Proove that √5 is irrational.
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The proof that the square root of 5 is irrational is exactly the same as the well-known proof that the square root of 2 is irrational - except using 5 in place of 2. We can prove a more general result: the square root of any prime is irrational.
First of all, we require the lemma:
for any prime p, and integer x,
p|x2 ⇒ p|x
That is, if x2 is divisible by p, then so is x.
Proof:
The prime factorization of x2 necessarily contains p at least once, since it is divisible by p. But it also has to contain an even power of every prime, since it is the prime factorization of a square. Therefore, it contains p at least twice, and its square root, x, contains p at least once: that is, x is divisible by p.
Now, given a prime p, assume that its square root is rational. Then, it may be written in the form a/b, where a and b have no common factors (that is, the fraction a/b is in lowest terms). This is always possible for any nonzero rational number. Since this quantity is the square root of p, its square equals p, that is
(a/b)2 = p
a2/b2 = p
a2 = pb2
Now, pb2 is a multiple of p, so a2 must be too. And, using the result above, this means that a must be a multiple of p also. Thus, there exists an integer c such that
a = PC
Then,
(PC)2 = pb2
p2c2 = pb2.
Since p is not zero, we may divide both sides by p to obtain
PC2 = b2
That is, b2 is divisible by p also, and thus b is divisible by p.
Since a and b were both divisible by p, the fraction a/b could not have been in lowest terms, which contradicts our initial assumption. Therefore, the square root of p cannot possibly be a rational number. Since 5 is prime, the proof is complete.
First of all, we require the lemma:
for any prime p, and integer x,
p|x2 ⇒ p|x
That is, if x2 is divisible by p, then so is x.
Proof:
The prime factorization of x2 necessarily contains p at least once, since it is divisible by p. But it also has to contain an even power of every prime, since it is the prime factorization of a square. Therefore, it contains p at least twice, and its square root, x, contains p at least once: that is, x is divisible by p.
Now, given a prime p, assume that its square root is rational. Then, it may be written in the form a/b, where a and b have no common factors (that is, the fraction a/b is in lowest terms). This is always possible for any nonzero rational number. Since this quantity is the square root of p, its square equals p, that is
(a/b)2 = p
a2/b2 = p
a2 = pb2
Now, pb2 is a multiple of p, so a2 must be too. And, using the result above, this means that a must be a multiple of p also. Thus, there exists an integer c such that
a = PC
Then,
(PC)2 = pb2
p2c2 = pb2.
Since p is not zero, we may divide both sides by p to obtain
PC2 = b2
That is, b2 is divisible by p also, and thus b is divisible by p.
Since a and b were both divisible by p, the fraction a/b could not have been in lowest terms, which contradicts our initial assumption. Therefore, the square root of p cannot possibly be a rational number. Since 5 is prime, the proof is complete.
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Answer.
let root 5 be rational
then it must in the form of p/q [q is not equal to 0][p and q are co-prime]
root 5=p/q
=> root 5 * q = p
squaring on both sides
=> 5*q*q = p*p ------> 1
p*p is divisible by 5
p is divisible by 5
p = 5c [c is a positive integer] [squaring on both sides ]
p*p = 25c*c --------- > 2
sub p*p in 1
5*q*q = 25*c*c
q*q = 5*c*c
=> q is divisble by 5
thus q and p have a common factor 5
there is a contradiction
as our assumsion p &q are co prime but it has a common factor
so √5 is an irrational
let root 5 be rational
then it must in the form of p/q [q is not equal to 0][p and q are co-prime]
root 5=p/q
=> root 5 * q = p
squaring on both sides
=> 5*q*q = p*p ------> 1
p*p is divisible by 5
p is divisible by 5
p = 5c [c is a positive integer] [squaring on both sides ]
p*p = 25c*c --------- > 2
sub p*p in 1
5*q*q = 25*c*c
q*q = 5*c*c
=> q is divisble by 5
thus q and p have a common factor 5
there is a contradiction
as our assumsion p &q are co prime but it has a common factor
so √5 is an irrational
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