Question

prove root 2 is irrational​

Answer 1

hey mate!

Assume $\sqrt{2}$ is rational, i.e. it can be expressed as a rational fraction of the form $\frac{b}{a}$, where $a$ and $b$ are two relatively prime integers. Now, since $\sqrt{2}=\frac{b}{a}$, we have $2=\frac{b^2}{a^2}$, or $b^2=2a^2$. Since $2a^2$ is even, $b^2$ must be even, and since $b^2$ is even, so is $b$. Let $b=2c$. We have $4c^2=2a^2$ and thus $a^2=2c^2$. Since $2c^2$ is even, $a^2$ is even, and since $a^2$ is even, so is a. However, two even numbers cannot be relatively prime, so $\sqrt{2}$ cannot be expressed as a rational fraction; hence $\sqrt{2}$ is irrational. $\blacksquare$

Answer 2

Hey...here's your answer:

➡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 = 2

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.