Question

root2 irriation number

Answer 1

let us assume √2 is rational.
if it is rational, then there must exist two integers 'r'and 's'such that ,
√2 = r/s
if 'r'and 's'have a common factor other than 1. then we divide by the common factor to get
√2=a/b
where a and b are co-primes .
so b√2=a .
on squaring on both sides we get,
b²2=a²
therefore 2 divides a²
if two divides a² it also divides 'a'.
so we can write a=2c for some integer c.
2b²=(2c)²
2b²=4c²
b²=2c²
this means that 2 divides b² ,and 2 divides b
Therefore both a and b have 2 as common factor.
but this contradicts the fact that a and b are co-primes.
this contradiction has arisen because of our assumption that √2 is rational.
but our assumption is false.
so we can conclude that √2 is irrational.


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Answer 2

Let √2 be a rational number.

A rational number can be written in the form of p/q where p,q are co primes.

√2=p/q

p = √2q

Squaring on both sides,

p² = 2q²

2 divides p² then 2 also divides p,

So, we can write p = 2a {a is any integer)

Put p = 2a

p² = 2q²

(2a)² = 2q²

4a² = 2q²

q² = 2a²

2 divides q² then 2 also divides q.

So, we can write q = 2b {b is any integer}

As we observe, we find that both p and q have 2 as their common factor.

But this contradicts the fact that p and q are co-primes.

Therefore, our supposition is false.

So, √2 is an irrational number.

Hence proved.

Hope it helps …