How to Prove √2 and √3 Are Irrational, Class 10 Maths

In Class IX you met the irra­tional num­bers: num­bers that sit on the num­ber line along­side the frac­tions but can never be writ­ten as a frac­tion them­selves. You were told that num­bers such as $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{15}$ are irra­tional and took it on trust. In Class 10 we finally prove it, using the Fun­da­men­tal The­o­rem of Arith­metic and a style of argu­ment called proof by con­tra­dic­tion. The same method then shows that num­bers such as $5 - \sqrt{3}$ and $3\sqrt{2}$ are irra­tional too. These proofs are stan­dard board-exam ques­tions, so this les­son sets out every step in the order you should write it.

Ratio­nal and irra­tional num­bers

Def­i­n­i­tions

A num­ber $s$ is ratio­nal if it can be writ­ten as $\displaystyle \frac{p}{q}$, where $p$ and $q$ are inte­gers and $q \neq 0$. A num­ber $s$ is irra­tional if it can­not be writ­ten in this form.

Exam­ples of irra­tional num­bers include $\sqrt{2}$, $\sqrt{3}$, $\sqrt{15}$, $\pi$, and non-ter­mi­nat­ing, non-repeat­ing dec­i­mals such as $0.10110111011110\ldots$

Irra­tional num­bers on the num­ber line

Irra­tional num­bers are real points on the num­ber line, not vague ideas. A right-angled tri­an­gle with legs 1 and 1 has hypotenuse $\sqrt{1^2 + 1^2} = \sqrt{2}$, and swing­ing that length down with a com­pass marks $\sqrt{2}$ exactly. Build­ing a sec­ond right tri­an­gle on it, with legs $\sqrt{2}$ and 1, gives a hypotenuse of $\sqrt{3}$.

A step­ping-stone: a prime that divides a square

The result

The­o­rem. Let $p$ be a prime num­ber and $a$ a pos­i­tive inte­ger. If $p$ divides $a^2$, then $p$ divides $a$.

Why it is true

By the Fun­da­men­tal The­o­rem of Arith­metic, write $a$ as a prod­uct of primes: $a = p_1 p_2 \cdots p_n$, where the primes need not be dif­fer­ent. Then

$$a^2 = (p_1 p_2 \cdots p_n)(p_1 p_2 \cdots p_n) = p_1^2 p_2^2 \cdots p_n^2.$$

Now sup­pose the prime $p$ divides $a^2$. Then $p$ appears in the prime fac­tori­sa­tion of $a^2$. Because this fac­tori­sa­tion is unique, the only primes in it are $p_1, p_2, \ldots, p_n$, so $p$ must be one of them. But $p_1, p_2, \ldots, p_n$ are exactly the primes that make up $a$, so $p$ divides $a$.

In short: squar­ing a num­ber never brings in a new prime fac­tor; it only dou­bles the pow­ers of the primes already there.

Why "prime" mat­ters

The result can fail when $p$ is not prime. For exam­ple, 4 divides $6^2 = 36$, but 4 does not divide 6. This is why every proof below uses a prime.

Proof by con­tra­dic­tion

To prove a state­ment by con­tra­dic­tion, you assume that it is false and then rea­son cor­rectly until you reach some­thing impos­si­ble. Since cor­rect rea­son­ing can­not lead from a true assump­tion to an impos­si­bil­ity, the assump­tion must be wrong, and the orig­i­nal state­ment is true.

Method for prov­ing that $\sqrt{p}$ is irra­tional ($p$ prime)

1. Assume, to the con­trary, that $\sqrt{p}$ is ratio­nal, so $\displaystyle \sqrt{p} = \frac{a}{b}$ with $a$, $b$ coprime inte­gers and $b \neq 0$.
2. Square and rearrange to get $p b^2 = a^2$.
3. Con­clude that $p$ divides $a^2$, hence $p$ divides $a$; write $a = pc$.
4. Sub­sti­tute to get $b^2 = p c^2$, so $p$ divides $b^2$, hence $p$ divides $b$.
5. State the con­tra­dic­tion: $p$ is a com­mon fac­tor of $a$ and $b$, yet they are coprime.
6. Con­clude that $\sqrt{p}$ is irra­tional.

Worked exam­ples

Exam­ple 1: $\sqrt{2}$ is irra­tional

This is one of the most famous proofs in math­e­mat­ics.

Assume, to the con­trary, that $\sqrt{2}$ is ratio­nal. Then there are inte­gers $r$ and $s$, with $s \neq 0$, such that $\displaystyle \sqrt{2} = \frac{r}{s}$. If $r$ and $s$ have a com­mon fac­tor other than 1, divide it out. So we may write

$$\displaystyle \sqrt{2} = \frac{a}{b},$$

where $a$ and $b$ are coprime (their only com­mon fac­tor is 1) and $b \neq 0$.

Mul­ti­ply­ing both sides by $b$ gives $b\sqrt{2} = a$. Squar­ing both sides:

$$2b^2 = a^2.$$

So 2 divides $a^2$. Since 2 is prime, the step­ping-stone the­o­rem tells us that 2 divides $a$. Write $a = 2c$ for some inte­ger $c$ and sub­sti­tute:

$$\begin{aligned} 2b^2 &= (2c)^2 = 4c^2 \\ b^2 &= 2c^2 \end{aligned}$$

So 2 divides $b^2$, and by the same the­o­rem 2 divides $b$.

Now both $a$ and $b$ have the fac­tor 2. This con­tra­dicts the fact that $a$ and $b$ are coprime. The con­tra­dic­tion arose from the assump­tion that $\sqrt{2}$ is ratio­nal, so that assump­tion is false.

Hence $\sqrt{2}$ is irra­tional.

Exam­ple 2: $\sqrt{3}$ is irra­tional

The same argu­ment works with 3 in place of 2.

Assume, to the con­trary, that $\sqrt{3}$ is ratio­nal. Then $\displaystyle \sqrt{3} = \frac{a}{b}$ for some coprime inte­gers $a$ and $b$ with $b \neq 0$. So $b\sqrt{3} = a$, and squar­ing gives

$$3b^2 = a^2.$$

There­fore 3 divides $a^2$, and since 3 is prime, 3 divides $a$. Write $a = 3c$ for some inte­ger $c$. Sub­sti­tut­ing:

$$\begin{aligned} 3b^2 &= 9c^2 \\ b^2 &= 3c^2 \end{aligned}$$

So 3 divides $b^2$, and hence 3 divides $b$. Now 3 is a com­mon fac­tor of $a$ and $b$, which con­tra­dicts their being coprime. So the assump­tion is false, and $\sqrt{3}$ is irra­tional.

Noth­ing in this argu­ment used any prop­erty of 3 except that it is prime. The same steps prove that $\sqrt{p}$ is irra­tional for every prime $p$.

Com­bin­ing ratio­nals and irra­tionals

In Class IX you also met two gen­eral facts:

• the sum or dif­fer­ence of a ratio­nal num­ber and an irra­tional num­ber is irra­tional;
• the prod­uct or quo­tient of a non-zero ratio­nal num­ber and an irra­tional num­ber is irra­tional.

The key tool behind both is that ratio­nal num­bers are closed under addi­tion, sub­trac­tion, mul­ti­pli­ca­tion and divi­sion by a non-zero num­ber. The exam­ples below prove par­tic­u­lar cases by con­tra­dic­tion.

Exam­ple 3: $5 - \sqrt{3}$ is irra­tional

Assume, to the con­trary, that $5 - \sqrt{3}$ is ratio­nal. Then there are coprime inte­gers $a$ and $b$, with $b \neq 0$, such that

$$\displaystyle 5 - \sqrt{3} = \frac{a}{b}.$$

Rear­rang­ing,

$$\displaystyle \sqrt{3} = 5 - \frac{a}{b} = \frac{5b - a}{b}.$$

Since $a$ and $b$ are inte­gers, $5b - a$ is an inte­ger and $b \neq 0$, so $\displaystyle \frac{5b - a}{b}$ is ratio­nal. This says $\sqrt{3}$ is ratio­nal, which con­tra­dicts Exam­ple 2. Hence $5 - \sqrt{3}$ is irra­tional.

Exam­ple 4: $3\sqrt{2}$ is irra­tional

Assume, to the con­trary, that $3\sqrt{2}$ is ratio­nal. Then there are coprime inte­gers $a$ and $b$, with $b \neq 0$, such that

$$\displaystyle 3\sqrt{2} = \frac{a}{b}.$$

Divid­ing both sides by 3,

$$\displaystyle \sqrt{2} = \frac{a}{3b}.$$

Since $a$ and $3b$ are inte­gers and $3b \neq 0$, $\displaystyle \frac{a}{3b}$ is ratio­nal. So $\sqrt{2}$ would be ratio­nal, con­tra­dict­ing Exam­ple 1. Hence $3\sqrt{2}$ is irra­tional.

Exam­ple 5: $\displaystyle \dfrac{2}{\sqrt{3}}$ is irra­tional

Assume, to the con­trary, that $\displaystyle \frac{2}{\sqrt{3}} = \frac{a}{b}$ with $a$, $b$ inte­gers and $b \neq 0$. Since $\displaystyle \frac{2}{\sqrt{3}} \neq 0$, also $a \neq 0$. Tak­ing rec­i­p­ro­cals and mul­ti­ply­ing by 2,

$$\displaystyle \sqrt{3} = \frac{2b}{a},$$

which is ratio­nal because $2b$ and $a$ are inte­gers and $a \neq 0$. This con­tra­dicts Exam­ple 2, so $\displaystyle \frac{2}{\sqrt{3}}$ is irra­tional.

What made all of this work

Every proof in this les­son has the same shape: assume the oppo­site, carry out cor­rect alge­bra, and arrive at some­thing impos­si­ble. For $\sqrt{p}$, the impos­si­bil­ity is that two num­bers cho­sen with no com­mon fac­tor turn out to share the fac­tor $p$. For com­bi­na­tions such as $5 - \sqrt{3}$, the impos­si­bil­ity is that a num­ber already proved irra­tional turns out to be ratio­nal. Learn to recog­nise this pat­tern; it appears again and again in math­e­mat­ics.

Com­mon mis­takes

• For­get­ting to say that $a$ and $b$ are coprime. With­out it there is no con­tra­dic­tion at the end.
• Writ­ing "$2$ divides $a^2$, so $2$ divides $a$" with­out giv­ing the rea­son: 2 is prime, so the step­ping-stone the­o­rem applies.
• Try­ing the same proof for a non-prime such as 4. Since 4 divides 36 but not 6, the key step fails, and in fact $\sqrt{4} = 2$ is ratio­nal.
• In proofs such as Exam­ple 3, not explain­ing why the rearranged expres­sion is ratio­nal. Say that it is a quo­tient of inte­gers with a non-zero denom­i­na­tor.
• Assum­ing that the sum of two irra­tionals is irra­tional. It need not be: $\sqrt{2} + (-\sqrt{2}) = 0$.
• Using the dec­i­mal $1.414$ as if it were exactly $\sqrt{2}$. It is only an approx­i­ma­tion, and a proof can­not rely on it.

Try these

Each answer below gives the key step; write out the full con­tra­dic­tion as in the exam­ples.

1. Prove that $\sqrt{5}$ is irra­tional. Answer: $\displaystyle \sqrt{5} = \frac{a}{b}$ gives $5b^2 = a^2$, so 5 divides $a$; with $a = 5c$, $b^2 = 5c^2$, so 5 divides $b$, con­tra­dict­ing copri­mal­ity.
2. Prove that $3 + 2\sqrt{5}$ is irra­tional. Answer: $\displaystyle 3 + 2\sqrt{5} = \frac{a}{b}$ gives $\displaystyle \sqrt{5} = \frac{a - 3b}{2b}$, which would be ratio­nal, con­tra­dict­ing ques­tion 1.
3. Prove that $\displaystyle \frac{1}{\sqrt{2}}$ is irra­tional. Answer: $\displaystyle \frac{1}{\sqrt{2}} = \frac{a}{b}$ with $a \neq 0$ gives $\displaystyle \sqrt{2} = \frac{b}{a}$, which would be ratio­nal.
4. Prove that $7\sqrt{5}$ is irra­tional. Answer: $\displaystyle 7\sqrt{5} = \frac{a}{b}$ gives $\displaystyle \sqrt{5} = \frac{a}{7b}$, which would be ratio­nal.
5. Prove that $6 + \sqrt{2}$ is irra­tional. Answer: $\displaystyle 6 + \sqrt{2} = \frac{a}{b}$ gives $\displaystyle \sqrt{2} = \frac{a - 6b}{b}$, which would be ratio­nal.

Key terms

Ratio­nal num­ber

A num­ber that can be writ­ten as $\displaystyle \frac{p}{q}$ with $p$, $q$ inte­gers and $q \neq 0$.

Irra­tional num­ber

A real num­ber that can­not be writ­ten in the form $\displaystyle \frac{p}{q}$, such as $\sqrt{2}$.

Coprime inte­gers

Inte­gers whose only com­mon pos­i­tive fac­tor is 1, such as 8 and 15.

Proof by con­tra­dic­tion

A proof that assumes a state­ment is false and shows that this leads to an impos­si­bil­ity.

Con­tra­dic­tion

A pair of state­ments that can­not both be true, such as "$a$ and $b$ are coprime" and "2 divides both $a$ and $b$".

Fun­da­men­tal The­o­rem of Arith­metic

Every com­pos­ite num­ber is a prod­uct of primes in exactly one way, apart from order.

Clo­sure

The prop­erty that adding, sub­tract­ing, mul­ti­ply­ing or divid­ing (by a non-zero num­ber) two ratio­nals gives a ratio­nal.

Com­mon ques­tions

Why do we assume that $a$ and $b$ are coprime?

Any frac­tion can be reduced to low­est terms by can­celling com­mon fac­tors, so this costs noth­ing. It is what makes the end­ing a con­tra­dic­tion: find­ing a com­mon fac­tor 2 is impos­si­ble for a frac­tion already in low­est terms.

Does the proof work for $\sqrt{6}$ or $\sqrt{8}$?

Both are irra­tional, but the argu­ment needs a prime. For $\sqrt{6}$, from $6b^2 = a^2$ use the prime 2: it divides $a^2$, so it divides $a$. Writ­ing $a = 2c$ gives $3b^2 = 2c^2$, so $3b^2$ is even; if $b$ were odd, $3b^2$ would be odd, so $b$ is even, and 2 divides both $a$ and $b$. For $\sqrt{8} = 2\sqrt{2}$, use the result for $\sqrt{2}$ and the rule for prod­ucts.

Is $\sqrt{2}$ exactly $1.414$?

No. $1.414$ is an approx­i­ma­tion; $1.414^2 = 1.999396$. Since $\sqrt{2}$ is irra­tional, its dec­i­mal expan­sion never ends and never repeats.

Can the prod­uct of two irra­tional num­bers be ratio­nal?

Yes. For exam­ple, $\sqrt{2} \times \sqrt{2} = 2$. The rule about prod­ucts only cov­ers a non-zero ratio­nal times an irra­tional.

What must a com­plete writ­ten proof con­tain?

Five things: the assump­tion that the num­ber is ratio­nal, the con­di­tion that $a$ and $b$ are coprime with $b \neq 0$, the use of the prime-divides-square the­o­rem with its rea­son, the con­tra­dic­tion, and the final con­clu­sion.

Ref­er­ences

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Hardy, G. H. and Wright, E. M. An Intro­duc­tion to the The­ory of Num­bers. Oxford Uni­ver­sity Press.
3. Bur­ton, D. M. Ele­men­tary Num­ber The­ory. McGraw-Hill Edu­ca­tion.
4. Heath, T. L. (trans.) The Thir­teen Books of Euclid's Ele­ments. Dover Pub­li­ca­tions.