prove that √6 is not a rational number
Answers
Answer:
Let √6 be a rational no.
√6=a/b ( where a and b are not equal to zero)
√6b=a
Square both side
(√6b)^2=(a)^2
6b^2=a^2….............(1)
Therefore a^2 is divisible by 6 and also a is also divisible by 6.
a=6c .......................(2)
Substitute equation 1 in 2
6b^2= 36 c^2
b^2= 6 c^2
This means b^2 is also divisible by 6 and b is also divisible by 6 according to the theorem
[ Let p be a prime no. If p divides a^2 then p divides a where a is a positive integer]
Therefore and b have at least 6 as a common factor other than 1 .
But this contradicts the fact that a and b are co primes.
Do our assumption is wrong that √6 is rational no.
So, we can conclude that √6 is an irrational no.
The following proof is a proof by contradiction.
Let us assume that
6
is rational number.
Then it can be represented as fraction of two integers.
Let the lowest terms representation be:
6
=
b
a
where b
=0
Note that this representation is in lowest terms and hence, a and b have no common factors
a
2
=6b
2
From above a
2
is even. If a
2
is even, then a should also be even.
⟹a=2c
4c
2
=6b
2
2c
2
=3b
2
From above 3b
2
is even. If 3b
2
is even, then b
2
should also be even and again b is even.
But a and b were in lowest form and both cannot be even. Hence, assumption was wrong and hence,
6
is an irrational number.