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Q) Prove that 4 – 2 √5 is an irrational number given that √5 is irrational.

(Q 23 – 30/5/2 – CBSE 2026 Question Paper)

Ans:

Let’s start by considering (4 – 2 √5) is a rational number (by the method of contradiction)

Step 1: ∵ (4 – 2 √5)  is a rational number

∴ then it can be expressed in the form of 23. Prove that 4 – 2

(here p and q are integers and q ≠  0)

∴ (4 – 2 √5)  = 23. Prove that 4 – 2

∴ – 2 √5   = 23. Prove that 4 – 2 – 4

∴ 2 √5   = 4 – 23. Prove that 4 – 2

∴ √5   = 2 – 23. Prove that 4 – 2 ……..(i)

Step 2: Since p and q are integers, so 23. Prove that 4 – 2 is also a rational number.

Since, in above equation (i), LHS = RHS.

∴ if RHS is rational, then LHS is also rational.

∴ √5 is a rational number.

Step 3: But it is given that √5 is an irrational number – this contradicts our finding

∴ It means that our assumption that “(4 – 2 √5) is a rational number” is wrong.

Therefore, it is confirmed that (4 – 2 √5) is an irrational number.

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