Question

prove that under root 6 is an irrational number​

Answer 1

Bonjour!

You asked to prove under root 6 is an irrational number.

Prove that √6 is an irrational number.

This problem can be solved by a contradiction method i.e assuming it is a rational number. But a and b were in lowest form and both cannot be even. NOTE: √6=ab , this representation is in lowest terms and hence, a and b have no common factors.So it is an irrational number.

Au revoir, mon ami!

Answer 2

Suppose we consider ,√6 is a rational number .

Then we can express it in the form of a/b

∴√6=a/b, where a and b are positive integer and they are co-prime ,i.e.HCF(a,b)=1

∴√6=a/b

=>b√6=a

=>(b√6)²=a² [squaring both sides]

=>6b²=a²………..(1)

here,a² is divided by 6

∴a is also divided by 6. [we know that if p divides

a²,then p divides a]

∴6|a

=>a=6c [c∈ℤ]

=>a²=(6c)²

=>6b²=36c² [from (1)]

=>b²=6c²

here,b² is divided by 6,

∴b is also divided by 6.

∴6|a and 6|b

we observe that a and b have at least 6 as a common factor .But this contradicts that “a amd b are co-prime .”

It means that our consideration of “√6 is a rational number” is not true.

Hence,√6 is a irrational number.