Chapter 1 · Real Numbers

1.3 Revisiting Irrational Numbers

Assume that $\sqrt2$ is a fraction in lowest terms and follow the logic until it breaks: a proof by contradiction, powered by the Fundamental Theorem of Arithmetic, shows that $\sqrt2$, $\sqrt3$, $\sqrt5$ and numbers such as $3+2\sqrt5$ are irrational.

Draw a square floor tile with sides of exactly $1$ metre. By Pythagoras, its diagonal is $\sqrt{1^2+1^2}=\sqrt2$ metres. A calculator says $\sqrt2=1.414213562\ldots$, and a tiler would happily call it $141.4$ cm. But is there a fraction that is exactly $\sqrt2$? Perhaps a cleverer fraction such as $\tfrac{41}{29}$ or $\tfrac{577}{408}$ does the job? (Squaring them gives $1.99881\ldots$ and $2.0000060\ldots$ — very close, but not $2$.)

In Class 9 you were told that $\sqrt2$ is irrational: no fraction equals it. Trying fractions one by one can never settle this, because there are infinitely many to try. In this section we prove it, using a powerful style of argument called proof by contradiction and the Fundamental Theorem of Arithmetic from Section 1.1. The same method then shows that $\sqrt3$, $\sqrt5$, $5\sqrt2$, $3+2\sqrt5$ and many other numbers are irrational. These proofs are asked in board examinations every year, so we write each one out step by step.

Rational and irrational numbers, and proof by contradiction

📘 Definition 1.4 — Rational and irrational numbers

A number $r$ is called rational if it can be written as $r=\dfrac pq$, where $p$ and $q$ are integers and $q\ne0$. The set of rational numbers is written $\mathbb{Q}$.
A real number that is not rational is called irrational: it cannot be written as $\dfrac pq$ with integers $p$, $q$ and $q\ne0$.
Every rational number can be written in lowest terms, that is, with $p$ and $q$ co-prime ($\operatorname{HCF}(p,q)=1$): just divide $p$ and $q$ by their HCF. For example $\dfrac{84}{120}=\dfrac{84\div12}{120\div12}=\dfrac{7}{10}$.

Examples of rational numbers: $\tfrac34$, $-5=\tfrac{-5}{1}$, $0=\tfrac01$, $\sqrt{49}=7$, $\sqrt{\tfrac{9}{16}}=\tfrac34$. To show a number is rational, it is enough to write it as a fraction. To show a number is irrational, we must show that no fraction at all can equal it. That is where proof by contradiction comes in.

🧭 Problem-solving strategy — proof by contradiction

Assume the opposite of what you want to prove. (To prove "$\sqrt2$ is irrational", assume "$\sqrt2$ is rational".)
Write down what the assumption gives you in symbols (for example $\sqrt2=\tfrac ab$ with $a$, $b$ co-prime).
Reason correctly, one small step at a time, using facts you know are true.
Reach a statement that is impossible — it contradicts the assumption or a known fact.
Conclude: the assumption must be false, so the statement you wanted is true.

An everyday version: "If my friend had come home, her shoes would be at the door. Her shoes are not at the door. So she has not come home." Here the assumption "she has come home" led to something that is not true, so we reject it.

✏️ Example 1 — A first proof by contradiction

Let $a$ be a positive integer. Prove that if $a^2$ is even, then $a$ is even.

💡 Show solution▶

Assume Suppose $a^2$ is even but $a$ is not even. Then $a$ is odd, so $a=2k+1$ for some whole number $k$.

Reason $a^2=(2k+1)^2=4k^2+4k+1=2(2k^2+2k)+1$. This is $1$ more than an even number, so $a^2$ is odd.

Contradiction We were told $a^2$ is even, but we have shown it is odd. A number cannot be both.

Conclude The assumption "$a$ is odd" is false, so $a$ is even. Check with numbers: $36=6^2$ and $6$ is even; $144=12^2$ and $12$ is even. ✓

When a prime divides a square

Example 1 is the case $p=2$ of a more general fact. It works for every prime, and it is the key step in all the proofs of this section.

📐 Theorem 1.4 — A prime dividing a square

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

Proof. By the Fundamental Theorem of Arithmetic (Theorem 1.1), write $a$ as a product of primes, $a=p_1\times p_2\times\cdots\times p_n$, where some of the primes may be repeated. (If $a=1$ there are no primes, $a^2=1$, and no prime divides $1$, so there is nothing to prove.) Squaring,

$$a^2=p_1\times p_1\times p_2\times p_2\times\cdots\times p_n\times p_n .$$

This is a prime factorisation of $a^2$, and by uniqueness it is the only one. Now $p$ divides $a^2$, so $a^2=p\times m$ for some positive integer $m$; factorising $m$ gives a prime factorisation of $a^2$ that contains $p$. By uniqueness, $p$ must be one of the primes $p_1, p_2, \dots, p_n$. But each of these divides $a$. So $p$ divides $a$. $\blacksquare$

✏️ Example 2 — Using Theorem 1.4, and where it fails

(a) $3$ divides $225$. Write $225$ as a square and check Theorem 1.4.
(b) A positive integer $a$ is such that $5$ divides $a^2$. Show that $25$ divides $a^2$.
(c) $4$ divides $36=6^2$. Does $4$ divide $6$? Why does this not contradict Theorem 1.4?

💡 Show solution▶

(a) $225=15^2$, and $15=3\times5$, so $3$ divides $15$, as the theorem promises. ($225=3^2\times5^2$: the prime $3$ appears in $225$ because it appears in $15$.)

(b) $5$ is prime and $5$ divides $a^2$, so by Theorem 1.4, $5$ divides $a$. Write $a=5c$ for a positive integer $c$. Then $a^2=25c^2$, which is a multiple of $25$.

(c) No: $6\div4=1.5$. This does not contradict the theorem, because $4$ is not prime. Theorem 1.4 is only about primes. (In $36=2^2\times3^2$ the two $2$s come one from each copy of $6$.)

⚠️ Theorem 1.4 needs a prime

"$p$ divides $a^2$, so $p$ divides $a$" is true when $p$ is prime, but can fail when $p$ is composite: $4$ divides $6^2$ but not $6$; $9$ divides $3^2=9$ but not $3$; $8$ divides $4^2$ but not $4$. In every proof below, check that the number you apply the theorem to is a prime ($2, 3, 5, 7, 11, \dots$).

$\sqrt2$, $\sqrt3$ and $\sqrt5$ are irrational

Now we can prove the famous result. Read the proof slowly; every later proof in this section copies its shape.

📐 Theorem 1.5 — $\sqrt2$ is irrational

There is no fraction $\dfrac ab$, with integers $a$, $b$ and $b\ne0$, such that $\dfrac ab=\sqrt2$. In other words, $\sqrt2$ is irrational.

Proof.

Assume the opposite: suppose $\sqrt2$ is rational. Since $\sqrt2\gt0$, we can write $\sqrt2=\dfrac ab$ with positive integers $a$, $b$ that are co-prime (lowest terms).
Then $\sqrt2\,b=a$. Squaring both sides gives $2b^2=a^2$. So $2$ divides $a^2$.
$2$ is prime, so by Theorem 1.4, $2$ divides $a$. Write $a=2c$ for some positive integer $c$.
Substitute in $2b^2=a^2$: $2b^2=4c^2$, so $b^2=2c^2$. So $2$ divides $b^2$, and by Theorem 1.4 again, $2$ divides $b$.
Contradiction: $2$ divides both $a$ and $b$, so $a$ and $b$ have the common factor $2$. But we chose $a$ and $b$ co-prime.
The contradiction came from assuming that $\sqrt2$ is rational. So $\sqrt2$ is irrational. $\blacksquare$

Another way to see why step 2 is impossible uses prime "fingerprints". In $a^2$ the prime $2$ appears an even number of times (each copy of $a$ gives the same number of $2$s). In $2b^2$ it appears an even number of times from $b^2$, plus one — an odd number of times. By uniqueness of prime factorisation, $a^2$ and $2b^2$ can never be the same number. Interactive 1 shows this fingerprint test live, while it hunts for a fraction that hits $\sqrt2$.

For each denominator $q$ the lab finds the fraction $\tfrac pq$ closest to $\sqrt n$ (the purple marker). The middle line zooms in so the miss is visible. If $\tfrac pq$ were exactly $\sqrt n$, then $p^2$ and $n\times q^2$ would be equal — compare the two bars and the count of the prime $n$ in each. Press Play to try $q=1, 2, 3, \dots$ Choose $n=4$ or $n=9$ to see what a hit looks like.

$n$:$q$:5

closest fraction $\tfrac pq$—

$\sqrt n$—

$p^2$—

$n\times q^2$—

difference $p^2-nq^2$—

Nothing in the proof of Theorem 1.5 used anything special about $2$ except that it is a prime. Replace $2$ by any prime $p$ and every step still works.

📐 Theorem 1.6 — $\sqrt p$ is irrational for every prime $p$

If $p$ is a prime number, then $\sqrt p$ is irrational. In particular $\sqrt3$, $\sqrt5$, $\sqrt7$, $\sqrt{11}$, … are irrational.

✏️ Example 3 — Prove that $\sqrt3$ is irrational

Prove that $\sqrt3$ is irrational. Write the proof as you would in a board examination.

💡 Show solution▶

Assume Suppose, on the contrary, that $\sqrt3$ is rational. Then $\sqrt3=\dfrac ab$, where $a$ and $b$ are co-prime positive integers.

Square $\sqrt3\,b=a$, so $3b^2=a^2$. Hence $3$ divides $a^2$, and since $3$ is prime, $3$ divides $a$ (Theorem 1.4).

Substitute Let $a=3c$. Then $3b^2=9c^2$, so $b^2=3c^2$. Hence $3$ divides $b^2$, and so $3$ divides $b$.

Contradiction $3$ is a common factor of $a$ and $b$, which contradicts the fact that $a$ and $b$ are co-prime.

Conclude Our assumption that $\sqrt3$ is rational is wrong. Therefore $\sqrt3$ is irrational. (The proof for $\sqrt5$ is the same with $5$ in place of $3$: $5b^2=a^2$, $a=5c$, $b^2=5c^2$.)

⚠️ Common mistakes in these proofs

1. Forgetting "co-prime". The contradiction is "$a$ and $b$ have a common factor". If you never said $a$ and $b$ are co-prime, there is nothing to contradict.

2. Skipping the reason. Write "$3$ divides $a^2$, so $3$ divides $a$ because $3$ is prime (Theorem 1.4)". Examiners look for that reason.

3. Stopping half-way. Showing that $3$ divides $a$ is not enough; you must substitute $a=3c$ and show that $3$ divides $b$ too.

4. Using a calculator. "$\sqrt3=1.7320508\ldots$ does not look like a fraction" is not a proof. A calculator shows only a few digits, and the fraction $\tfrac{1351}{780}=1.7320512\ldots$ already shares its first six digits with $\sqrt3=1.7320508\ldots$

5. Trying it on a perfect square. The method does not work for $\sqrt4$ or $\sqrt9$: $4$ and $9$ are not prime, and indeed $\sqrt4=2$ and $\sqrt9=3$ are rational.

Numbers built from irrationals: $5\sqrt2$, $3+2\sqrt5$, …

Once we know $\sqrt2$, $\sqrt3$, $\sqrt5$ are irrational, many more numbers follow quickly. The idea: if the bigger number were rational, we could "undo" the arithmetic and make $\sqrt5$ (say) rational, which we know is false.

📐 Theorem 1.7 — Rational and irrational together

Let $r$ be a rational number and $s$ an irrational number. Then

$r+s$ and $r-s$ are irrational;
if $r\ne0$, then $r\times s$ and $\dfrac sr$ are irrational.

Why. Suppose $r+s=t$ were rational. Then $s=t-r$, a difference of two rational numbers, which is rational — but $s$ is irrational. Contradiction. In the same way, if $r\times s=t$ were rational with $r\ne0$, then $s=\dfrac tr$ would be rational. (The condition $r\ne0$ matters: $0\times\sqrt2=0$ is rational.)

⚠️ Two irrationals can make a rational

Theorem 1.7 says nothing about two irrational numbers, and for good reason: $\sqrt2+(-\sqrt2)=0$, $\;\sqrt2\times\sqrt8=\sqrt{16}=4$ and $(3+\sqrt5)+(3-\sqrt5)=6$ are all rational. Before you decide, simplify the expression to the form $a+b\sqrt n$.

✏️ Example 4 — Prove that $5\sqrt2$ is irrational

Given that $\sqrt2$ is irrational, prove that $5\sqrt2$ is irrational.

💡 Show solution▶

Assume Suppose, on the contrary, that $5\sqrt2$ is rational. Then $5\sqrt2=\dfrac ab$ for integers $a$, $b$ with $b\ne0$.

Rearrange Dividing by $5$: $$\sqrt2=\frac{a}{5b}.$$ Since $a$ and $5b$ are integers and $5b\ne0$, the right-hand side is rational.

Contradiction So $\sqrt2$ would be rational, which contradicts the fact that $\sqrt2$ is irrational.

Conclude Hence $5\sqrt2$ is irrational. (Here we do not need $a$, $b$ co-prime: the contradiction is with a known fact, not with lowest terms.)

✏️ Example 5 — Prove that $3+2\sqrt5$ is irrational

Given that $\sqrt5$ is irrational, prove that $3+2\sqrt5$ is irrational.

💡 Show solution▶

Assume Suppose $3+2\sqrt5$ is rational: $3+2\sqrt5=\dfrac ab$ with integers $a$, $b$, $b\ne0$.

Isolate √5 $2\sqrt5=\dfrac ab-3=\dfrac{a-3b}{b}$, so $$\sqrt5=\frac{a-3b}{2b}.$$

Contradiction $a-3b$ and $2b$ are integers with $2b\ne0$, so $\dfrac{a-3b}{2b}$ is rational. Then $\sqrt5$ would be rational — but $\sqrt5$ is irrational.

Conclude So $3+2\sqrt5$ is irrational. Check of the algebra with numbers: $3+2\sqrt5\approx7.4721$, and $\tfrac{7.4721-3}{2}\approx2.2361\approx\sqrt5$. ✓

✏️ Example 6 — Rational or irrational?

Simplify each number and decide whether it is rational or irrational: (a) $(2+\sqrt3)(2-\sqrt3)$ (b) $(1+\sqrt5)^2$ (c) $\dfrac{6}{2\sqrt3}$ (d) $\sqrt{50}-\sqrt{18}-\sqrt8$.

💡 Show solution▶

(a) $(2+\sqrt3)(2-\sqrt3)=2^2-(\sqrt3)^2=4-3=1$. Rational, although both factors are irrational.

(b) $(1+\sqrt5)^2=1+2\sqrt5+5=6+2\sqrt5$. This is rational $6$ plus the irrational $2\sqrt5$ (Theorem 1.7 twice), so irrational.

(d) $\sqrt{50}=5\sqrt2$, $\sqrt{18}=3\sqrt2$, $\sqrt8=2\sqrt2$, so the number is $(5-3-2)\sqrt2=0$. Rational.

Lesson Always simplify first. Theorem 1.7 applies only to "rational with irrational", never to two irrationals.

Start with a segment of length $1$. At its end, attach a perpendicular segment of length $1$ and join back to the centre: by Pythagoras the new spoke has length $\sqrt2$. Repeat to get $\sqrt3, \sqrt4, \sqrt5, \dots$ Spokes are green when the length is rational and red when it is irrational; the list on the right gives the reason. Press Play to grow the spiral.

Spokes up to $\sqrt m$, $m$ =17

newest spoke—

length—

rational spokes so far—

irrational spokes so far—

Exercises

Simplify each number, then say whether it is rational or irrational. Give a reason.

1 $\sqrt{81}$
2 $\sqrt{12}$
3 $\tfrac27+\sqrt3$
4 $\sqrt5\times\sqrt{20}$
5 $(3-\sqrt2)+(3+\sqrt2)$
6 $(3-\sqrt2)(3+\sqrt2)$
7 $(2+\sqrt5)^2$
8 $\dfrac{\sqrt{72}}{\sqrt2}$
9 $\sqrt{\tfrac{9}{16}}$
10 $\sqrt2+\sqrt8$
11 $(\sqrt3+1)^2-2\sqrt3$

Check: write your simplified answer as $a+b\sqrt n$ and type $a$, $b$ and $n$

$a$:$b$:$n$:

number—

approximate value—

rational?—

💡 Show solutions▶

1 $\sqrt{81}=9=\tfrac91$. Rational.

2 $\sqrt{12}=\sqrt{4\times3}=2\sqrt3$: the non-zero rational $2$ times the irrational $\sqrt3$. Irrational (Theorem 1.7).

3 Rational $\tfrac27$ plus irrational $\sqrt3$. Irrational (Theorem 1.7).

4 $\sqrt5\times\sqrt{20}=\sqrt{100}=10$. Rational.

5 The $\sqrt2$ terms cancel: $3+3=6$. Rational.

6 $3^2-(\sqrt2)^2=9-2=7$. Rational.

7 $4+4\sqrt5+5=9+4\sqrt5$: rational $9$ plus irrational $4\sqrt5$. Irrational.

8 $\sqrt{\tfrac{72}{2}}=\sqrt{36}=6$. Rational.

9 $\tfrac{\sqrt9}{\sqrt{16}}=\tfrac34$. Rational.

10 $\sqrt8=2\sqrt2$, so the sum is $3\sqrt2$. Irrational.

11 $(\sqrt3+1)^2=3+2\sqrt3+1=4+2\sqrt3$; subtracting $2\sqrt3$ leaves $4$. Rational.

Proofs. Write each proof in full, as in Examples 3–5. You may use that $\sqrt2$, $\sqrt3$ and $\sqrt5$ are irrational.

1 Prove that $\sqrt5$ is irrational.
2 Prove that $\sqrt7$ is irrational.
3 Prove that $3\sqrt2$ is irrational.
4 Prove that $2+\sqrt3$ is irrational.
5 Prove that $4-3\sqrt2$ is irrational.
6 Prove that $\dfrac{1}{\sqrt3}$ is irrational.
7 Prove that $\dfrac{\sqrt2+5}{3}$ is irrational.
8 Prove that $\sqrt6$ is irrational. (Hint: start as for $\sqrt2$ and use the prime $2$.)
9 Challenge. Prove that $\sqrt2+\sqrt3$ is irrational. (Hint: if $r=\sqrt2+\sqrt3$ were rational, square $\sqrt3=r-\sqrt2$.)

💡 Show solutions▶

1 Suppose $\sqrt5=\tfrac ab$ with $a$, $b$ co-prime positive integers. Then $5b^2=a^2$, so $5$ divides $a^2$, hence $5$ divides $a$ ($5$ is prime). Put $a=5c$: $5b^2=25c^2$, so $b^2=5c^2$ and $5$ divides $b$. Then $5$ is a common factor of $a$ and $b$ — contradiction. So $\sqrt5$ is irrational.

2 Suppose $\sqrt7=\tfrac ab$, $a$, $b$ co-prime. Then $7b^2=a^2$, so $7$ divides $a$ (as $7$ is prime); $a=7c$ gives $b^2=7c^2$, so $7$ divides $b$. Common factor $7$ — contradiction. So $\sqrt7$ is irrational.

3 Suppose $3\sqrt2=\tfrac ab$ ($a$, $b$ integers, $b\ne0$). Then $\sqrt2=\tfrac{a}{3b}$ is rational — contradiction, since $\sqrt2$ is irrational. So $3\sqrt2$ is irrational.

4 Suppose $2+\sqrt3=\tfrac ab$. Then $\sqrt3=\tfrac ab-2=\tfrac{a-2b}{b}$, which is rational — contradiction. So $2+\sqrt3$ is irrational.

5 Suppose $4-3\sqrt2=\tfrac ab$. Then $3\sqrt2=4-\tfrac ab=\tfrac{4b-a}{b}$, so $\sqrt2=\tfrac{4b-a}{3b}$, rational — contradiction. So $4-3\sqrt2$ is irrational.

6 Suppose $\tfrac1{\sqrt3}=\tfrac ab$. The number is positive, so $a\ne0$, and $\sqrt3=\tfrac ba$ is rational — contradiction. So $\tfrac1{\sqrt3}$ is irrational.

7 Suppose $\tfrac{\sqrt2+5}{3}=\tfrac ab$. Then $\sqrt2+5=\tfrac{3a}{b}$ and $\sqrt2=\tfrac{3a-5b}{b}$, rational — contradiction. So the number is irrational.

8 Suppose $\sqrt6=\tfrac ab$ with $a$, $b$ co-prime. Then $6b^2=a^2$, so $2$ divides $a^2$ and hence $2$ divides $a$. Put $a=2c$: $6b^2=4c^2$, so $3b^2=2c^2$. Then $3b^2$ is even; $3$ is odd, so $b^2$ must be even (odd $\times$ odd is odd), and by Example 1 $b$ is even. Common factor $2$ — contradiction. So $\sqrt6$ is irrational.

9 Suppose $r=\sqrt2+\sqrt3$ is rational; note $r\gt0$. Then $\sqrt3=r-\sqrt2$. Squaring: $3=r^2-2r\sqrt2+2$, so $2r\sqrt2=r^2-1$ and $\sqrt2=\dfrac{r^2-1}{2r}$. The right side is rational (made from the rational $r\ne0$), so $\sqrt2$ would be rational — contradiction. So $\sqrt2+\sqrt3$ is irrational.

III

Using Theorems 1.4 and 1.7.

1 A positive integer $a$ is such that $7$ divides $a^2$. Show that $49$ divides $a^2$.
2 Find a positive integer $a$ such that $8$ divides $a^2$ but $8$ does not divide $a$. Why does this not contradict Theorem 1.4?
3 True or false? "The sum of two irrational numbers is always irrational." Give a reason or a counter-example.
4 True or false? "The product of a non-zero rational number and an irrational number is irrational." Give a reason.
5 Find two different irrational numbers whose product is rational but whose sum is irrational.
6 If $x=2+\sqrt3$, show that $x^2-4x$ is rational, and find its value.

💡 Show solutions▶

1 $7$ is prime and divides $a^2$, so $7$ divides $a$ (Theorem 1.4). Write $a=7c$; then $a^2=49c^2$, a multiple of $49$.

2 $a=4$: $a^2=16=2\times8$, but $4\div8$ is not whole. ($a=12$, with $a^2=144=18\times8$, also works.) No contradiction: $8$ is not prime.

3 False. $\sqrt2$ and $-\sqrt2$ are irrational, but $\sqrt2+(-\sqrt2)=0$ is rational. (Also $(5+\sqrt3)+(5-\sqrt3)=10$.)

4 True (Theorem 1.7). If $r\ne0$ is rational, $s$ irrational and $rs=t$ were rational, then $s=\tfrac tr$ would be rational — contradiction.

5 For example $\sqrt2$ and $\sqrt8=2\sqrt2$: product $\sqrt{16}=4$ (rational), sum $3\sqrt2$ (irrational). Many other answers are possible, e.g. $\sqrt3$ and $2\sqrt3$ (product $6$, sum $3\sqrt3$).

6 $x^2=(2+\sqrt3)^2=7+4\sqrt3$ and $4x=8+4\sqrt3$, so $x^2-4x=7+4\sqrt3-8-4\sqrt3=-1$. Rational, value $-1$.

Irrational numbers in the real world. Use $\sqrt2\approx1.414$ where a decimal is needed.

1 A square floor tile has side $30$ cm. Find its diagonal exactly and to $2$ decimal places. Is the exact length rational or irrational?
2 A square garden plot has area $50\ \text{m}^2$. Find its side exactly and to $2$ decimal places, and say whether it is rational. For which whole-number areas from $40\ \text{m}^2$ to $60\ \text{m}^2$ is the side a rational number of metres?
3 Modelling. A sheet of A4 paper measures $210$ mm by $297$ mm. The A-series is designed so that long side $\div$ short side $=\sqrt2$ (then folding a sheet in half gives the same shape). Show that $\tfrac{297}{210}$ is not exactly $\sqrt2$ by comparing $297^2$ with $2\times210^2$, and explain why no rectangle whose sides are whole numbers of millimetres can have ratio exactly $\sqrt2$.
4 A right triangle has legs (the two shorter sides) of whole-number lengths. Find the hypotenuse and say whether it is rational, when the legs are (a) $1$ cm and $2$ cm (b) $6$ cm and $8$ cm (c) $2$ cm and $2$ cm (d) $5$ cm and $12$ cm.

💡 Show solutions▶

1 Diagonal $=\sqrt{30^2+30^2}=\sqrt{1800}=30\sqrt2$ cm $\approx30\times1.414=42.42$ cm. Exactly $30\sqrt2$ cm, which is irrational (non-zero rational $\times$ irrational).

2 Side $=\sqrt{50}=5\sqrt2$ m $\approx7.07$ m, irrational. The side $\sqrt A$ is rational only when $A$ is a perfect square, and the only perfect square from $40$ to $60$ is $49$ (side $7$ m).

3 $297^2=88209$ and $2\times210^2=88200$. They differ by $9$, so $\left(\tfrac{297}{210}\right)^2=\tfrac{88209}{44100}\ne2$ and $\tfrac{297}{210}\ne\sqrt2$ (it is about $1.41429$, very close). If sides of $p$ mm and $q$ mm had $\tfrac pq=\sqrt2$, then $\sqrt2$ would be rational, contradicting Theorem 1.5. So every real A4 sheet is only an approximation of the ideal shape.

4 (a) $\sqrt{1+4}=\sqrt5$ cm, irrational. (b) $\sqrt{36+64}=\sqrt{100}=10$ cm, rational. (c) $\sqrt{4+4}=\sqrt8=2\sqrt2$ cm, irrational. (d) $\sqrt{25+144}=\sqrt{169}=13$ cm, rational.

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case study

How this is examined: almost every board paper has a 2- or 3-mark proof that $\sqrt p$ ($p$ prime) or a number such as $a+b\sqrt p$ is irrational, plus 1-mark questions that test whether you simplify before deciding. In a proof, the marks are for the assumption (with "co-prime" where needed), the squaring step with its reason, and the stated contradiction.

E1. Board · 1 mark Which of the following is irrational? (A) $\sqrt{\tfrac{49}{64}}$ (B) $(2-\sqrt3)(2+\sqrt3)$ (C) $\sqrt2\times\sqrt{18}$ (D) $\sqrt3+\sqrt{12}$

💡 Show solution▶

Simplify (A) $\tfrac78$; (B) $4-3=1$; (C) $\sqrt{36}=6$; (D) $\sqrt3+2\sqrt3=3\sqrt3$.

Trap (B) and (C) look irrational because they contain square roots, but they simplify to whole numbers. Only (D) stays a non-zero rational times $\sqrt3$.

Answer (D) $3\sqrt3$.

E2. Board · 1 mark Assertion (A): $\sqrt2\times\sqrt8$ is a rational number. Reason (R): The product of two irrational numbers is always irrational.
(A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.

💡 Show solution▶

Check A $\sqrt2\times\sqrt8=\sqrt{16}=4$. True.

Check R False — A itself is a counter-example: two irrationals with a rational product.

Answer (C). A student who trusts R without testing it is led to (D).

E3. Board · 2 marks Given that $\sqrt3$ is irrational, prove that $5-2\sqrt3$ is irrational.

💡 Show solution▶

Assume Suppose $5-2\sqrt3$ is rational: $5-2\sqrt3=\dfrac ab$ with integers $a$, $b$ and $b\ne0$. (½ mark)

Rearrange $2\sqrt3=5-\dfrac ab=\dfrac{5b-a}{b}$, so $$\sqrt3=\frac{5b-a}{2b}.$$ (1 mark)

Conclude The right side is rational ($5b-a$ and $2b$ are integers, $2b\ne0$), so $\sqrt3$ would be rational. This contradicts the given fact. Hence $5-2\sqrt3$ is irrational. (½ mark)

E4. Board · 3 marks Prove that $\sqrt{11}$ is irrational.

💡 Show solution▶

Assume Suppose $\sqrt{11}$ is rational: $\sqrt{11}=\dfrac ab$ with $a$, $b$ co-prime positive integers. (1 mark)

Square $11b^2=a^2$, so $11$ divides $a^2$; since $11$ is prime, $11$ divides $a$. Write $a=11c$: $11b^2=121c^2$, so $b^2=11c^2$, and $11$ divides $b^2$, hence $11$ divides $b$. (1 mark)

Contradiction $11$ divides both $a$ and $b$, contradicting that they are co-prime. So $\sqrt{11}$ is irrational. (1 mark)

E5. Board · Case study A square lawn. The gardeners of a school in Kochi mark out a square lawn of area $72\ \text{m}^2$ and plan a straight path along one diagonal.
(i) Find the side of the lawn in simplest form. Is it rational or irrational? (1 mark)
(ii) Find the length of the diagonal path. Is it rational or irrational? (1 mark)
(iii) Fencing is needed round the whole lawn. Given that $\sqrt2$ is irrational, prove that the length of fencing is irrational. (2 marks)

💡 Show solution▶

(i) Side $=\sqrt{72}=\sqrt{36\times2}=6\sqrt2$ m: a non-zero rational times $\sqrt2$, so irrational.

(ii) Diagonal $=\sqrt{72+72}=\sqrt{144}=12$ m (or $\sqrt2\times6\sqrt2=12$ m). Rational — two irrational numbers multiplied to give a rational one.

(iii) Perimeter $=4\times6\sqrt2=24\sqrt2$ m. Suppose $24\sqrt2=\tfrac ab$ with integers $a$, $b$, $b\ne0$. Then $\sqrt2=\tfrac{a}{24b}$ would be rational — a contradiction. So the fencing length $24\sqrt2$ m ($\approx33.94$ m) is irrational.

🔬 Research–Teaching NexusMini project · optional, about 45 minutes

Research question. No fraction equals $\sqrt2$ — but how close can a fraction get, and how close do real A-size paper sheets come to the ideal ratio $\sqrt2$?

Why it matters. A clay tablet from ancient Babylon (about 1800 BCE, now at Yale University) gives $\sqrt2$ correct to five decimal places, and the Indian Śulba-sūtras, written for laying out fire altars, give the approximation $1+\tfrac13+\tfrac{1}{3\times4}-\tfrac{1}{3\times4\times34}=\tfrac{577}{408}$. Today the international paper standard ISO 216 (A4, A3, A5…) uses the ratio $\sqrt2$, so that enlarging A4 to A3 on a photocopier needs no trimming. Engineers, printers and computer programmers all work with good rational approximations of irrational numbers, because measurements and computers can store only fractions.

What to do.

In a spreadsheet, put $p=1$, $q=1$ in the first row. In each new row use the rule $p_{\text{new}}=p+2q$ and $q_{\text{new}}=p+q$ (in cells: =A2+2*B2 and =A2+B2). Make $10$ rows.
In two more columns compute $\tfrac pq$ and $p^2-2q^2$. What do you notice about $p^2-2q^2$? Why can it never be $0$?
Measure an A4 sheet (and an A5 or A3 sheet if you have one) with a ruler, to the nearest millimetre. Compute long side $\div$ short side and compare with $\sqrt2\approx1.41421$.
Which fraction from your table is closest to your measured ratio? Is the paper more accurate than your ruler?
Fold the A4 sheet in half. Measure again and check the new ratio.

Tools: Ruler, A4 sheet, spreadsheet (Google Sheets / Excel) or calculator

Hand in: Your table of $10$ fractions with the column $p^2-2q^2$, your paper measurements and ratios, and a paragraph (4–5 sentences) explaining why a sheet of paper can be made close to $\sqrt2$ but never exactly equal to it.

Going further. Change the rule to $p_{\text{new}}=p+3q$, $q_{\text{new}}=p+q$ and look at $p^2-3q^2$. Do the fractions approach $\sqrt3$? Try to find a rule for $\sqrt5$.

Teacher/parent note: the table starts $\tfrac11, \tfrac32, \tfrac75, \tfrac{17}{12}, \tfrac{41}{29}, \tfrac{99}{70}, \tfrac{239}{169}, \tfrac{577}{408}, \tfrac{1393}{985}, \tfrac{3363}{2378}$, and $p^2-2q^2$ alternates $-1, +1, -1, \dots$ — never $0$, exactly as Theorem 1.5 requires. A4 is specified as $210\times297$ mm, ratio $\approx1.41429$. For $\sqrt3$ the rule $p+3q$, $p+q$ gives $\tfrac11, \tfrac42, \tfrac{10}{6}, \tfrac{28}{16},\dots$; in lowest terms $\tfrac11, \tfrac21, \tfrac53, \tfrac74, \tfrac{19}{11}, \tfrac{26}{15},\dots$, with $p^2-3q^2$ equal to $-2$ or $1$ — never $0$.

🧪 Self-check quiz

0/5 correct

Q1. Which of these numbers is rational?

Q2. To prove that $\sqrt3$ is irrational, we begin by assuming $\sqrt3=\tfrac ab$, where $a$ and $b$ are

Q3. Let $p$ be a prime and $a$ a positive integer. If $p$ divides $a^2$, then

Q4. Which statement is always true?

Q5. The number $(3+\sqrt5)(3-\sqrt5)$ is

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