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
Open in App
Verified by Toppr
p : "If x is a real number such that x3+4x=0 then x is 0" Let q:x is a real number such that x3+4x=0 r:x is 0 (i) To show that statement p is true we assume that q is true and then show that r is true. Therefore let statement q be true. ∴x3+4x=0 x(x2+4)=0 ⇒x=0orx2+4=0 However sincex is real it is 0. Thus statement r is true Therefore the given statement is true. (ii) To show statement p to be true by contradiction we assume that p is not true. Let x be a real number such that x3+4x=0 and let x is not 0. Therefore x3+4x=0 x(x2+4)=0 x=0 or x2+4=0 x=0 or x2=−4 However x is real. Therefore x=0 which is a contradiction since we have assumed that x is not 0. Thus the given statement p is true. (iii) To prove statement p to be true by contrapositive method we assume that r is false and prove that q must be false. Here r is false implies that it is required to consider the negation of statement r. This obtains the following statement, ∼r:xisnot0 It can be seen that (x2+4) will always be positive. x=0 implies that the product of any positive real number with x is not zero. Let us consider the product of x with (x2+4) ∴x(x2+4)=0 ⇒x3+4x=0 This shows that statement q is not true. Thus it has been proved that ∼r⇒∼q Therefore the given statement p is true.