Try to prove the irrationality of 3 using the approach of proof by contradiction. Class 9

Try to prove the irrationality of 3 using the approach of proof by contradiction. Class 9

Question 1.

Try to prove the irrationality of [latex]\sqrt{3}[/latex] using the approach of proof by contradiction. Will the same approach work for [latex]\sqrt{5}[/latex], [latex]\sqrt{7}[/latex], or [latex]\sqrt{10}[/latex] ? Class 9

Solution:

Step 1: Assumption

Assume that [latex]\sqrt{3}[/latex] is a rational number.

∴ [latex]\sqrt{3}[/latex] = [latex]\frac{p}{q}[/latex] , where p and q are co-prime numbers and q ≠ 0.

Step 2: Square both sides

Step 3: Multiply both sides by q²

3q² = p² ...(i)

Step 4: Deduction for p

Since p² = 3q², p² is divisible by 3.

∴ p is also divisible by 3.

Hence, let p=3k,

where k is an integer.

Step 5: Substitute p = 3k in equation (i),

3q² = (3k)²

∴ 3q² = 9k²

Step 6: Divide both sides by 3

q² = 3k²

Step 7: Deduction for q

Since q² = 3k², q² is divisible by 3.

Therefore, q is also divisible by 3.

Step 8: Contradiction

We have shown that both p and q are divisible by 3.

Therefore, both have a common factor 3.

But this contradicts our assumption that p and q have no common factor other than 1.

Hence, our assumption that [latex]\sqrt{3}[/latex] is a rational number is wrong.

∴ [latex]\sqrt{3}[/latex] is an irrational number.

Yes, the same approach will work for [latex]\sqrt{5}[/latex], [latex]\sqrt{7}[/latex], and [latex]\sqrt{10}[/latex].