Prove that square root of 2 belongs to the set of irrational numbers.?

Question

Prove that square root of 2 belongs to the set of irrational numbers.?

Answer 1

If not then rt 2 is rational. If rt2 = a/b in the least form [ No common factor between a and b]

Then rt2 = (2b-a)/(a-b) now the numerators 2b-a<a and a-b<b. so rt2 is

another fraction whose Nr < a and Deno<b. This can further be extended and we get several rational factions equal to rt2 and nr are decreasing and deno are decreasing. which certainly contradicts the assumption.

Hence rt2 cannot be rational.

You will have to read this and understand properly. Don't show this to your teacher befor that. He may cross question you otherwise.

Answer 2

Proof by contradiction.

Assume √2 is a rational number

√2 = m/n where m and n are integers and relatively prime (definition of rational numbers and reduction of fractions)

m = √2n

Square both sides:

m² = 2n²

m² must be even and hence m must be even

Since m is even, m can be written as 2u where u is an integer

(2u)² = 2n²

4u² = 2n²

2u² = n²

2u² is certainly even

Hence n² is even

Hence n is even

If m and n are both even the fraction cannot have been reduced. 'm' and 'n' cannot be relatively prime. Hence √2 cannot be written as the ratio of two integers. Hence √2 is irrational.

Answer 3

it cant be written into a fraction and it has repeating decimals