You have proved $\sqrt{2}$ and $\sqrt{3}$ are irrational.

You can see the template.

Now we formalise it for any prime $p$ and, just as importantly, figure out where the template hits its limits.

Item	Focus
General Case	Any prime $p$
The Boundary	Where the proof fails

Not every number under the square root is prime, and the board exam does ask about:

$\sqrt{6}$
$\sqrt{8}$
$\sqrt{12}$

You need to know what to do in those cases.

1. Filter -- General Proof and Boundary Awareness

The Ultimate Test 🎯

You've proved $\sqrt{2}$ is irrational. You've proved $\sqrt{3}$ is irrational. You've seen the template in action.

Now comes the real challenge:

Can you write the general proof for $\sqrt{p}$ where $p$ is any prime?
Can you recognise where this template breaks down?

Understanding the limits of a proof is just as important as knowing the proof itself.

📋 Given Info

What you have available:

Theorem 1: If $p$ is a prime and $p ∣ a^{2}$ , then $p ∣ a$ .

This theorem is your key tool — but remember, it only works when $p$ is prime.

✍️ Question

Your Challenge ✍️

(a) Write the general proof that $\sqrt{p}$ is irrational for any prime $p$ .

(b) Explain why the same template cannot be used directly for $\sqrt{12}$ . What approach would you use instead?

General proof for $\sqrt{p}$ :

Let $p$ be any prime number. Assume $\sqrt{p}$ is rational. Let $\sqrt{p} = \frac{m}{n}$ , where $m$ and $n$ are positive integers, $HCF (m, n) = 1$ , $n \neq 0$ .

✍️ Yes/No

Yes or No?

If

\text{HCF}(m, n) = 1

, can

m

and

n

both be even numbers?

Square both sides: $p = \frac{m^{2}}{n^{2}}$ , so $p n^{2} = m^{2}$ ...(i)

From (i): $p$ divides $m^{2}$ . Since $p$ is prime, by Theorem 1, $p$ divides $m$ .

✍️ MCQ

Choose one

If

p

divides

m

, which of these is the correct way to express

m

algebraically?

Let $m = p k$ for some positive integer $k$ .

Substitute in (i): $p n^{2} = (p k)^{2} = p^{2} k^{2}$ . Divide by $p$ : $n^{2} = p k^{2}$ .

So $p$ divides $n^{2}$ . Since $p$ is prime, by Theorem 1, $p$ divides $n$ .

$p$ divides $m$ and $p$ divides $n$ , contradicting $HCF (m, n) = 1$ .

Hence $\sqrt{p}$ is irrational.

For $\sqrt{12}$ :

$12 = 2^{2} \times 3$ . It is not prime, so the template does not apply directly.

✍️ MCQ

Choose one

Is

12

a prime number or a composite number?

The cleanest approach: simplify the square root first.

✍️ FIB

Fill in the blank

What is the largest perfect square that is a factor of

12

?

$\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2 \times \sqrt{3} = 2 \sqrt{3}$

Now:

$\sqrt{3}$ is irrational (proved by our template, since $3$ is prime).
$2$ is a non-zero rational number.

✍️ MCQ

Choose one

If you multiply a non-zero rational number by an irrational number, the result is always:

By the property 'non-zero rational $\times$ irrational = irrational' (this is Theorem 8 from CL-11), $2 \sqrt{3}$ is irrational.

Hence $\sqrt{12}$ is irrational.

This approach works for many non-prime radicands: factor out perfect squares, reduce to $\sqrt{prime}$ , then use the closure property.

✍️ Yes/No

Yes or No?

To prove

\sqrt{8}

is irrational, we write it as

2\sqrt{2}

. Since

2

is rational and

\sqrt{2}

is irrational, then

2\sqrt{2}

is irrational. Is this the same logic?

2. Writing the General Proof with Correct Variables

Writing the General Proof ✍️

You've already proved that $\sqrt{2}$ and $\sqrt{3}$ are irrational. You've seen the pattern — now it's time to write the general proof that works for any prime $p$ .

The general proof uses:

$p$ for the prime number
$m$ and $n$ for the fraction (where $\sqrt{p} = \frac{m}{n}$ )
$q$ for the substitution variable

Let's see if you can write the complete proof without mixing up the variables.

✍️ Question

Your Task 📝

Write the complete general proof that $\sqrt{p}$ is irrational for any prime $p$ .

Use the variables as specified:

$p$ for the prime
$m$ and $n$ for the fraction
$q$ for the substitution variable

Write out all the steps, from the assumption to the contradiction.

Here is the proof with careful variable management. Notice how each variable has a clear role.

✍️ MCQ

Choose one

Which variable represents the constant prime number we are testing?

Variables:

$p$ = the prime number (fixed throughout, never changes)
$m$ = numerator of the fraction (the first 'target' for Theorem 1)
$n$ = denominator of the fraction (the second 'target' for Theorem 1)

✍️ MCQ

Choose one

Why are

m

and

n

called 'targets' for Theorem 1?

$q$ = the quotient when $m$ is divided by $p$ (introduced during substitution)

✍️ MCQ

Choose one

If

m = p \times q

, then:

The General Proof for $\sqrt{p}$

Let $p$ be any prime number. To prove $\sqrt{p}$ is irrational, we begin by assuming the opposite.

The Assumption: Assume $\sqrt{p} = \frac{m}{n}$ , where:

$m$ and $n$ are integers ( $n \neq 0$ )
The fraction is in its simplest form, meaning $HCF (m, n) = 1$ .

✍️ Yes/No

Yes or No?

If

\text{HCF}(m, n) = 1

, can

m

and

n

both be even numbers?

Step 1: Squaring and Rearranging

Square both sides of the equation: $p = \frac{m^{2}}{n^{2}}$

Rearrange to isolate $m^{2}$ : $p n^{2} = m^{2} ...(i)$

✍️ MCQ

Choose one

Based on the equation

pn^2 = m^2

, which of these must be true?

Step 2: Applying Theorem 1

From Equation (i), we see that $p$ divides $m^{2}$ .

Since $p$ is a prime number, we apply Theorem 1: $If p ∣ m^{2}, then p ∣ m$

Let $m = p q$ . ( $q$ is a NEW variable — not $m$ , not $n$ , not $p$ .)

✍️ MCQ

Choose one

If

m = pq

, then

m^2

is equal to:

Substitute in (i): $p n^{2} = (p q)^{2} = p^{2} q^{2}$ .

Divide by $p$ : $n^{2} = p q^{2}$ .

✍️ MCQ

Choose one

Based on the equation

n^2 = pq^2

, which statement is true?

From this: $p ∣ n^{2}$ . [ $p$ is prime] $\Rightarrow$ by Theorem 1, $p ∣ n$ .

So $p ∣ m$ and $p ∣ n$ .

✍️ MCQ

Choose one

If

p

divides both

m

and

n

, then

\text{HCF}(m, n)

is:

But $HCF (m, n) = 1$ means no common factor $> 1$ .

✍️ FIB

Fill in the blank

What is the mathematical term for this logical 'clash'?

Contradiction. Hence $\sqrt{p}$ is irrational.

3. Boundary Cases -- When the Template Does Not Apply

You've now seen how the template works beautifully for proving $\sqrt{2}$ and $\sqrt{3}$ are irrational — and you understand the general pattern for any prime $p$ .

But here's the thing: CBSE doesn't only ask about primes.

You'll see questions about:

$\sqrt{6}$
$\sqrt{8}$
$\sqrt{12}$
$\sqrt{15}$

Numbers that are not prime.

So you need to know: When does the template apply directly, and when do you need a different approach?

✍️ Question

Diagnostic Check 🔍

For each of the following, state whether the direct template applies and, if not, how you would prove irrationality:

(a) $\sqrt{7}$
(b) $\sqrt{8}$
(c) $\sqrt{15}$
(d) $\sqrt{9}$

(a) $\sqrt{7}$

7 is prime. The template applies directly.

✍️ Yes/No

Yes or No?

Does the standard proof template apply directly to

\sqrt{7}

?

Standard proof: assume $\sqrt{7} = \frac{a}{b}$ (simplest form), square to get $7 b^{2} = a^{2}$ , apply Theorem 1 twice, reach contradiction.

✍️ MCQ

Choose one

If

7b^2 = a^2

, which prime number must divide

a^2

?

(b) $\sqrt{8}$ : $8 = 2^{3} = 2 \times 2 \times 2$ . Not prime.

✍️ Yes/No

Yes or No?

Can we use the standard prime proof template directly for

\sqrt{8}

?

Simplify: $\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2 \sqrt{2}$ .

✍️ MCQ

Choose one

In the expression

2\sqrt{2}

, which part is irrational?

Since $\sqrt{2}$ is irrational (proved by our template) and 2 is a non-zero rational, the product $2 \sqrt{2}$ is irrational.

The Property: 'non-zero rational $\times$ irrational = irrational' (Theorem 8).

Case (c): $\sqrt{15}$

$15 = 3 \times 5$ .

Notice that $15$ is not prime, so just like we saw with $\sqrt{12}$ and $\sqrt{8}$ , we need to be careful with how we apply our logic.

✍️ MCQ

Choose one

Is

15

a prime number or a composite number?

Approach 1: Using Divisibility

Suppose we assume $\sqrt{15}$ is rational, which leads to the equation: $15 b^{2} = a^{2}$

This implies $3 ∣ a^{2}$ , so by Theorem 1, $3 ∣ a$ .
Let $a = 3 c$ . Substituting this in: $15 b^{2} = (3 c)^{2} = 9 c^{2}$ .
Simplifying gives $5 b^{2} = 3 c^{2}$ .
Now, $3$ must divide the left side $5 b^{2}$ . Since $\gcd (3, 5) = 1$ , it must be that $3 ∣ b$ .

Since $3$ divides both $a$ and $b$ , we have a contradiction.

✍️ MCQ

Choose one

In the equation

5b^2 = 3c^2

, why does

3

divide

b^2

?

Approach 2: Decomposition

We can also use the properties of irrational numbers we already know: $\sqrt{15} = \sqrt{3} \times \sqrt{5}$

If $\sqrt{15}$ were rational, then the ratio $\frac{\sqrt{15}}{\sqrt{5}}$ would have to be rational (since the ratio of two rationals is rational).

However, $\frac{\sqrt{15}}{\sqrt{5}} = \sqrt{3}$ , and we already know that $\sqrt{3}$ is irrational. This is a contradiction.

(d) $\sqrt{9}$

Notice something different here: $9 = 3^{2}$ .

9 is not prime. And more importantly, $\sqrt{9} = 3$ , which is clearly a rational number!

✍️ MCQ

Choose one

Is

9

a prime number?

Let's see what happens if we try to force this into our irrationality template:

Assume $\sqrt{9} = \frac{a}{b}$ is rational.
Square both sides: $9 = \frac{a^{2}}{b^{2}} \Rightarrow 9 b^{2} = a^{2}$ .
Since 3 divides 9, it must divide $a^{2}$ , so $3 ∣ a$ (Theorem 1).
Let $a = 3 c$ . Substitute it back: $9 b^{2} = (3 c)^{2} = 9 c^{2}$ .
Divide by 9: $b^{2} = c^{2} \Rightarrow b = c$ .

Calculating the ratio: $\frac{a}{b} = \frac{3 c}{c} = 3$ .

✍️ Yes/No

Yes or No?

Did the template produce a contradiction for

\sqrt{9}

?

No contradiction! The algebra just confirms that $\sqrt{9} = 3$ .

This is exactly what should happen: the template should fail for perfect squares because their square roots are actually rational. The logic is consistent!

Summary: When to Use What

Radicand is prime (2, 3, 5, 7, 11, 13, ...): use template directly.

✍️ MCQ

Choose one

Which approach should you use to prove that

\sqrt{11}

is irrational?

Radicand has a perfect-square factor (8, 12, 18, 20, ...): simplify first, then use template on the remaining prime.

✍️ MCQ

Choose one

For

\sqrt{20}

, what would be your very first step in a proof?

Radicand is a product of distinct primes (6, 10, 15, ...): use the decomposition approach or argue from prime factors.
Radicand IS a perfect square (4, 9, 16, 25, ...): $\sqrt{}$ is rational, template correctly fails.

✍️ T/F

True or False?

If we try to prove

\sqrt{16}

is irrational using our template, we will eventually reach a logical contradiction.

The General sqrt(p) Proof and Its Boundaries

1. Filter -- General Proof and Boundary Awareness

The Ultimate Test 🎯

What you have available:

Your Challenge ✍️

General proof for p\sqrt{p}p​:

For 12\sqrt{12}12​:

2. Writing the General Proof with Correct Variables

Writing the General Proof ✍️

Your Task 📝

The General Proof for p\sqrt{p}p​

3. Boundary Cases -- When the Template Does Not Apply

Diagnostic Check 🔍

(a) 7\sqrt{7}7​

Case (c): 15\sqrt{15}15​

(d) 9\sqrt{9}9​

Summary: When to Use What

General proof for $\sqrt{p}$ :

For $\sqrt{12}$ :

The General Proof for $\sqrt{p}$

(a) $\sqrt{7}$

Case (c): $\sqrt{15}$

(d) $\sqrt{9}$