NCERT Class 10 Exercise 1.2 Solutions: Irrational Numbers

Proof by con­tra­dic­tion, step by step#

1. Assume the oppo­site of what you want: the num­ber is ratio­nal.
2. Write it as $\displaystyle \frac{a}{b}$ with inte­gers $a$ and $b \neq 0$ (for a square root, also take $a$ and $b$ coprime, which is always pos­si­ble by can­celling com­mon fac­tors).
3. Rearrange or square, and fol­low the alge­bra.
4. Reach some­thing impos­si­ble: a com­mon fac­tor of coprime num­bers, or a known irra­tional num­ber equal to a ratio­nal one.
5. Con­clude that the assump­tion was false, so the num­ber is irra­tional.

For Ques­tions 2 and 3 you also use the fact that the sum, dif­fer­ence, prod­uct and quo­tient (with non-zero divi­sor) of ratio­nal num­bers are ratio­nal.

Prov­ing a square root is irra­tional#

Ques­tion 1#

Prove that $\sqrt{5}$ is irra­tional.

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

$$\displaystyle \sqrt{5} = \frac{a}{b}$$

Sup­pose $a$ and $b$ have a com­mon fac­tor other than 1. Divid­ing both by it, we may assume that $a$ and $b$ are coprime.

Then $b\sqrt{5} = a$. Squar­ing both sides,

$$5b^2 = a^2 \qquad \text{(1)}$$

So 5 divides $a^2$. Since 5 is prime, by the the­o­rem above 5 divides $a$. So we can write $a = 5c$ for some inte­ger $c$. Sub­sti­tut­ing in (1),

$$\begin{aligned} 5b^2 &= 25c^2 \\ b^2 &= 5c^2 \end{aligned}$$

So 5 divides $b^2$, and again by the the­o­rem, 5 divides $b$.

There­fore $a$ and $b$ both have 5 as a com­mon fac­tor. This con­tra­dicts the fact that $a$ and $b$ are coprime.

The con­tra­dic­tion arose from assum­ing that $\sqrt{5}$ is ratio­nal. So that assump­tion is false.

Answer: $\sqrt{5}$ is irra­tional.

Why this works: in low­est terms, $a$ and $b$ can­not both be mul­ti­ples of 5, but the equa­tion $a^2 = 5b^2$ forces both to be. As a numer­i­cal check, $\sqrt{5} = 2.2360679\ldots$, a dec­i­mal that nei­ther ter­mi­nates nor repeats.

Num­bers built from a known irra­tional#

Ques­tion 2#

Prove that $3 + 2\sqrt{5}$ is irra­tional.

Assume, to the con­trary, that $3 + 2\sqrt{5}$ is ratio­nal. Then there are inte­gers $a$ and $b$ ($b \neq 0$) such that

$$\displaystyle 3 + 2\sqrt{5} = \frac{a}{b}$$

Rear­rang­ing,

$$\displaystyle \begin{aligned} 2\sqrt{5} &= \frac{a}{b} - 3 = \frac{a - 3b}{b} \\ \sqrt{5} &= \frac{a - 3b}{2b} \end{aligned}$$

Since $a$ and $b$ are inte­gers, $a - 3b$ is an inte­ger and $2b$ is a non-zero inte­ger. So $\displaystyle \frac{a - 3b}{2b}$ is ratio­nal, and there­fore $\sqrt{5}$ would be ratio­nal.

This con­tra­dicts the fact, proved in Ques­tion 1, that $\sqrt{5}$ is irra­tional. So our assump­tion is false.

Answer: $3 + 2\sqrt{5}$ is irra­tional.

Check of the alge­bra: mul­ti­ply­ing $\displaystyle \sqrt{5} = \frac{a - 3b}{2b}$ by $2b$ gives $2b\sqrt{5} = a - 3b$, so $a = b(3 + 2\sqrt{5})$, which is the start­ing equa­tion. Numer­i­cally, $3 + 2\sqrt{5} = 7.4721359\ldots$

Ques­tion 3 (i)#

Prove that the fol­low­ing are irra­tional: (i) $\displaystyle \frac{1}{\sqrt{2}}$

Assume, to the con­trary, that $\displaystyle \frac{1}{\sqrt{2}}$ is ratio­nal. Then there are inte­gers $a$ and $b$ ($b \neq 0$) with

$$\displaystyle \frac{1}{\sqrt{2}} = \frac{a}{b}$$

Since $\displaystyle \frac{1}{\sqrt{2}} \neq 0$, also $a \neq 0$. Tak­ing rec­i­p­ro­cals,

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

Here $b$ and $a$ are inte­gers with $a \neq 0$, so $\displaystyle \frac{b}{a}$ is ratio­nal, which would make $\sqrt{2}$ ratio­nal.

But $\sqrt{2}$ is irra­tional (proved in the text­book by exactly the method of Ques­tion 1, with 2 in place of 5). This con­tra­dic­tion shows the assump­tion is false.

Answer: $\displaystyle \frac{1}{\sqrt{2}}$ is irra­tional.

Another way to see it: $\displaystyle \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$, so if it were ratio­nal, twice it, which is $\sqrt{2}$, would be ratio­nal too. Numer­i­cally, $\displaystyle \frac{1}{\sqrt{2}} = 0.7071067\ldots$

Ques­tion 3 (ii)#

(ii) $7\sqrt{5}$

Assume, to the con­trary, that $7\sqrt{5}$ is ratio­nal. Then there are inte­gers $a$ and $b$ ($b \neq 0$) with

$$\displaystyle 7\sqrt{5} = \frac{a}{b}, \qquad \text{so} \qquad \sqrt{5} = \frac{a}{7b}$$

Since $a$ and $7b$ are inte­gers and $7b \neq 0$, the right-hand side is ratio­nal, which would make $\sqrt{5}$ ratio­nal.

This con­tra­dicts Ques­tion 1. So the assump­tion is false.

Answer: $7\sqrt{5}$ is irra­tional.

Numer­i­cally, $7\sqrt{5} = 15.6524758\ldots$

Ques­tion 3 (iii)#

(iii) $6 + \sqrt{2}$

Assume, to the con­trary, that $6 + \sqrt{2}$ is ratio­nal. Then there are inte­gers $a$ and $b$ ($b \neq 0$) with

$$\displaystyle 6 + \sqrt{2} = \frac{a}{b}$$

Rear­rang­ing,

$$\displaystyle \sqrt{2} = \frac{a}{b} - 6 = \frac{a - 6b}{b}$$

Since $a - 6b$ and $b$ are inte­gers with $b \neq 0$, the right-hand side is ratio­nal, which would make $\sqrt{2}$ ratio­nal.

But $\sqrt{2}$ is irra­tional. This con­tra­dic­tion shows that the assump­tion is false.

Answer: $6 + \sqrt{2}$ is irra­tional.

Check of the alge­bra: $b\sqrt{2} = a - 6b$ gives $a = b(6 + \sqrt{2})$, the start­ing equa­tion. Numer­i­cally, $6 + \sqrt{2} = 7.4142135\ldots$

Why Ques­tions 2 and 3 work: in each one we iso­late the square root. The other side is then built from inte­gers using only addi­tion, sub­trac­tion, mul­ti­pli­ca­tion and divi­sion by a non-zero num­ber, so it is ratio­nal. A known irra­tional num­ber can­not equal a ratio­nal one. This is the gen­eral fact that a non-zero ratio­nal num­ber added to, sub­tracted from, mul­ti­plied by or divided into an irra­tional num­ber always gives an irra­tional num­ber.

Key terms#

Ratio­nal num­ber

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

Irra­tional num­ber

A real num­ber that can­not be writ­ten in the form $\displaystyle \frac{p}{q}$; its dec­i­mal expan­sion nei­ther ter­mi­nates nor repeats. Exam­ples: $\sqrt{2}$, $\sqrt{5}$.

Coprime inte­gers

Inte­gers whose only com­mon fac­tor is 1, as in a frac­tion writ­ten in its low­est terms.

Proof by con­tra­dic­tion

A proof that assumes the state­ment is false and shows that this assump­tion leads to some­thing impos­si­ble.

Prime divi­sor the­o­rem

If a prime $p$ divides $a^2$ for a pos­i­tive inte­ger $a$, then $p$ divides $a$.

Fun­da­men­tal The­o­rem of Arith­metic

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

Com­mon ques­tions#

Why can we assume that $a$ and $b$ are coprime?#

Any frac­tion can be reduced to low­est terms by divid­ing numer­a­tor and denom­i­na­tor by their HCF. So if $\sqrt{5}$ were a frac­tion at all, it would also be a frac­tion in low­est terms.

Why do Ques­tions 2 and 3 not need the coprime con­di­tion?#

Their con­tra­dic­tion does not come from a com­mon fac­tor. It comes from show­ing that $\sqrt{5}$ or $\sqrt{2}$ would be ratio­nal, which is already known to be false. Adding the coprime con­di­tion is harm­less but not needed.

Is it enough to say the dec­i­mal of $\sqrt{5}$ never ends?#

No. A cal­cu­la­tor shows only finitely many dig­its, so it can­not prove that the dec­i­mal never ends or never repeats. The dec­i­mal is a use­ful check, but the proof must be the alge­braic argu­ment.

Can I use the same proof for $\sqrt{4}$?#

No, because 4 is not prime and the the­o­rem needs a prime. In fact $\sqrt{4} = 2$ is ratio­nal. The method works for $\sqrt{p}$ when $p$ is prime.

Why must $a \neq 0$ in Ques­tion 3 (i)?#

We divide by $a$ when we take rec­i­p­ro­cals. That is only allowed because $\displaystyle \frac{1}{\sqrt{2}}$ is not zero, so $a$ can­not be zero.

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. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.