Prove that √2 is an irrational number by contradictory method
Answers
Step-by-step explanation:
To prove that √2 is an irrational number, we will use the contradiction method.
Let us assume that √2 is a rational number with p and q as co-prime integers and q ≠ 0
⇒ √2 = p/q
On squaring both sides we get,
⇒ 2q2 = p2
⇒ p2 is an even number that divides q2. Therefore, p is an even number that divides q.
Let p = 2x where x is a whole number.
By substituting this value of p in 2q2 = p2, we get
⇒ 2q2 = (2x)2
⇒ 2q2 = 4x2
⇒ q2 = 2x2
⇒ q2 is an even number that divides x2. Therefore, q is an even number that divides x.
Since p and q both are even numbers with 2 as a common multiple which means that p and q are not co-prime numbers as their HCF is 2.
This leads to the contradiction that root 2 is a rational number in the form of p/q with p and q both co-prime numbers and q ≠ 0.
Thus, √2 is an irrational number by the contradiction method