Question

# Show that the statementp : "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

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#### 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=0r: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=0x(x2+4)=0⇒x=0orx2+4=0However sincex is real it is 0.Thus statement r is trueTherefore 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=0x(x2+4)=0x=0 or x2+4=0x=0 or x2=−4However 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:xisnot0It 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≠0This shows that statement q is not true.Thus it has been proved that ∼r⇒∼qTherefore the given statement p is true.

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