Proof that √5 is Irrational

 

❓ Question

Prove that:

$$\sqrt{5}$$

is irrational.

---

🖼️ Solution Image

---

✍️ Short Explanation

We prove this using the contradiction method.

We assume that:

$$\sqrt{5}$$

is rational and then arrive at a contradiction 💯

---

🔹 Step 1 — Assume $\sqrt{5}$ is Rational

Let:

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

where:

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

---

🔹 Step 2 — Square Both Sides

$$5=\frac{a^2}{b^2}$$
$$a^2=5b^2$$

Thus,

$$a^2$$

is divisible by $5$.

Using Theorem 1.2:

$$5 \mid a$$

So $a$ is divisible by $5$.

Let:

$$a=5c$$

for some integer $c$.

---

🔹 Step 3 — Substitute Value of $a$

$$a^2=5b^2$$
$$(5c)^2=5b^2$$
$$25c^2=5b^2$$
$$b^2=5c^2$$

Thus,

$$b^2$$

is divisible by $5$.

Again using Theorem 1.2:

$$5 \mid b$$

So $b$ is also divisible by $5$.

---

🔹 Step 4 — Contradiction

Both $a$ and $b$ are divisible by $5$.

So they have a common factor $5$.

But we assumed:

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

This is a contradiction.

Hence, our assumption is wrong.

---

✅ Final Answer

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

---

⭐ Key Insight

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

🧠 Memory Line:

Common factor appearing again means the rational assumption fails

---

📚 Related Topics

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