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Question

Question: Consider the following two statements: P: If \(7\) is an odd number, then \(7\) is divisible by \(...

Consider the following two statements:
P: If 77 is an odd number, then 77 is divisible by 22.
Q: If 77 is prime number, then 77 is an odd number.
If V1{V_1} is the truth value of the contrapositive of P and V2{V_2} is the truth value of contrapositive of Q, then the ordered pair (V1,V2)({V_1},{V_2}) equals:
A)(F,F) B)(T,T) C)(T,F) D)(F,T)  A)\,(F,F) \\\ B)\,(T,T) \\\ C)\,(T,F) \\\ D)\,(F,T) \\\

Explanation

Solution

At first find out the contrapositive of the given statements. Then find the truth value of these statements. Hence you will get the value of (V1,V2)({V_1},{V_2}).

Complete step-by-step answer:
We know we have to find the contrapositive of the given statements P and Q. but what does contrapositive mean? Just the opposite of a statement ‘is’ is changed to ‘is not’ in the statements.
For example: 77 is prime number
It’s contrapositive is 77 is not a prime number
Contrapositive statement of P:
If 77 is not an odd number, then 77 is not divisible by 22.
Contrapositive statement of Q:
If 77 is not a prime number, then 77 is not an odd number.
Now we will find their truth values. Truth values are determined by the following table where P1&P2{P_1}\,\&\, {P_2} are the first and second halves of the student:

Hence for statement P, not P2{P_2} is true and not P1{P_1} is false.
Not P2{P_2}:77 is not divisible by 22
Not P1{P_1}: 77 is not an odd number
Hence truth table of P is False
Similarly for statement Q, not Q2{Q_2} is false and not Q1{Q_1} is false
Hence Q is True
V1=F&V2=T\Rightarrow {V_1} = F\,\,\,\,\& \,\,\,{V_2} = T
Therefore ordered pair is (F,T)(F,T)

So, the correct answer is “Option D”.

Note: A truth table is a mathematical table used to determine if a compound statement is true or false.Remembering the table is very much necessary as it is the most important part. If you don’t remember, you won’t be able to find the truth value and hence cannot solve the question.Also, P1P2{P_1} \Rightarrow {P_2} is same as notP1notP2not{P_1} \Rightarrow \,not{P_2}