Question

show that root 2 is irrational​

Answer 1

Given √2 is irrational number.

Let √2 = a / b wher a,b are integers b ≠ 0

we also suppose that a / b is written in the simplest form

Now √2 = a / b ⇒ 2 = a2 / b2 ⇒ 2b2 = a2

∴ 2b2 is divisible by 2

⇒ a2 is divisible by 2

⇒ a is divisible by 2

∴ let a = 2c

a2 = 4c2 ⇒ 2b2 = 4c2 ⇒ b2 = 2c2

∴ 2c2 is divisible by 2

∴ b2 is divisible by 2

∴ b is divisible by 2

∴a are b are divisible by 2 .

this contradicts our supposition that a/b is written in the simplest form

Hence our supposition is wrong

∴ √2 is irrational number.

Answer 2

Step-by-step explanation:

assume that √2 is rational

√2=a/b

squaring on both sides

so, 2=a^2/b^2

2b^2=a^2

so, 2dividesa^2

2divides a.

put a=2c

so, 2b^2=(2c) ^2

2b^2=4c^2

b^2=2c^2

so, 2 divides b^2

2 divides b

therefore,2 is the common factor of a and b

but we assume that 1 is the common factor of a and b

by contradiction ,

√2 is irrational.

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