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# IT1101 Mathematics for Computing II 2001

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Preparing for BIT – IT1101 Mathematics for Computing II, on Rupavahini: Gihan Wikramanayake with Prof. VK Samaranayake and Mr. JKB Abeyesinghe, 08th March 2001, 2230-2300 hrs.

Preparing for BIT – IT1101 Mathematics for Computing II, on Rupavahini: Gihan Wikramanayake with Prof. VK Samaranayake and Mr. JKB Abeyesinghe, 08th March 2001, 2230-2300 hrs.

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• 1. PREPARING FOR THE PART II - Mathematics for Computing I
• 2. Mathematics for Computing I
• Today we will discuss some
• Model Paper Questions on Logic
• namely: 26, 28, 30, 32, 38
• 4.
• How to clarify any doubts?
• E-mail us, the address is
• 5. When can we have the Model Paper?
• Model Papers are available in the web :
• www.ict.cmb.ac.lk/bit.htm
• Also they have been posted to you.
• 6.
• Model Question 26
• Consider the following:
• 2 is not an integer
• Is 2 a positive integer?
• The presidential system in Sri Lanka was abolished in the year 2000
• x 2 > 10
• 7.
• Which of the following are (is) correct?
• (a) ( i ) , ( ii ) , ( iii ) are propositions
• (b)   ( i ) , ( iii ) , ( iv ) are propositions
• ( c)   ( i ) , ( iii ) are propositions
• ( d)   None of ( i ) ( ii ) , ( iii ) , ( iv ) are propositions
• ( e)    All ( i ) , ( ii ) , ( iii ) ( iv ) are propositions .
• 8.
• Objectives
• To recognise sentences which are propositions and those which are not propositions .
• 9.
• Solution
• Definition:
• Something written is called a proposition if it is either true or else it is false.
• 10.
• 2 is not an integer
• Is a proposition . It is false.
• Is 2 a positive integer?
• Is not a proposition as it is a question and we cannot talk of it being true or it being false.
• The presidential system in Sri Lanka was abolished in the year 2000
• Is a proposition . It is false.
Solution...
• 11.
• x 2 > 10
• Is not a proposition. In x 2 > 10 we do not know the value of x. So we cannot say whether x 2 > 10 is true or whether it is false.
• E.g. if x takes the value 2 it is false, but if x takes the value 4 it is true.
• So, we have ( i ) , ( iii ) are propositions and
• ( ii ) , ( iv ) are not propositions.
• So, (c) is the only correct choice.
Solution...
• 12. Which is tested in Mathematics - the Theory or it’s application ?
• Both.
• The students have to know the theory as well as their applications.
• Refer module objectives or URL
• www.ict.cmb.ac.lk/bit.htm
• 13. Are we allowed to use calculators for the Mathematics Paper ?
• No.
• We are not testing the calculating knowledge of the students. You need to study the theory and it’s various applications.
• 14.
• Model Question 30
• Out of the 8 possible truth values for p, q, r
• ( p  q )  r
• is true only for:
• (a) 7 set values (b) 6 set values
• (c) 5 set values (d) 4 set values
• (e) 3 set values
• 15.
• Objectives
• To know the truth tables in summary and thus to apply them fairly quickly.
• 16.
• Solution
• ( p  q )  r is F only when p  q is T with r is F
• i.e. it is F only in the following cases
• p is T, q is T, r is F
• p is T, q is F, r is F
• p is F, q is T, r is F
• i.e. it is F only for 3 sets of values
•  It is T only for 5 sets of values
• 17. Announcement
• Tel. 074-720511
• BIT ID cards and admission cards are being posted.
• 18.
• Model Question 28
• Consider the following.
• (i) p  ~p (ii) p  (~(p  q)) (iii) p  (p  q)
• Which of the following is correct?
• (a ) ( i ) , ( ii ) , ( iii ) are all tautologies .
• (b )  ( i ) , ( ii ) are tautologies but ( iii ) is not a tautology .
• ( c)  ( i ) , ( iii ) are tautologies but ( ii ) is not a tautology .
• ( d)  ( i ) is a tautology but ( ii ) , ( iii ) are not tautologies .
• ( e)   None of ( i ) ( ii ) , ( iii ) , ( iv ) is a tautology .
• 19.
• Objectives
• To know the truth tables in summary so that truth values of compound propositions are found out very quickly .
• 20.
• Solution
• Definition
• A proposition which in all circumstances is true, is called a tautology.
• 21.
• It is immediately seen that
• i.e. ( i ) ( p  ~p ) is a tautology , since one of p , ~p is T . (In a truth table for ‘  ’ if at least one of the components is T then the ‘  ’ proposition is T ).
• p  ( p  q ) is F when p is T and q is F . (Since we get p – T, p  q – F ).
• So ( iii ) is not a tautology .
Solution...
• 22.
• When p is T , p  ( ~ ( p  q )) is T .
• When p is F , p  q is F and so ~ ( p  q ) is T .
•  when p is F , p  ( ~ ( p  q )) is T
•  In all circumstances p  ( ~ ( p  q )) is T
•  ( ii ) is a tautology .
• So, the only correct choice is (b)
Solution...
• 23.
• Although, what is written on the previous slides is long, the thinking behind it is short and fast.
• There is another long but sure way of answering this question. This is to draw the truth tables of (i), (ii), (iii).
Solution...
• 24.
• It is seen from this truth tables that only (i) and (ii) are tautologies. So the only correct choice is (b).
Solution... T F T T p  ~p p T F F T T F T F T T T T p  ( ~ ( p  q )) q p T F F T T F F F T T T T p  ( p  q ) q p
• 25.
• Model Question 32
• The following four propositions are got by substitutions p, q, r, ~p, ~q in ( i ) , ( ii ) , ( iii ) , ( iv ) for the propositions there which are connected by ‘and’, ‘or’, ‘but’:
• (  ) p  r (  ) r  q
• (  ) ( ~q )  p (  ) r  ~p
• 26.
• Model Question 32...
• W hich of the following is correct?
• (i) It rained yesterday and there is no play at the Royal_ Thomain match today,
• (ii) The ground is wet or it rained yesterday.
• (iii) It rained yesterday and the ground is dry.
• (iv) There is play at the Royal_Thomian match today but the ground is wet.
(  ) p  r (  ) r  q (  ) ( ~q )  p (  ) r  ~p
• 27.
• Model Question 32...
• Choices
• (a) (i) is (  ) , (ii) is (  ), (iii) is (  ) and (iv) is (  )
• (b) (i) is (  ) , (ii) is (  ), (iii) is (  ) and (iv) is (  )
• (c) (i) is (  ) , (ii) is (  ), (iii) is (  ) and (iv) is (  )
• (d) (i) is (  ) , (ii) is (  ), (iii) is (  ) and (iv) is (  )
• (e) (i) is (  ) , (ii) is (  ), (iii) is (  ) and (iv) is (  )
• 28.
• Objectives
• To recognise the logical form of sentences given in ordinary English
• To learn to think logically in solving problems.
• 29.
• ( ii ) is (  )
• (This is seen by the fact that only sentence with an ‘ or ’ is ( ii ) . This is also seen by the fact that in all the choices we have, ‘ ( ii ) is (  ) ’
• r appears in (  ) .  r appears in ( ii )
•  r is one of ‘the ground is wet’, ‘it rained yesterday’
Solution (  ) p  r ( ii ) The ground is wet or it rained yesterday.
• 30.
• r appears again twice and in both places there is ‘ and ’.
•  r must be, ‘it rained yesterday’
• Since (  ) is p  r , p must be ‘the ground is wet’
• Since we know p, r we get that ( iii ) is (  )
Solution... (i) It rained yesterday and there is no ... (ii) The ground is wet or it rained yesterday. (iii) It rained yesterday and the ground is dry. (iv) There is ... but the ground is wet. (  ) r  q (  ) r  ~p
• 31.
• ‘ the ground is wet’ appears in ( iv ) .  ( iv ) is (  )
• (Note: ‘ but ’ means here the same as ‘and’. In English ‘but’ means a bit more than ‘and’ . Example : ‘He is young but his hair is gray. We can say ‘He is young and his hair is gray but earlier sentence is saying more than this)
•  the correct choice is ( c )
Solution... (  ) ( ~q )  p (iv) There is ... but the ground is wet. (c) (i) is (  ) , (ii) is (  ), (iii) is (  ) and (iv) is (  )
• 32.
• Model Question 38
• It is given that p  q and p  ~q are true. Now which of the following must necessarily be true,
• (a) p (b) ~p (c) q
• (d) ~q (e) q  r
• 33.
• Objectives
• Learn logical arguments.
• i.e. given premises ( i.e. propositions taken to be true ) to derive valid conclusions
• 34.
• Solution
• One of q, ~q is F. Since p  q, p  ~q are taken to be true, p must be F
• i.e. ~ p must be true
• with p is F, p  q, p  ~q is T both when q is T and when q is F ( i.e. ~q is T )
• 35.
• Solution
• when q is F, q  r is T
• when q is T and r is F, q  r is F
•  only , ~p is necessarily true
•  (b) is the only correct choice
• 36.
• Solution
• we can also do this by drawing truth tables in the following manner
T F F T T F F F T T T T p  q q p T T T F p  ~q
• 37.
• Solution
• From the truth table we get that P must be F
• i.e. ~p must be T
• q could either be T or it could be F
• So, we cannot say that q must be T
• 38.
• Solution
• Also we cannot say, ~q must be T
• we can have q is T, r is T. When this happens q  r is T
• We can also have, q is T, r is F. When this happens q  r is F
• 39.
• Solution
•  only ~p is necessarily true
•  ( b ) is the only correct choice
• Note: From what we have written in the first method, it appears long. However, this can be done mentally and then it is quite short.
• Also, importantly, that method involves logical thinking