Verify by the method of contradiction √2 is irrational

1.	Verify by the method of contradiction √2 is irrational
Answer» Let us assume that √2 is rational i.e.. the gives statement is false. ∴ √2 = 1/2 ≠ 0 a and b have no conunon factor. Squaring ∴ 2 = a²/b²⇒ [a² = 2b²] ⇒ 2 divides a Again put a = 2C(C ∈ Z) ⇒ ∴ (2c)² = 2b² ∴ 4c² = 2b² or [b² = 2c²]² divide b i.e., 2 divides both a and b hence our assumption is i.e.. a and b does flot have common contradict out statement √2 is rational is false. ∴ √2 is irrational.

Answer»

Let us assume that √2 is rational i.e.. the gives statement is false.

∴ √2 = 1/2 ≠ 0 a and b have no conunon factor.

Squaring ∴ 2 = a²/b²⇒ [a² = 2b²] ⇒ 2 divides a

Again put a = 2C(C ∈ Z) ⇒ ∴ (2c)² = 2b²

∴ 4c² = 2b² or [b² = 2c²]² divide b

i.e., 2 divides both a and b hence our assumption is i.e.. a and b does flot have common contradict out statement √2 is rational is false.

∴ √2 is irrational.

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Verify by the method of contradiction √2 is irrational

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