Let us assume the contrary that root 3 is rational. Then √3 = p/q, where p, q are the integers i.e., p, q ∈ Z and co-primes, i.e., GCD (p,q) = 1.
√3 = p/q
⇒ p = √3 q
By squaring both sides, we get,
p2 = 3q2
p2 / 3 = q2 ------- (1)
(1) shows that 3 is a factor of p. (Since we know that by theorem, if a is a prime number and if a divides p2, then a divides p, where a is a positive integer)
Here 3 is the prime number that divides p2, then 3 divides p and thus 3 is a factor of p.
Since 3 is a factor of p, we can write p = 3c (where c is a constant). Substituting p = 3c in (1), we get,
(3c)2 / 3 = q2
9c2/3 = q2
3c2 = q2
c2 = q2 /3 ------- (2)
Hence 3 is a factor of q (from 2)
Equation 1 shows 3 is a factor of p and Equation 2 shows that 3 is a factor of q. This is the contradiction to our assumption that p and q are co-primes. So, √3 is not a rational number. Therefore, the root of 3 is irrational.
Let us assume the contrary that root 3 is rational. Then √3 = p/q, where p, q are the integers i.e., p, q ∈ Z and co-primes, i.e., GCD (p,q) = 1.
√3 = p/q
⇒ p = √3 q
By squaring both sides, we get,
p2 = 3q2
p2 / 3 = q2 ------- (1)
(1) shows that 3 is a factor of p. (Since we know that by theorem, if a is a prime number and if a divides p2, then a divides p, where a is a positive integer)
Here 3 is the prime number that divides p2, then 3 divides p and thus 3 is a factor of p.
Since 3 is a factor of p, we can write p = 3c (where c is a constant). Substituting p = 3c in (1), we get,
(3c)2 / 3 = q2
9c2/3 = q2
3c2 = q2
c2 = q2 /3 ------- (2)
Hence 3 is a factor of q (from 2)
Equation 1 shows 3 is a factor of p and Equation 2 shows that 3 is a factor of q. This is the contradiction to our assumption that p and q are co-primes. So, √3 is not a rational number. Therefore, the root of 3 is irrational.
Answers & Comments
Answer:
Let us assume the contrary that root 3 is rational. Then √3 = p/q, where p, q are the integers i.e., p, q ∈ Z and co-primes, i.e., GCD (p,q) = 1.
√3 = p/q
⇒ p = √3 q
By squaring both sides, we get,
p2 = 3q2
p2 / 3 = q2 ------- (1)
(1) shows that 3 is a factor of p. (Since we know that by theorem, if a is a prime number and if a divides p2, then a divides p, where a is a positive integer)
Here 3 is the prime number that divides p2, then 3 divides p and thus 3 is a factor of p.
Since 3 is a factor of p, we can write p = 3c (where c is a constant). Substituting p = 3c in (1), we get,
(3c)2 / 3 = q2
9c2/3 = q2
3c2 = q2
c2 = q2 /3 ------- (2)
Hence 3 is a factor of q (from 2)
Equation 1 shows 3 is a factor of p and Equation 2 shows that 3 is a factor of q. This is the contradiction to our assumption that p and q are co-primes. So, √3 is not a rational number. Therefore, the root of 3 is irrational.
Given: Number 3
To Prove: Root 3 is irrational
Proof:
Let us assume the contrary that root 3 is rational. Then √3 = p/q, where p, q are the integers i.e., p, q ∈ Z and co-primes, i.e., GCD (p,q) = 1.
√3 = p/q
⇒ p = √3 q
By squaring both sides, we get,
p2 = 3q2
p2 / 3 = q2 ------- (1)
(1) shows that 3 is a factor of p. (Since we know that by theorem, if a is a prime number and if a divides p2, then a divides p, where a is a positive integer)
Here 3 is the prime number that divides p2, then 3 divides p and thus 3 is a factor of p.
Since 3 is a factor of p, we can write p = 3c (where c is a constant). Substituting p = 3c in (1), we get,
(3c)2 / 3 = q2
9c2/3 = q2
3c2 = q2
c2 = q2 /3 ------- (2)
Hence 3 is a factor of q (from 2)
Equation 1 shows 3 is a factor of p and Equation 2 shows that 3 is a factor of q. This is the contradiction to our assumption that p and q are co-primes. So, √3 is not a rational number. Therefore, the root of 3 is irrational.