Prove that root 2 is an irrational number.

By Shivank Goel|Updated : August 5th, 2022

Rational numbers are in the form p/q, where p and q can be any integer and q ≠ 0.

To prove that √2 is an irrational number, let us use the contradiction method

Assume that √2 is a rational number with p and q as coprime integers and q ≠ 0

√2 = p/q

By squaring on both sides

2q2 = p2

p2 is an even number that divides q2

So p is an even number that divides q

Consider p = 2x where x is a whole number

Now substitute the value of p in 2q2 = p2,

2q2 = (2x)2

2q2 = 4x2

q2 = 2x2

q2 is an even number that divides x2.

So q is an even number that divides x

As both p and q are even numbers with a common multiple 2, which means that p and q are not coprime as the HCF is 2

It leads to a contradiction that root 2 is a rational number of the form p/q where p and q are coprime and q ≠ 0.

Therefore, root 2 is an irrational number using the contradiction method.

Summary:

Prove that root 2 is an irrational number.

Root 2 is an irrational number.

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