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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

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Solution

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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.

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