Prove that $\sqrt{3}$ is an irrational number using the method of contradiction.

Q: Prove that $\sqrt{3}$ is an irrational number using the method of contradiction.

Proof that $\sqrt{3}$ is Irrational Step 1: Assumption\nLet us assume, to the contrary, that $\sqrt{3}$ is a rational number. \nTherefore, we can find integers $a$ and $b$ ($b \neq 0$) such that: $$\sqrt{3} = \frac{a}{b}$$\nwhere $a$ and $b$ are co-prime (they have no common factor other than 1). Step 2: Squaring both sides\nRearranging the equation, we get $a = \sqrt{3}b$.\nSquaring both sides, we get: $$a^2 = 3b^2 \quad \text{--- (Equation 1)}$$ Step 3: Analysis of $a$\nFrom Equation 1, it follows that $a^2$ is divisible by 3. \nAccording to the theorem, if a prime number divides the square of a positive integer, it also divides the integer itself. \nTherefore, 3 divides $a$.\nSo, we can write $a = 3c$ for some integer $c$. Step 4: Analysis of $b$\nSubstituting $a = 3c$ in Equation 1, we get: $$(3c)^2 = 3b^2$$ $$9c^2 = 3b^2$$ $$3c^2 = b^2$$\nThis means that $b^2$ is divisible by 3, and therefore, $b$ is also divisible by 3. Step 5: Conclusion\nFrom the above steps, we see that both $a$ and $b$ have at least 3 as a common factor. \nBut this contradicts our initial assumption that $a$ and $b$ are co-prime (have no common factors other than 1).\nThis contradiction has arisen because of our incorrect assumption that $\sqrt{3}$ is rational. Hence, we conclude that $\sqrt{3}$ is irrational.

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Proof that $\sqrt{3}$ is Irrational Step 1: Assumption
Let us assume, to the contrary, that $\sqrt{3}$ is a rational number. 
Therefore, we can find integers $a$ and $b$ ($b 
eq 0$) such that: $$\sqrt{3} = \frac{a}{b}$$
where $a$ and $b$ are co-prime (they have no common factor other than 1). Step 2: Squaring both sides
Rearranging the equation, we get $a = \sqrt{3}b$.
Squaring both sides, we get: $$a^2 = 3b^2 \quad 	ext{--- (Equation 1)}$$ Step 3: Analysis of $a$
From Equation 1, it follows that $a^2$ is divisible by 3. 
According to the theorem, if a prime number divides the square of a positive integer, it also divides the integer itself. 
Therefore, 3 divides $a$.
So, we can write $a = 3c$ for some integer $c$. Step 4: Analysis of $b$
Substituting $a = 3c$ in Equation 1, we get: $$(3c)^2 = 3b^2$$ $$9c^2 = 3b^2$$ $$3c^2 = b^2$$
This means that $b^2$ is divisible by 3, and therefore, $b$ is also divisible by 3. Step 5: Conclusion
From the above steps, we see that both $a$ and $b$ have at least 3 as a common factor. 
But this contradicts our initial assumption that $a$ and $b$ are co-prime (have no common factors other than 1).
This contradiction has arisen because of our incorrect assumption that $\sqrt{3}$ is rational. Hence, we conclude that $\sqrt{3}$ is irrational.

NCERT · Class 10 · Mathematics · Real NumbersProve that $\sqrt{3}$ is an irrational number using the method of contradiction.

Proof that $\sqrt{3}$ is Irrational