how to prove root 5 is an irrirational number
Answers
Answered by
1
If √5 is rational, then it can be expressed by some number a/b (in lowest terms). This would mean:
(a/b)² = 5. (Squaring),
a² / b² = 5. (Multiplying by b²),
a² = 5b².
If a and b are in lowest terms (as supposed), their squares would each have an even number of prime factors. 5b² has one more prime factor than b², meaning it would have an odd number of prime factors.
Every composite has a unique prime factorization and can't have both an even and odd number of prime factors. This contradiction forces the supposition wrong, so √5 cannot be rational. It is, therefore, irrational.
(a/b)² = 5. (Squaring),
a² / b² = 5. (Multiplying by b²),
a² = 5b².
If a and b are in lowest terms (as supposed), their squares would each have an even number of prime factors. 5b² has one more prime factor than b², meaning it would have an odd number of prime factors.
Every composite has a unique prime factorization and can't have both an even and odd number of prime factors. This contradiction forces the supposition wrong, so √5 cannot be rational. It is, therefore, irrational.
Answered by
2
by contradiction method
let √5 be rational
√5 = p/q
squaring both sides 5= p^2/ q^2
p^2= 5q^2
p^2 is multiple of 5
p is also multiple of
let p =5a
on squaring
p^2=25a^2
25a^2=5q^2
q^2=5a^2
q is also multiple of 5
since p and q both are multiple of 5
hence our contradiction is wrong
√5 is irrational
let √5 be rational
√5 = p/q
squaring both sides 5= p^2/ q^2
p^2= 5q^2
p^2 is multiple of 5
p is also multiple of
let p =5a
on squaring
p^2=25a^2
25a^2=5q^2
q^2=5a^2
q is also multiple of 5
since p and q both are multiple of 5
hence our contradiction is wrong
√5 is irrational
Similar questions