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


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