Show that the following numbers are irrational.
(i)
(ii) ![]()
(iii) 6 +
(iv) 3 - ![]()
(i) Let assume that
is rational
Therefore it can be expressed in the form of
, where p and q are integers and q≠0
Therefore we can write
= ![]()
√2 = ![]()
is a rational number as p and q are integers. Therefore √2 is rational which contradicts the fact that √2 is irrational.
Hence, our assumption is false and we can say that
is irrational.
(ii) Let assume that 7√5 is rational therefore it can be written in the form of ![]()
where p and q are integers and q≠0. Moreover, let p and q have no common divisor > 1.
7√5 =
for some integers p and q
Therefore √5 = ![]()
is rational as p and q are integers, therefore √5 should be rational.
This contradicts the fact that √5 is irrational.
Therefore our assumption that 7√5 is rational is false. Hence 7√5 is irrational.
(iii) Let assume that 6 + √2 is rational therefore it can be written in the form of ![]()
where p and q are integers and q≠0. Moreover, let p and q have no common divisor > 1.
6 + √2 =
for some integers p and q
Therefore √2 =
- 6
Since p and q are integers therefore
- 6 is rational, hence √2 should be rational. This contradicts the fact that √2 is irrational. Therefore our assumption is false. Hence, 6 + √2 is irrational.
(iv) Let us assume that 3 -
is rational
Therefore 3 -
can be written in the form of
where p and q are integers and q≠0
3 -
=
⇒
-3 = ![]()
⇒
= ![]()
Since p, q and 3 are integers therefore
is rational number
But we know
is irrational number, Therefore it is a contradiction.
Hence 3 -
is irrational