Using method of contradiction verify "√2 is irrational.”

1.	Using method of contradiction verify "√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 = 2b²or [b² = 2c²]² divide b i.e., 2 divides both a and b hence our assumption is i.e.. a and b does f lot 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 = 2b²or [b² = 2c²]² divide b

i.e., 2 divides both a and b hence our assumption is

i.e.. a and b does f lot have common contradict out statement √2 is rational is false.

∴ √2 is irrational.

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Using method of contradiction verify "√2 is irrational.”

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