Real Numbers

Take any composite number — say $32760$ — and start pulling it apart into prime factors with a factor tree. No matter which prime you peel off first, you always land on exactly the same collection of primes at the end: $2^3 \times 3^2 \times 5 \times 7 \times 13$ . This is not a coincidence for this one number; it is a law that governs every composite number that exists, and it is the first pillar of this chapter — the Fundamental Theorem of Arithmetic. Once you know that every number has one and only one prime "fingerprint," a whole toolkit opens up: reading off the HCF and LCM of two numbers by comparing their prime powers, deciding whether an expression like $4^n$ can ever end in the digit zero, and proving that certain sums are always composite. The second pillar looks completely different on the surface but is built on the very same fingerprint idea. In Class IX you met irrational numbers like $\sqrt{2}$ and were told, without proof, that their decimal expansions never terminate and never repeat. This chapter finally proves it — for $\sqrt{2}$ , $\sqrt{3}$ , $\sqrt{5}$ , and combinations like $5 - \sqrt{3}$ or $3\sqrt{2}$ — using a technique called proof by contradiction, powered by a small but sharp lemma: if a prime number divides a perfect square, it must divide the number being squared. That lemma is itself a direct consequence of unique prime factorisation. By the end of this chapter you will be able to move fluently between a number and its prime fingerprint, use that fingerprint to compute HCF and LCM without long division, know exactly when the HCF-times-LCM shortcut is valid, and construct a clean, rigorous contradiction proof to show that a given real number cannot be written as a ratio of integers. These are not isolated tricks — they are the two load-bearing ideas that the rest of your number theory, from polynomials to coordinate geometry, quietly leans on.

Key takeaways

The Fundamental Theorem of Arithmetic says every composite number can be expressed as a product of primes, and this factorisation is unique except for the order of the factors.
A factor tree is the standard tool for finding a number's prime factorisation, for example $32760 = 2^3 \times 3^2 \times 5 \times 7 \times 13$ .
By the prime factorisation method, HCF of two or more numbers is the product of the smallest power of each common prime factor, and LCM is the product of the greatest power of every prime factor involved.
For any two positive integers $a$ and $b$ : $\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b$ — but this product relation does not extend to three or more numbers.
Theorem 1.2: if $p$ is a prime and $p$ divides $a^2$ , then $p$ divides $a$ — this works only because $p$ is prime; it is the engine behind every irrationality proof in the chapter.
Theorem 1.3: $\sqrt{2}$ is irrational, proved by assuming the opposite (a fraction in lowest terms) and reaching a contradiction — the same method proves $\sqrt{3}$ , $\sqrt{5}$ , and any $\sqrt{p}$ for prime $p$ , irrational.
The sum, difference, product or quotient of a nonzero rational number and an irrational number is always irrational — so $5 - \sqrt{3}$ and $3\sqrt{2}$ are irrational.
The Fundamental Theorem of Arithmetic can also settle divisibility questions directly, such as showing $4^n$ can never end in the digit $0$ because its only prime factor is $2$ , never $5$ .
Key idea: because every composite number has one unique prime fingerprint (the Fundamental Theorem of Arithmetic), that fingerprint both computes HCF and LCM by comparing prime powers and, through the lemma "prime p divides a squared implies p divides a," powers a contradiction proof that numbers like root 2 are irrational.

The core insight

Every composite number breaks down into primes in exactly one way (the Fundamental Theorem of Arithmetic), and this uniqueness both gives HCF and LCM by comparing prime powers directly and, through the lemma that a prime dividing a square must divide the number itself, drives the proof-by-contradiction that $\sqrt{2}$ , $\sqrt{3}$ , $\sqrt{5}$ and similar numbers are irrational.

What this chapter covers

The chapter rests on two connected pillars. It opens by stating the Fundamental Theorem of Arithmetic — every composite number factorises into primes uniquely — and shows how to build that factorisation with a factor tree. That single fact is then turned into a practical tool: the prime factorisation method for finding the HCF and LCM of two or more positive integers, and the useful shortcut $\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b$ , together with a warning about exactly when that shortcut fails. The second half of the chapter revisits the irrational numbers met in Class IX. It proves a key divisibility lemma — if a prime $p$ divides $a^2$ then $p$ divides $a$ — and uses it, through proof by contradiction, to prove $\sqrt{2}$ is irrational, then extends the same method to $\sqrt{3}$ , $\sqrt{5}$ , and combinations such as $5 - \sqrt{3}$ and $3\sqrt{2}$ .

The core ideas

Prime factorisation and the Fundamental Theorem of Arithmetic

Every natural number greater than $1$ is either prime (its only factors are $1$ and itself) or composite (it has at least one other factor). Composite numbers can always be broken down into primes using a factor tree — repeatedly pulling out a prime factor until only primes remain. The central result of this section is:

x = p_1 \times p_2 \times \cdots \times p_n, \qquad p_1 \le p_2 \le \cdots \le p_n \text{ all prime.} \qquad (1)

where: $x$ is any composite number and $p_1, p_2, \ldots, p_n$ are its prime factors, written in ascending order and combined into powers when they repeat. This is the Fundamental Theorem of Arithmetic: every composite number can be expressed as a product of primes, and this factorisation is unique, apart from the order in which the factors are written. So $32760$ is always $2^3 \times 3^2 \times 5 \times 7 \times 13$ , however you choose to start splitting it.

Figure

A factor tree cascading down from 32760 through repeated prime splits to the final prime factorisation two cubed times three squared times five times seven times thirteen

HCF and LCM by the prime factorisation method

Once two numbers are written as products of prime powers, their HCF and LCM can be read off directly, without any long division. Compare the exponent of each prime that appears in either number:

\text{HCF} = \prod p_i^{\min(x_i, y_i)}, \qquad \text{LCM} = \prod p_i^{\max(x_i, y_i)}. \qquad (2)

where: $p_i$ runs over every prime that appears in either number, $x_i$ is its exponent in the first number and $y_i$ its exponent in the second (using exponent $0$ if a prime is absent). In words: the HCF takes the smallest power of each common prime factor, and the LCM takes the greatest power of every prime factor involved in either number. For $6 = 2^1 \times 3^1$ and $20 = 2^2 \times 5^1$ , the only common prime is $2$ , with smaller power $2^1$ , so $\text{HCF}(6,20) = 2$ ; every prime involved anywhere — $2, 3, 5$ — taken at its greatest power gives $\text{LCM}(6,20) = 2^2 \times 3^1 \times 5^1 = 60$ .

Figure

Two overlapping circles for 6 equals two times three and 20 equals two squared times five, with the overlap showing the shared prime 2 giving the HCF and the full union giving the LCM

This prime-power view also explains a shortcut that saves a great deal of arithmetic. For any two positive integers,

\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b. \qquad (3)

where: $a$ and $b$ are the two positive integers whose HCF and LCM are being compared. The relation is proved below, and it is genuinely useful: if the HCF of two numbers is easy to find but the LCM looks tedious, equation $(3)$ recovers the LCM instantly as $a \times b$ divided by the HCF. A sharp warning belongs right next to this formula: it works only for two numbers at a time. For $6, 72$ and $120$ together, $\text{HCF} = 6$ and $\text{LCM} = 360$ , so $\text{HCF} \times \text{LCM} = 2160$ , while the product $6 \times 72 \times 120 = 51840$ — nowhere close. Correct three-number formulas exist but involve the pairwise HCFs and LCMs of every pair among $p, q, r$ as well, and lie beyond what this chapter expects you to reproduce.

Revisiting irrational numbers

A number is rational if it can be written as $\dfrac{p}{q}$ for integers $p, q$ with $q \neq 0$ ; a number that cannot be written this way is irrational. In Class IX you accepted, without proof, that numbers like $\sqrt{2}$ , $\sqrt{3}$ and $0.10110111011110\ldots$ are irrational. This chapter supplies the missing proofs, and they all lean on one small lemma:

p \mid a^2 \implies p \mid a, \qquad p \text{ prime}, \ a \text{ a positive integer.} \qquad (4)

where: the symbol $\mid$ means "divides," so $p \mid a^2$ reads " $p$ divides $a^2$ ." This is Theorem 1.2 of the chapter, proved in full below using the Fundamental Theorem of Arithmetic. It depends critically on $p$ being prime: it is false for a composite divisor such as $4$ , since $4 \mid 2^2$ but $4 \nmid 2$ . With this lemma, the standard method for proving a square root irrational is proof by contradiction: assume the number is rational, written as a fraction in lowest terms, and show that assumption forces the numerator and denominator to share a common factor greater than $1$ — directly contradicting "lowest terms," so the original assumption must be false.

Two to three worked derivations

Derivation AProving that a prime dividing a square must divide the number itself

This is Theorem 1.2 of the chapter, and although it looks like a small technical fact about divisibility, it is the single most important tool in the second half of the chapter: every proof that a square root is irrational calls on this lemma at least once. The goal here is to show, using nothing beyond the Fundamental Theorem of Arithmetic, that if a prime number $p$ divides $a^2$ for some positive integer $a$ , then $p$ must already divide $a$ itself — and to see exactly where the argument breaks down if $p$ is composite instead of prime.

Step 1. Write the prime factorisation of $a$ : $a = p_1 p_2 \cdots p_n$ , where $p_1, p_2, \ldots, p_n$ are primes, not necessarily distinct from one another (repeated primes are allowed, exactly as $32760$ repeats the prime $2$ three times).

Step 2. Square both sides: $a^2 = (p_1 p_2 \cdots p_n)(p_1 p_2 \cdots p_n) = p_1^2 p_2^2 \cdots p_n^2$ . So the prime factorisation of $a^2$ consists of exactly the same primes as $a$ , each with its exponent doubled — no new prime can appear.

Step 3. We are given that $p$ divides $a^2$ . By the Fundamental Theorem of Arithmetic, $p$ must be one of the prime factors of $a^2$ , because prime factorisation is unique — a prime cannot divide a number without appearing in that number's one and only prime fingerprint. But by Step 2, the only primes appearing in $a^2$ are $p_1, p_2, \ldots, p_n$ , so $p$ must equal one of them.

Step 4. Since $p$ is one of $p_1, p_2, \ldots, p_n$ and $a = p_1 p_2 \cdots p_n$ , $p$ appears as one of the factors multiplying together to give $a$ . Hence $p$ divides $a$ . Notice exactly where primality of $p$ was essential: in Step 3, uniqueness of the prime factorisation of $a^2$ forced $p$ to coincide with one of $p_1, \ldots, p_n$ . If $p$ were composite instead — say $p = 4$ — this step fails, because $4$ is not itself one of the primes in the factorisation even though it might still divide $a^2$ ; indeed $4 \mid 2^2 = 4$ but $4 \nmid 2$ , the counterexample that shows why the lemma needs $p$ prime.

Go deeper · Derivation B · 1 minProving that root 2 is irrational by contradiction

Derivation BProving that root 2 is irrational by contradiction

This derivation carries out Theorem 1.3 of the chapter in full, turning the informal claim " $\sqrt{2}$ 's decimal expansion never terminates or repeats" into a rigorous proof. The goal is to show no fraction of integers can ever equal $\sqrt{2}$ exactly, using proof by contradiction: assume the opposite of what we want to prove, follow the logic, and show it collides with something already known to be true — here, the very meaning of "lowest terms."

Step 1. Assume, to the contrary, that $\sqrt{2}$ is rational. Then we can write $\sqrt{2} = \dfrac{a}{b}$ for integers $a$ and $b \neq 0$ that share no common factor other than $1$ — that is, $a$ and $b$ are coprime. (If they did share a common factor, we could divide it out first, so assuming coprimality costs nothing.)

Step 2. Square both sides of $\sqrt{2} = \dfrac{a}{b}$ to remove the square root: $2 = \dfrac{a^2}{b^2}$ , and rearrange to clear the denominator: $a^2 = 2b^2$ . This tells us $2$ divides $a^2$ (since $a^2$ is exactly $2$ times the integer $b^2$ ).

Step 3. By the lemma proved in the first derivation (Theorem 1.2, with $p = 2$ ), since $2$ divides $a^2$ , it follows that $2$ divides $a$ . So we can write $a = 2c$ for some integer $c$ . Substitute this back into $a^2 = 2b^2$ : $(2c)^2 = 2b^2$ , so $4c^2 = 2b^2$ , which simplifies to $b^2 = 2c^2$ .

Step 4. This new equation says $2$ divides $b^2$ , so applying the same lemma again (Theorem 1.2, $p = 2$ , this time on $b$ ) gives $2$ divides $b$ . But now both $a$ and $b$ are divisible by $2$ — directly contradicting Step 1, where $a$ and $b$ were assumed to share no common factor other than $1$ . This contradiction arose purely from assuming $\sqrt{2}$ was rational, so that assumption must be false. Therefore $\sqrt{2}$ is irrational, and the identical four-step argument, with $2$ replaced by any prime $p$ , proves $\sqrt{p}$ is irrational for every prime $p$ — in particular $\sqrt{3}$ and $\sqrt{5}$ .

Go deeper · Derivation C · 1 minProving that HCF times LCM equals the product of two positive integers

Derivation CProving that HCF times LCM equals the product of two positive integers

NCERT's worked examples verify $\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b$ numerically for a few pairs, such as $6$ and $20$ , but a single verified case is not a proof. This derivation establishes the relation in full generality for any two positive integers $a$ and $b$ , using the prime-power description of HCF and LCM from equation $(2)$ , and reveals exactly why the same identity breaks down once a third number is introduced.

Step 1. Write $a$ and $b$ using every prime $p_i$ that appears in either number, allowing exponent $0$ where a prime is absent from one of them: $a = \prod_i p_i^{x_i}$ and $b = \prod_i p_i^{y_i}$ . For $a = 6 = 2^1 3^1 5^0$ and $b = 20 = 2^2 3^0 5^1$ , this just lists the exponents of $2$ , $3$ , $5$ in both numbers.

Step 2. By the prime factorisation method, $\text{HCF}(a,b) = \prod_i p_i^{\min(x_i,y_i)}$ and $\text{LCM}(a,b) = \prod_i p_i^{\max(x_i,y_i)}$ , exactly as in equation $(2)$ .

Step 3. Multiply the two together: $\text{HCF}(a,b) \times \text{LCM}(a,b) = \prod_i p_i^{\min(x_i,y_i) + \max(x_i,y_i)}$ . Now use a simple fact about any two numbers $x_i, y_i$ : whichever is smaller and whichever is larger, their min plus their max always equals their sum, $\min(x_i,y_i) + \max(x_i,y_i) = x_i + y_i$ — if $x_i \le y_i$ this gives $x_i + y_i$ directly, and if $x_i > y_i$ the roles just swap, giving the same total.

Step 4. Substituting prime by prime gives $\text{HCF}(a,b) \times \text{LCM}(a,b) = \prod_i p_i^{x_i + y_i} = \left(\prod_i p_i^{x_i}\right)\left(\prod_i p_i^{y_i}\right) = a \times b$ , exactly equation $(3)$ , holding for every pair. This also exposes why three numbers differ: for three exponents $1, 2, 3$ , the minimum plus maximum is $1+3=4$ , not the true sum $1+2+3=6$ — the middle value is simply dropped by taking only min and max. That missing middle term is exactly why $\text{HCF}(p,q,r) \times \text{LCM}(p,q,r) \neq p \times q \times r$ in general, as confirmed by $6, 72, 120$ .

Worked examples

Board level - HCF and LCM of 96 and 404 by the prime factorisation method (Example 3 pattern)

Find the HCF of $96$ and $404$ by the prime factorisation method. Hence find their LCM.

Step 1. Prime factorise both numbers using a factor tree: $96 = 2^5 \times 3^1$ and $404 = 2^2 \times 101^1$ .

Step 2. Identify the only common prime factor: it is $2$ , appearing to the power $5$ in $96$ and to the power $2$ in $404$ . By equation $(2)$ , the HCF takes the smaller of these two powers, $2^2$ .

Step 3. So $\text{HCF}(96, 404) = 2^2 = 4$ . Now use the two-number shortcut from equation $(3)$ instead of recomputing the LCM from scratch: $\text{LCM}(96, 404) = \dfrac{96 \times 404}{\text{HCF}(96,404)} = \dfrac{38784}{4} = 9696$ .

Step 4. Quick check: every prime involved is $2$ , $3$ , $101$ , at their greatest powers $2^5$ , $3^1$ , $101^1$ ; multiplying these directly gives $32 \times 3 \times 101 = 9696$ , matching Step 3. Answer: HCF(96, 404) = 4 and LCM(96, 404) = 9696.

Board level - showing an expression is composite using the Fundamental Theorem of Arithmetic (Exercise 1.1 pattern)

Explain why $7 \times 11 \times 13 + 13$ is a composite number.

Step 1. Do not evaluate by brute arithmetic first — look for structure instead. Every term in $7 \times 11 \times 13 + 13$ contains a factor of $13$ : the first term is $7 \times 11 \times 13$ , the second is $13$ itself.

Step 2. Factor out the common $13$ : $7 \times 11 \times 13 + 13 = 13 \times (7 \times 11 + 1) = 13 \times 78$ .

Step 3. Both factors, $13$ and $78$ , are integers greater than $1$ . A number expressible as a product of two integers, each greater than $1$ , is by definition composite.

Step 4. For completeness, $78 = 2 \times 3 \times 13$ , so $7 \times 11 \times 13 + 13 = 2 \times 3 \times 13^2$ , three distinct prime factors, confirming it is composite. Answer: 7 × 11 × 13 + 13 = 2 × 3 × 13² is composite, since it factors as 13 × 78 with both factors greater than 1.

Board HOTS / foundation JEE-NEET - proving 2 minus 3 root 5 is irrational (Examples 6 and 7 combined)

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

Step 1. Assume, to the contrary, that $2 - 3\sqrt{5}$ is rational. Then it can be written as $\dfrac{a}{b}$ for coprime integers $a, b$ with $b \neq 0$ : $2 - 3\sqrt{5} = \dfrac{a}{b}$ .

Step 2. Isolate the term containing the square root: $3\sqrt{5} = 2 - \dfrac{a}{b} = \dfrac{2b - a}{b}$ .

Step 3. Divide both sides by $3$ to isolate $\sqrt{5}$ completely: $\sqrt{5} = \dfrac{2b - a}{3b}$ . The right-hand side is a ratio of the integers $(2b-a)$ and $3b$ , which makes it a rational number by definition, so this equation claims $\sqrt{5}$ is rational.

Step 4. This directly contradicts the known fact that $\sqrt{5}$ is irrational (proved by the same four-step method as $\sqrt{2}$ , using Theorem 1.2 with $p = 5$ ). The contradiction means our assumption in Step 1 was false. Answer: 2 - 3√5 is irrational. Timing note: budget about two to three minutes. This problem chains two separate NCERT results — "rational minus irrational is irrational" and "rational times irrational is irrational" — into a single contradiction proof, exactly the HOTS twist examiners like to add. The efficient route is to isolate the irrational term completely on one side before dividing away any leftover rational coefficient, so the final line reads "irrational number equals a ratio of integers" in one clean step, ready to contradict.

Common traps

The most damaging mistake in this chapter is applying Theorem 1.2 to a divisor that is not prime. The lemma " $p \mid a^2 \implies p \mid a$ " absolutely requires $p$ to be prime; for a composite number it can fail outright, as with $4 \mid 2^2$ while $4 \nmid 2$ . A closely related slip is forgetting why the coprimality (lowest-terms) assumption matters — students write "let $\sqrt{2} = a/b$ " without insisting $a, b$ are coprime, so the final "contradiction" (both divisible by $2$ ) proves nothing, since sharing a factor was never ruled out. Another common error is reaching for $\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b$ when three or more numbers are involved — it is strictly a two-number identity. Finally, many students memorise "irrational plus irrational is irrational" as a blanket rule, when it is false in general ( $\sqrt{2} + (-\sqrt{2}) = 0$ is rational); the results proved here are narrower: a nonzero rational combined with an irrational by addition, subtraction, multiplication or division always gives an irrational result.

Common trapThe lemma p divides a squared implies p divides a only works when p is prime, never for a composite divisor

The single biggest misconception in this chapter is treating "if a number divides $a^2$ , it divides $a$ " as a general divisibility fact rather than a special property of primes. The entire proof depends, in its very first line, on the Fundamental Theorem of Arithmetic guaranteeing that a prime dividing $a^2$ must already be sitting inside $a$ 's unique prime fingerprint. A composite divisor is not a single entry in that fingerprint, so the argument does not go through, and the counterexample $4 \mid 2^2$ yet $4 \nmid 2$ shows the conclusion can genuinely be false. Whenever you use Theorem 1.2 in an irrationality proof, explicitly check that the number you are calling $p$ is prime.

You should be able to…

State the Fundamental Theorem of Arithmetic precisely, including the uniqueness clause, and build a factor tree for any composite number.
Compute HCF and LCM of two or more positive integers by the prime factorisation method, using smallest and greatest prime powers.
Apply $\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b$ for two positive integers, and explain why this fails for three or more numbers.
State and prove that if a prime $p$ divides $a^2$ , then $p$ divides $a$ , and explain why the proof needs $p$ to be prime.
Prove that $\sqrt{2}$ , $\sqrt{3}$ , $\sqrt{5}$ , or $\sqrt{p}$ for any prime $p$ , is irrational, using proof by contradiction.
Prove expressions like $5 - \sqrt{3}$ or $3\sqrt{2}$ are irrational, combining a known irrational result with rational-irrational rules.
Use the Fundamental Theorem of Arithmetic to answer divisibility questions, such as whether $a^n$ can end in a particular digit.
Use common-term factoring to show an expression such as $7 \times 11 \times 13 + 13$ is composite without evaluating it fully.
Distinguish a true general result (rational times nonzero irrational is irrational) from a false over-generalisation (irrational times irrational is always irrational).

Ask the tutor: Want to walk through proving that √5 (or another prime's square root) is irrational, step by step, in your own words?

Why does the proof that root 2 is irrational insist on writing the fraction in lowest terms?

The whole proof is a contradiction argument, and the contradiction it reaches is precisely " $a$ and $b$ share a common factor of $2$ ." That statement is only a contradiction if we had earlier assumed $a$ and $b$ share no common factor other than $1$ — in other words, that the fraction $a/b$ is already in lowest terms, or that $a$ and $b$ are coprime. Skipping this assumption at the start means the final step, discovering both are even, proves nothing at all, since sharing a common factor was never ruled out. Every rational number can always be written in lowest terms by cancelling common factors first, so assuming coprimality costs nothing and is essential to make the logic airtight.

Is 1 a prime number, and why doesn't it show up in the Fundamental Theorem of Arithmetic?

No, $1$ is neither prime nor composite; it is a special case on its own. A prime number is defined to have exactly two distinct factors, $1$ and itself, but $1$ has only one factor, itself, so it fails the definition. This exclusion is deliberate and important: if $1$ were allowed as a "prime," the Fundamental Theorem of Arithmetic would instantly lose its uniqueness, because you could always multiply in extra copies of $1$ without changing the value, giving infinitely many different-looking factorisations of the same number, such as $6 = 2 \times 3 = 1 \times 2 \times 3 = 1 \times 1 \times 2 \times 3$ , and so on.

Why doesn't HCF times LCM equal the product for three numbers when it works perfectly for two?

The two-number identity $\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b$ works because, for any two exponents, the smaller one plus the larger one always adds back up to their total — nothing is lost. With three numbers, the HCF only keeps the smallest of three exponents and the LCM only keeps the largest, so the middle exponent is dropped entirely by both formulas. For exponents $1, 2, 3$ , the minimum is $1$ and the maximum is $3$ , giving $1 + 3 = 4$ , while the true sum of all three exponents is $1+2+3=6$ — the middle value $2$ never gets counted. That is exactly why $6, 72, 120$ gives $\text{HCF} \times \text{LCM} = 2160$ while $6 \times 72 \times 120 = 51840$ .

If root 2 and root 3 are both irrational, is root 2 times root 3 automatically irrational too?

Not automatically — you cannot assume "irrational times irrational is irrational" as a blanket rule, because it is false in general; for instance $\sqrt{2} \times \sqrt{2} = 2$ , a rational number, even though $\sqrt{2}$ is irrational. What this chapter actually proves is the narrower, correct statement: the product of a nonzero rational number and an irrational number is always irrational (as used for $3\sqrt{2}$ ). For $\sqrt{2} \times \sqrt{3} = \sqrt{6}$ , you would need a separate argument — here it happens to be irrational too, since $6$ is not a perfect square, but that requires its own proof-by-contradiction, not a blind general rule about "irrational times irrational."