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