Irrational Numbers Concept Using Contradiction Method

 

❓ Question

Show that:

$$3\sqrt{2}$$

is irrational.

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🖼️ Solution Image

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✍️ Short Explanation

This proof is based on the fact that:

$$\sqrt{2}$$

is irrational.

We use contradiction to show that:

$$3\sqrt{2}$$

cannot be rational 💯

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🔹 Step 1 — Assume $3\sqrt{2}$ is Rational

Let:

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

where:

• $a$ and $b$ are integers
• $b \ne 0$
  

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🔹 Step 2 — Rearrange the Equation

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

Since:

• $a$ and $b$ are integers
• $3b$ is also an integer

Therefore,

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

is a rational number.

This implies:

$$\sqrt{2}$$

is rational.

But this is false because:

$$\sqrt{2}$$

is irrational.

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🔹 Step 3 — Contradiction

Our assumption is wrong.

Hence,

$$3\sqrt{2}$$

is irrational.

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✅ Final Answer

$$\boxed{3\sqrt{2}\text{ is irrational}}$$

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⭐ Key Insight

• Non-zero rational × irrational = irrational
• Contradiction method helps prove irrationality

🧠 Memory Line:

Rational number se multiply karne par irrational number irrational hi rehta hai

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📚 Related Topics

• Irrational Numbers Proof Using Contradiction Method
• Irrational Numbers Proof 
• Rational and Irrational Numbers Concept Explained