Question

Prove √2 is an irrational number in simple language

Answer 1

To prove root 2 is irrational, let us assume root2 is rational.
Since root2 is rational by our assumption, root2 can be written as p/q(all rational no.s can be written in a fractional form) where q is not equal to 0 and where p,q are coprimes( are in the simplest form)
Root2. = p/q
Squaring on both sides,
2 = p^2/q^2
Crossmultiplying we get,
2q^2 = p^2. ------1
Which implies, p^2 is divisible by 2. Which implies p is divisible by 2
Since p is divisible by 2, it can be written as,
p=2x
Squaring on both sides,
p^2 = 4x^2. --------2
From 1&2,
2q^2=4x^2
q^2. =. 2x^2
Which implies q^2 is divisible by 2. Which implies q is divisible by 2
Since both p and q are divisible by 2. This means the fraction p/q is not in its simplest form. This contradicts our assumption that p/q is in the simplest form. This in turn contradicts our assumption that root2 is rational. Since root2 is not rational it is irrational.

Answer 2

Answer :

To prove,

√2 is an irrational no.

Proof :

Let √2 be a rational number in the form of p / q where q is not equal to zero at p and q are co-prime integers.

√2 = p/q

Whole sqauring both sides of this equation :-

2 = p^2/q^2

p^2 = 2q^2 (I)

From (I),

2 divided p^2

So, p divides p. (a)

Now , let p= 2k where k is any integer.

Substituting the values , we get :-

(2k)^2 = 2q^2

4k^2 = 2q^2

q^2 = 2k^2 (ii)

From (ii),

2 divides q^2.

Therefore, 2 divides q also. (b)

From statements (a) and (b) , we can say that :-

p and q have a common factor namely 2.

Hence, our assumption that p and q are co-prime is wrong. Hence , √2 is an irrational no.

Hence proved.

This method is called contradiction method.