Prove √2 is an irrational number in simple language
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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 :
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.