Irrational Numbers Proof Using Contradiction Method

 

❓ Question

Prove that:

$$\sqrt{3}$$

is irrational.

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🖼️ Solution Image

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✍️ Short Explanation

This proof uses the contradiction method.

We assume that:

$$\sqrt{3}$$

is rational and then show that this assumption leads to a contradiction 💯

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🔹 Step 1 — Assume $\sqrt{3}$ is Rational

Let:

$$\sqrt{3} = \frac{a}{b}$$

where:

• $a$ and $b$ are coprime integers
• $\text{HCF}(a,b)=1$
  

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🔹 Step 2 — Square Both Sides

$$3 = \frac{a^2}{b^2}$$
$$a^2 = 3b^2$$

👉 Thus,

$$a^2$$

is divisible by $3$.

Using Theorem 1.2:

$$3 \mid a$$

So $a$ is divisible by $3$.

Let:

$$a = 3c$$

for some integer $c$.

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🔹 Step 3 — Substitute Value of $a$

$$a^2 = 3b^2$$
$$(3c)^2 = 3b^2$$
$$9c^2 = 3b^2$$
$$b^2 = 3c^2$$

👉 Therefore,

$$b^2$$

is also divisible by $3$.

Again using Theorem 1.2:

$$3 \mid b$$

So $b$ is also divisible by $3$.

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🔹 Step 4 — Contradiction

Both $a$ and $b$ are divisible by $3$.

So they have a common factor $3$.

But we assumed:

$$\text{HCF}(a,b)=1$$

This is a contradiction.

Hence, our assumption is wrong.

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✅ Final Answer

$$\boxed{\sqrt{3}\text{ is irrational}}$$

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⭐ Key Insight

• If numerator and denominator both become divisible by the same number, they are not coprime
• Contradiction proves the assumption false

🧠 Memory Line:

Common factor appearing again means the rational assumption fails

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📚 Related Topics

• Proof irrational number
• Prime Divisibility
• Unit digit patterns