Question

Show that the statement p : “If x is a real number such that x 3 + 4 x = 0, then x is 0” is true by (i) direct method (ii) method of contradiction (iii) method of contrapositive

Answer 1

The given statement is: p: “If x is a real number such that x 3 +4x=0 , then x is 0.” Break the given statement into parts.

Let q: x is a real number such that x 3 +4x=0 .

And r: x is 0.

(i) To show that p is true by direct method, consider that q is true, then show that r is true.

Since q is true, thus

x 3 +4x=0 x( x 2 +4 )=0

Either x=0 or x 2 +4=0 . If x 2 +4=0 occurs then the value of x will be imaginary because

x 2 +4=0 x 2 =−4 x= −4 x=±2i

As it is given that x is real, thus the only value of x is 0. Therefore, statement r; x is 0 is true.

Thus, the given statement is true.

(ii) To show that p is true by method of contradiction, consider that p is not true. Therefore, it can be said that “If x is a real number such that x 3 +4x=0 and let x is not 0.”

x 3 +4x=0 x( x 2 +4 )=0

Either x=0 or x 2 +4=0 .

x 2 +4=0 x 2 =−4 x= −4 x=±2i

As it is given that x is real, thus the only value of x is 0, which is a contradiction to the assumed statement.

Thus, the given statement is true.

(iii) To show that p is true by the method of contrapositive, consider that r is false then show that q must be false. Since the statement r is false, then it should be the negation of the statement r which will be denoted as ∼r .

∼r : x is not 0.

As x 2 +4 is always positive, then x≠0 implies that the product of x and any positive real number is not zero.

Consider the product of x with ( x 2 +4 ) .

x( x 2 +4 )≠0 x 3 +4x≠0

This shows that the statement q is not true. Therefore, ∼r implies ∼q .

Thus, the given statement is true.