prove that √5,√2,√3,√7 is an irrational number
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Answer:
Please check the explanation below !!!
Step-by-step explanation:
Proof of √5 is irrational.
Given: Number 5
To Prove: Root 5 is irrational
Proof: Let us assume that square root 5 is rational. Thus we can write, √5 = p/q, where p, q are the integers, and q is not equal to 0. The integers p and q are coprime numbers thus, HCF (p,q) = 1.
√5 = p/q
⇒ p = √5 q ------- (1)
On squaring both sides we get,
⇒ p2 = 5 q2
⇒ p2/5 = q2 ------- (2)
Assuming if p was a prime number and p divides a2, then p divides a, where a is any positive integer.
Hence, 5 is a factor of p2.
This implies that 5 is a factor of p.
Thus we can write p = 5a (where a is a constant)
Substituting p = 5a in (2), we get
(5a)2/5 = q2
⇒ 25a2/5 = q2
⇒ 5a2 = q2
⇒ a2 = q2/5 ------- (3)
Hence 5 is a factor of q (from 3)
(2) indicates that 5 is a factor of p and (3) indicates that 5 is a factor of q. This contradicts our assumption that √5 = p/q.
Therefore, the square root of 5 is irrational.
Proof of √2 is irrational.
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.
Proof 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.
Proof of √7 is irrational.
Given √7
To prove: √7 is an irrational number.
Proof:
Let us assume that √7 is a rational number.
So it t can be expressed in the form p/q where p,q are co-prime integers and q≠0
√7 = p/q
Here p and q are coprime numbers and q ≠ 0
Solving
√7 = p/q
On squaring both the side we get,
=> 7 = (p/q)2
=> 7q2 = p2……………………………..(1)
p2/7 = q2
So 7 divides p and p and p and q are multiple of 7.
⇒ p = 7m
⇒ p² = 49m² ………………………………..(2)
From equations (1) and (2), we get,
7q² = 49m²
⇒ q² = 7m²
⇒ q² is a multiple of 7
⇒ q is a multiple of 7
Hence, p,q have a common factor 7. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number
√7 is an irrational number.
hope this helped you !!!