Irrational Numbers Proof Using Contradiction Method

 

❓ Theorem

Prove that:

$$\sqrt{2}$$

is irrational.

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🖼️ Solution Image

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✍️ Short Explanation

This proof is done using the contradiction method.

We assume that:

$$\sqrt{2}$$

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

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

Let:

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

where:

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

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🔹 Step 2 — Square Both Sides

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

👉 So,

$$a^2$$

is divisible by $2$.

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🔹 Step 3 — Apply Theorem 1.2

Since $2$ divides $a^2$,

$$2 \mid a$$

Therefore, $a$ is even.

Let:

$$a = 2c$$

for some integer $c$.

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🔹 Step 4 — Substitute Value of $a$

$$a^2 = 2b^2$$
$$(2c)^2 = 2b^2$$
$$4c^2 = 2b^2$$
$$b^2 = 2c^2$$

👉 Thus,

$$b^2$$

is also divisible by $2$.

Again using Theorem 1.2:

$$2 \mid b$$

So $b$ is also even.

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🔹 Step 5 — Contradiction

Both $a$ and $b$ are divisible by $2$.

So they have a common factor $2$.

But we assumed:

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

This is a contradiction.

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🔹 Step 6 — Final Conclusion

Our assumption is wrong.

Therefore,

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

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⭐ Key Insight

• Rational numbers in lowest form cannot have a common factor
• Contradiction proves the assumption false

🧠 Memory Line:

If both numerator and denominator become even, the rational assumption fails