prove that 3+2√2 is irrational number .given √2 is a irrational no.
Answers
Answer:
Let us consider
2
be a rational number, then
2
=p/q, where ‘p’ and ‘q’ are integers, q
=0 and p, q have no common factors (except 1).
So,
2=p
2
/q
2
p
2
=2q
2
…. (1)
As we know, ‘2’ divides 2q
2
, so ‘2’ divides p
2
as well. Hence, ‘2’ is prime.
So 2 divides p
Now, let p=2k, where ‘k’ is an integer
Square on both sides, we get
p
2
=4k
2
2q
2
=4k
2
[Since, p
2
=2q
2
, from equation (1)]
q
2
=2k
2
As we know, ‘2’ divides 2k
2
, so ‘2’ divides q
2
as well. But ‘2’ is prime.
So 2 divides q
Thus, p and q have a common factor of 2. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).
We can say that
2
is not a rational number.
2
is an irrational number.
Now, let us assume 3−
2
be a rational number, ‘r’
So, 3−
2
=r
3–r=
2
We know that, ‘r’ is rational, ‘3−r’ is rational, so ‘
2
’ is also rational.
This contradicts the statement that
2
is irrational.
So, 3−
2
is an irrational number.
Hence proved.
Step-by-step explanation:
Let us consider
2
be a rational number, then
2
=p/q, where ‘p’ and ‘q’ are integers, q
=0 and p, q have no common factors (except 1).
So,
2=p
2
/q
2
p
2
=2q
2
…. (1)
As we know, ‘2’ divides 2q
2
, so ‘2’ divides p
2
as well. Hence, ‘2’ is prime.
So 2 divides p
Now, let p=2k, where ‘k’ is an integer
Square on both sides, we get
p
2
=4k
2
2q
2
=4k
2
[Since, p
2
=2q
2
, from equation (1)]
q
2
=2k
2
As we know, ‘2’ divides 2k
2
, so ‘2’ divides q
2
as well. But ‘2’ is prime.
So 2 divides q
Thus, p and q have a common factor of 2. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).
We can say that
2
is not a rational number.
2
is an irrational number.
Now, let us assume 3−
2
be a rational number, ‘r’
So, 3−
2
=r
3–r=
2
We know that, ‘r’ is rational, ‘3−r’ is rational, so ‘
2
’ is also rational.
This contradicts the statement that
2
is irrational.
So, 3−
2
is an irrational number.
Hence proved.