Q. 14.5( 15 Votes )

Show that the statement p: “If x is a real number such that x3 + 4x = 0, then x is 0” is true by

(i) direct method

(ii) method of contradiction

(iii) method of contrapositive

Answer :

(i) The given statement is of the form if p then q


Here


p: x is aa real number such that x3 + 4x = 0.


q: x is 0


In Direct Method,


we assume p is true, and prove q is true


So,


Let x be a real number such that x3 + 4x = 0, prove that x = 0


p is true


Consider


x3 + 4x = 0 where x is real


x(x2 + 4) = 0


x = 0 or x2 + 4 = 0


x = 0 or x2 = – 4


x = 0 or x =


x = is not possible because it is given that x is real.


Hence x = 0 only


q is true


Hence proved


(ii) p: If x is a real number such that x3 + 4x = 0, then x is 0


Let us assume


x3 + 4x = 0


but x ≠ 0


Solving x3 + 4x = 0 where x is real


x(x2 + 4) = 0


x = 0 or x2 + 4 = 0


x = 0 or x2 = – 4


x = 0 or x =


x = is not possible because it is given that x is real number


Hence, only solution is x = 0 but we take x ≠ 0


Hence we get a contradiction


Hence our assumption is wrong


Hence x is real number such that x3 + 4x = 0 then x is 0.


(iii) Let p: if x is real number such that x3 + 4x = 0


q: x is 0


The above statement is of the form if p then q


By method of Contrapositive


By assuming that q is false, prove that p must be false


Let q is false


i.e. x is not equal to 0


i.e. x ≠ 0


i.e. x × (positive number) ≠ 0 × (positive number)


i.e. x × (x2 + 4) ≠ 0 × (x2 + 4)


i.e. x3 + 4x ≠ 0


i.e. p is false


Hence x is real number such that x3 + 4x = 0 then x is 0.


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