Show that √2 is an irrational number. Hence, using this result, prove that 5 + 3√2 is an irrational
number.
Answers
Answer:
Let us consider
5
be a rational number, then
5
=p/q, where ‘p’ and ‘q’ are integers, q
=0 and p, q have no common factors (except 1).
So,
5=p
2
/q
2
p
2
=5q
2
…. (1)
As we know, ‘5’ divides 5q
2
, so ‘5’ divides p
2
as well. Hence, ‘5’ is prime.
So 5 divides p
Now, let p=5k, where ‘k’ is an integer
Square on both sides, we get
p
2
=25k
2
5q
2
=25k
2
[Since, p
2
=5q
2
, from equation (1)]
q
2
=5k
2
As we know, ‘5’ divides 5k
2
, so ‘5’ divides q
2
as well. But ‘5’ is prime.
So 5 divides q
Thus, p and q have a common factor of 5. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).
We can say that
5
is not a rational number.
5
is an irrational number.
Now, let us assume −3+2
5
be a rational number, ‘r’
So, −3+2
5
=r
−3–r=2
5
(−3–r)/2=
5
We know that, ‘r’ is rational, ‘(−3–r)/2’ is rational, so ‘
5
’ is also rational.
This contradicts the statement that √5 is irrational.
So, −3+2
5
is irrational number.
Hence proved.