Short Answer TypeShow, by method of contrapositive, that following statement is true.
“If x is an integer and x2 is odd, then x is also odd
Show that the statement
p: “If x is a real number such that x3 + 4x = 0, then x is 0” is true by direct method.
The component statements of the given statement are
q : x is a real number such that x3 + 4x = 0
r : x is 0.
To check the validity of statement, we assume that q is true.
Now, we have to show that r is also true.
Since q is true, therefore, x is a real number such that 
i.e, 
i.e, either x = 0 or ![<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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i.e, either x = 0 or 
Now, x is real means ![<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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Therefore, x = 0 i.e, x is 0
Therefore, r is true.
Hence, the given statement is true.
Show that the statement
p: “If x is a real number such that x3 + 4x = 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 + 4x = 0, then x is 0” is true by method of contrapositive.
Multiple Choice QuestionsThe statement ~ (p↔ ~q) is
equivalent to p ↔ q
equivalent to ~ p ↔q
a tautology
a tautology
Consider :
Statement − I : (p ∧ ~ q) ∧ (~ p ∧ q) is a fallacy.
Statement − II : (p → q) ↔ (~ q → ~ p) is a tautology.
Statement -I is True; Statement -II is True; Statement-II is a correct explanation for Statement-I
Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I
Statement -I is True; Statement -II is False.
Statement -I is True; Statement -II is False.
The negation of the statement “If I become a teacher, then I will open a school” is
I will become a teacher and I will not open a school
Either I will not become a teacher or I will not open a school
Neither I will become a teacher nor I will open a school
Neither I will become a teacher nor I will open a school
Consider the following statements
P: Suman is brilliant
Q: Suman is rich
R: Suman is honest. The negation of the statement ì Suman is brilliant and dishonest if and only if Suman is richî can be ex- pressed as
~ P ^ (Q ↔ ~ R)
~ (Q ↔ (P ^ ~R)
~ Q ↔ ~ P ^ R
~ Q ↔ ~ P ^ R
Switch
is equivalent to
