Answer:
Method of Contradiction:
If possible, let p be false.
P is not true
-p is true
-p (p => r) is true
q and –r is true
x is a real number such that x3+x = 0and x≠ 0
x =0 and x≠0
This is a contradiction.
p is true.
Show that the statement p : “If x is a real number such that x3 + x = 0, then x is 0” is true by method of contradiction
Show that the statement p : “If x is a real number such that x3 + x = 0, then x is 0” is true by method of contradiction