prove that√2 is irrational number
Answers
Let's suppose √2 is a rational number. Then we can write it √2 = a/b where a, b are whole numbers, b not zero.
We additionally assume that this a/b is simplified to lowest terms, since that can obviously be done with any fraction. Notice that in order for a/b to be in simplest terms, both of a and b cannot be even. One or both must be odd. Otherwise, we could simplify a/b further.
From the equality √2 = a/b it follows that 2 = a2/b2, or a2 = 2 · b2. So the square of a is an even number since it is two times something.
In symbols, a = 2k
If we substitute a = 2k into the original equation 2 = a2/b2, this is what we get:
2 = (2k)2/b2
2 = 4k2/b2
2*b2 = 4k2
b2 = 2k2
This means that b2 is even, from which follows again that b itself is even. And that is a contradiction!!!
We ended at a contradiction; thus our original assumption (that √2 is rational) is not correct. Therefore √2 cannot be rational.
Answer:
Sol: 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.