Prove that Given Numbers are Irrational

 

❓ Question

Prove that the following numbers are irrational:

(i)

$$\frac{1}{\sqrt{2}}$$

(ii)

$$7\sqrt{5}$$

(iii)

$$6+\sqrt{2}$$

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🖼️ Solution Image

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✍️ Short Explanation

These proofs use the contradiction method.

We assume each number is rational and then show that it leads to a contradiction 💯

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🔹 (i) Prove that $\frac{1}{\sqrt{2}}$ is Irrational

Step 1 — Assume Rationality

Let:

$$\frac{1}{\sqrt{2}}=\frac{a}{b}$$

where $a$ and $b$ are integers.

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Step 2 — Rearrange

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

Since $\frac{b}{a}$ is rational, this implies:

$$\sqrt{2}$$

is rational.

But this is false because:

$$\sqrt{2}$$

is irrational.

Hence, our assumption is wrong.

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Final Conclusion

$$\boxed{\frac{1}{\sqrt{2}}\text{ is irrational}}$$

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🔹 (ii) Prove that $7\sqrt{5}$ is Irrational

Step 1 — Assume Rationality

Let:

$$7\sqrt{5}=\frac{a}{b}$$

where $a$ and $b$ are integers.

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Step 2 — Rearrange

$$\sqrt{5}=\frac{a}{7b}$$

Since $\frac{a}{7b}$ is rational, this implies:

$$\sqrt{5}$$

is rational.

But this is false because:

$$\sqrt{5}$$

is irrational.

Hence, our assumption is wrong.

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Final Conclusion

$$\boxed{7\sqrt{5}\text{ is irrational}}$$

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🔹 (iii) Prove that $6+\sqrt{2}$ is Irrational

Step 1 — Assume Rationality

Let:

$$6+\sqrt{2}=\frac{a}{b}$$

where $a$ and $b$ are integers.

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Step 2 — Rearrange

$$\sqrt{2}=\frac{a-6b}{b}$$

Since $\frac{a-6b}{b}$ is rational, this implies:

$$\sqrt{2}$$

is rational.

But this is false because:

$$\sqrt{2}$$

is irrational.

Hence, our assumption is wrong.

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Final Conclusion

$$\boxed{6+\sqrt{2}\text{ is irrational}}$$

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✅ Final Answer

$$\boxed{\frac{1}{\sqrt{2}},\ 7\sqrt{5},\ 6+\sqrt{2}\ \text{are irrational numbers}}$$

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⭐ Key Insight

• Rational × irrational = irrational
• Rational + irrational = irrational
• Contradiction method proves irrationality

🧠 Memory Line:

Agar irrational number rational ban raha ho, toh assumption definitely wrong hai

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📚 Related Topics

• Proof that √5 is Irrational
• Irrational Numbers Proof Using Contradiction Method
• Rational and Irrational Numbers