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Question

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

Solution
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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=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.
x0 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+4x0
This shows that statement q is not true.
Thus it has been proved that
rq
Therefore the given statement p is true.

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Q1
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
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