

Build the candidate list from $a_0$ and $a_n$ .
Test candidates one at a time, integers before fractions and small before large. Direct substitution is fine for a single quick check. Synthetic division is worth the extra setup when a candidate looks likely to succeed: by the Remainder Theorem the number at the end of the row is exactly $P(c)$ , so one tableau both tests the candidate and, on success, hands you the quotient for free.
When a candidate $c$ gives remainder $0$ , the Factor Theorem converts the discovery into algebra: $x - c$ is a factor, and $P(x) = (x - c)\,Q(x)$ with $Q$ one degree lower. Keep hunting inside $Q$ , and test $c$ itself again, since roots can repeat.
When the quotient drops to a quadratic, stop hunting. Factor it directly or run the quadratic formula, which finds all remaining roots, rational or not.
When instead every remaining candidate fails and the quotient is still degree $3$ or higher, the theorem has done all it can: it proves that quotient has no rational roots, not that the polynomial is finished factoring. What is left may still factor over the irrationals or the complex numbers, by methods outside this theorem, or it may not reduce any further by hand at all. The candidate list only ever promised to find the rational roots, when the polynomial has any.

Worked example 1 Find every root of $2x^3 + 3x^2 - 8x + 3$

The candidate list, built above, is $\pm 1, \pm 3, \pm\tfrac{1}{2}, \pm\tfrac{3}{2}$ . Start with the easiest candidate, $x = 1$ , by direct substitution:

P(1) = 2 + 3 - 8 + 3 = 0.

A root on the first try. By the Factor Theorem $x - 1$ divides $P(x)$ , and synthetic division by $1$ produces the quotient. Bring down the $2$ ; multiply by $1$ and add to $3$ to get $5$ . Then multiply by $1$ and add to $-8$ to get $-3$ ; multiply by $1$ and add to $3$ to get the remainder $0$ :

\begin{array}{c|rrrr} 1 & 2 & 3 & -8 & 3 \\ & & 2 & 5 & -3 \\ \hline & 2 & 5 & -3 & \boxed{0} \end{array}

So $P(x) = (x - 1)\left(2x^2 + 5x - 3\right)$ . The quotient is a quadratic, so the hunt is over: factor it directly,

2x^2 + 5x - 3 = (2x - 1)(x + 3),

which you can confirm by expanding. The complete factorization is $P(x) = (x-1)(2x-1)(x+3)$ , and the roots are

x = 1, \qquad x = \tfrac{1}{2}, \qquad x = -3.

All three sit on the original candidate list, exactly as the theorem promised. Note the other side of that promise: five of the eight candidates turned out not to be roots. The list narrows the search; it never claims every entry succeeds.

Check your understanding

The number $3$ is a root of $P(x) = x^3 - 2x^2 - 2x - 3$ . Divide by $(x - 3)$ using synthetic division. What is the quotient?

x^2 + x + 1

x^2 - x + 1

x^2 + x - 1

x^2 - 5x + 13

Worked example 2 Find every root of $3x^3 - 5x^2 + 5x - 2$

The constant term $-2$ has divisors $1, 2$ and the leading coefficient $3$ has divisors $1, 3$ , so the candidates are $\pm 1, \pm 2, \pm\tfrac{1}{3}, \pm\tfrac{2}{3}$ .

Half the list can be dismissed by a sign observation. For any $x < 0$ , each of the four terms $3x^3$ , $-5x^2$ , $5x$ , $-2$ is negative, so $P(x) < 0$ and no negative number can be a root. Only the four positive candidates need testing:

P(1) = 3 - 5 + 5 - 2 = 1, \qquad P(2) = 24 - 20 + 10 - 2 = 12,

P\!\left(\tfrac{1}{3}\right) = \tfrac{1}{9} - \tfrac{5}{9} + \tfrac{15}{9} - \tfrac{18}{9} = -\tfrac{7}{9}.

Three failures. Test the last candidate, $\tfrac{2}{3}$ , by synthetic division, letting the remainder be the verdict:

\begin{array}{c|rrrr} \tfrac{2}{3} & 3 & -5 & 5 & -2 \\ & & 2 & -2 & 2 \\ \hline & 3 & -3 & 3 & \boxed{0} \end{array}

The remainder is $0$ , so $\tfrac{2}{3}$ is a root and the quotient row gives $P(x) = \left(x - \tfrac{2}{3}\right)\left(3x^2 - 3x + 3\right)$ . Pull the common factor $3$ out of the quadratic and fold it into the linear factor:

P(x) = (3x - 2)\left(x^2 - x + 1\right).

The quadratic ends the hunt. Its discriminant is $(-1)^2 - 4(1)(1) = -3 < 0$ , so by the quadratic formula its roots are the complex conjugate pair from the last chapter:

x = \frac{1 \pm i\sqrt{3}}{2}.

The full set of roots is $x = \tfrac{2}{3}$ together with $x = \tfrac{1 \pm i\sqrt{3}}{2}$ . Only the first of the three is rational, so the Rational Root Theorem was never going to find the other two. The candidate list found the one root it could, and the quadratic formula found the rest.

What the theorem does not say

Two misreadings account for nearly every error with this theorem, and both are worth staring at.

The list contains candidates, not roots. The implication runs from root to divisibility, never backwards. Consider $P(x) = x^2 - 2$ . The number $2$ passes both divisibility checks, since $p = 2$ divides the constant term $-2$ and $q = 1$ divides the leading coefficient. Yet $P(2) = 2 \ne 0$ . A place on the list is an invitation to test, nothing more. In practice most candidates fail, and a polynomial may pass through its entire list without a single success.

The theorem sees only rational roots. Finish the $x^2 - 2$ story: its candidates are $\pm 1, \pm 2$ , and $P(\pm 1) = -1$ , $P(\pm 2) = 2$ , so all four fail. The correct conclusion is that $x^2 - 2$ has no rational roots. It certainly has roots: $x^2 = 2$ holds for $x = \pm\sqrt{2}$ . Those roots are irrational, so no list of fractions could ever contain them. The theorem is structurally blind to those irrational roots, just as it was blind to the nonreal pair in Worked Example 2. An empty verdict from the candidate list never means “no roots”; it means “no rational ones”, and that blindness is a weapon: if every candidate fails, you have proved the polynomial has no rational roots, which is exactly how mathematicians prove numbers irrational.

Worked example 3 Prove that $\sqrt{5}$ is irrational

The number $\sqrt{5}$ is, by definition, a root of the integer-coefficient polynomial

P(x) = x^2 - 5.

Suppose $P$ had a rational root. The polynomial is monic, so any rational root would be an integer dividing $5$ , leaving only four candidates: $\pm 1$ and $\pm 5$ . Test all four; squaring kills the sign, so two computations cover them:

P(\pm 1) = 1 - 5 = -4, \qquad P(\pm 5) = 25 - 5 = 20.

No candidate is a root, so $x^2 - 5$ has no rational roots at all. But $\sqrt{5}$ is a root. A number that is a root of a polynomial with no rational roots cannot be rational, so $\sqrt{5}$ is irrational.

The same four lines generalize. For any positive integer $n$ , a rational root of $x^2 - n$ must be an integer $d$ with $d^2 = n$ . So when that positive integer $n$ is not a perfect square, $x^2 - n$ has no rational roots and $\sqrt{n}$ is irrational. One theorem disposes of $\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{6}, \ldots$ in a single stroke.

Check your understanding

You build the full candidate list for an integer-coefficient polynomial and test every entry. None is a root. What can you conclude?

A The polynomial has no roots at all B The polynomial has no rational roots C The polynomial has no real roots D The polynomial cannot be factored

Common mistakes

Flipping the fraction. Numerators come from the constant term and denominators from the leading coefficient. For $2x^3 + 3x^2 - 8x + 3$ the candidates are $\pm\tfrac{\text{divisor of } 3}{\text{divisor of } 2}$ , never the other way around. And every candidate carries both signs, even when all coefficients are positive.
Promoting candidates to roots. Passing the divisibility test proves nothing: for $x^2 - 2$ the candidate $2$ passes and still is not a root. Every candidate must be tested by substitution or synthetic division.
Overreading an empty result. When the whole list fails, the polynomial has no rational roots. It may still have irrational or nonreal roots, and usually does.
Applying the theorem to non-integer coefficients. For $x^2 - \tfrac{1}{2}x - \tfrac{1}{2}$ , first multiply by $2$ to get $2x^2 - x - 1$ , which has the same roots and integer coefficients the theorem can read.
Ignoring a zero constant term. If $a_0 = 0$ , the condition $p \mid 0$ holds for every integer, so the theorem filters nothing. Factor out the largest power of $x$ first, then apply the theorem to the cofactor.
Forgetting repeated roots. After dividing out $x - c$ , test $c$ again on the quotient. In $4x^3 - 8x^2 + 5x - 1 = (x - 1)(2x - 1)^2$ the root $\tfrac{1}{2}$ occurs twice.

Key takeaways

The Rational Root Theorem. If a polynomial with integer coefficients has a rational root $\tfrac{p}{q}$ in lowest terms, then $p \mid a_0$ (constant term) and $q \mid a_n$ (leading coefficient).
The candidate list is every $\pm\tfrac{p}{q}$ with $p$ a positive divisor of $a_0$ and $q$ a positive divisor of $a_n$ , reduced and deduplicated. It is finite whenever $a_0 \ne 0$ .
Workflow. List the candidates, test them by synthetic division (the remainder is $P(c)$ ), peel off each discovered factor with the Factor Theorem, and finish the final quadratic with the quadratic formula.
Monic case. If $a_n = 1$ , every rational root is an integer dividing the constant term.
One-way street. Candidates are not roots; most fail. The converse of the theorem is false.
Rational eyes only. Irrational and nonreal roots never appear on the list. A fully failed list proves the polynomial has no rational roots, which is the standard modern proof that numbers like $\sqrt{5}$ are irrational.

Practice

Multiple Choice Questions (MCQ)

Progressively harder sets of questions. Each opens on its own page.

Level 1 Foundational 12 questions Start → Level 2 Intermediate 12 questions Start → Level 3 Challenge 12 questions Start →

Core practice

Practice problems at the level of the course, to be worked out on paper. Hints one at a time, then the answer or the full worked solution, with your progress kept in this browser.

More practice (optional)

Extra sets, as hard as the Challenge set. Each one opens on its own page.

More resources (optional)

Other explanations of this lesson, if you want a second take.

Polynomial division (Algebra 2) (Khan Academy)
5.5 Zeros of Polynomial Functions (OpenStax)
The Rational Roots Test (Purplemath)
Solving Polynomials (Math is Fun)

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You can skip this and keep going. Read it if you want to know more.

The full proof, for a polynomial of any degree

The core proof above runs the argument on a cubic, where the pattern is easy to see written out. Nothing about it actually depends on the degree being $3$ . Here is the same argument for $P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$ , tracking the divisibility with prime factorizations instead of leaving it to a sentence.

Proof of the Rational Root Theorem, for any degree#

Suppose $P\!\left(\tfrac{p}{q}\right) = 0$ with $\gcd(p, q) = 1$ and $q \ge 1$ . Substituting $x = \tfrac{p}{q}$ into $P(x) = 0$ gives

a_n\left(\frac{p}{q}\right)^{\!n} + a_{n-1}\left(\frac{p}{q}\right)^{\!n-1} + \cdots + a_1\!\left(\frac{p}{q}\right) + a_0 = 0.

Multiply both sides by $q^n$ , which is not zero. Each power $\left(\tfrac{p}{q}\right)^{k} = \tfrac{p^k}{q^k}$ picks up exactly the factor $q^{n-k}$ it needs, and the equation becomes

a_n p^n + a_{n-1}p^{n-1}q + \cdots + a_1 p\,q^{n-1} + a_0 q^n = 0.

Every term here is an integer, because the coefficients, $p$ , and $q$ all are. This single equation yields both conclusions, one from each end.

One value of $p$ needs no such argument: $p = 0$ . Lowest terms forces $\gcd(0, q) = 1$ , and the only $q \ge 1$ satisfying that is $q = 1$ , so $q \mid a_n$ holds because $1$ divides every integer. And $p = 0$ makes $x = 0$ the root, so $a_0 = P(0) = 0$ , which makes $p \mid a_0$ read $0 \mid 0$ , true by definition. Both divisibilities hold without appeal to any prime factorization. What follows handles the remaining case, $p \ne 0$ , where $p$ has a well-defined prime factorization to compare against $q$ ‘s.

For the first, move the last term across. Every remaining term contains at least one factor of $p$ , so $p$ factors out of the whole bracket:

\begin{aligned} a_0 q^n &= -\left(a_n p^n + a_{n-1}p^{n-1}q + \cdots + a_1 p\,q^{n-1}\right) \\ &= -p\left(a_n p^{n-1} + a_{n-1}p^{n-2}q + \cdots + a_1 q^{n-1}\right). \end{aligned}

The bracket on the right is an integer, so $p$ divides the product $a_0 q^n$ . Now the lowest-terms hypothesis does its work. If $a_0 = 0$ then $p \mid a_0$ holds trivially, so assume $a_0 \ne 0$ and compare prime factorizations. Because $\gcd(p, q) = 1$ , no prime divides both $p$ and $q$ . The prime factors of $q^n$ are exactly the prime factors of $q$ , so by that coprimality no prime factor of $p$ appears in $q^n$ at all. Take any prime $r$ that appears in the factorization of $p$ , say to the exponent $e$ . Since $p$ divides $a_0 q^n$ , the factorization of $a_0 q^n$ contains $r$ at least $e$ times, and none of those copies can come from $q^n$ , which contains no $r$ whatsoever. All of them therefore sit inside $a_0$ . This holds for every prime power in $p$ , so the entire factorization of $p$ is contained in that of $a_0$ , which is precisely the statement $p \mid a_0$ . (Signs never disturb a divisibility claim, since a factor of $-1$ can be absorbed into the cofactor.)

For the second conclusion, run the same argument from the other end. Move the first term across; every remaining term contains at least one factor of $q$ :

\begin{aligned} a_n p^n &= -\left(a_{n-1}p^{n-1}q + \cdots + a_1 p\,q^{n-1} + a_0 q^n\right) \\ &= -q\left(a_{n-1}p^{n-1} + a_{n-2}p^{n-2}q + \cdots + a_1 p\,q^{n-2} + a_0 q^{n-1}\right). \end{aligned}

So $q$ divides $a_n p^n$ . By the identical prime-factor comparison, with the roles of $p$ and $q$ exchanged, no prime factor of $q$ appears in $p^n$ . So, since $q$ divides $a_n p^n$ , every prime power in $q$ must sit inside $a_n$ , and $q \mid a_n$ .

A bit of history (optional)

Draw a square one unit on a side and rule in the diagonal. The school of Pythagoras, teaching in Greece twenty-five centuries ago, insisted that every length is a ratio of two whole numbers. That diagonal refused. Worse, the failure was not confined to one awkward line.

Each new number then demanded a separate fight. A dialogue of Plato’s, the Theaetetus, shows the geometer Theodorus grinding through the square roots of three, five, six and the rest, as far as seventeen. Case by case, one proof apiece. Then, the dialogue reports, he stopped, and nobody knows why. What is plain is that he had no general argument, only a growing stack of particular ones.

The theorem you proved today is that general argument. Hand it the polynomial $x^2 - n$ . The polynomial is monic, so any rational root must be a whole number dividing $n$ , and its square must be $n$ exactly. When $n$ is not a perfect square, no such whole number exists, so $\sqrt{n}$ is irrational. One divisibility argument settles every case Theodorus fought individually. It also settles every case beyond seventeen that he never reached.

That is what a theorem is worth. It does not make the struggle cleverer. It ends the struggle.