assignment 14

course Mth 151

zŜdžefLassignment #014

014. Truth Tables

Liberal Arts Mathematics I

10-06-2008

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16:48:21

`q001. There are 8 questions in this set.

If each of the propositions p and q can be either true or false, what combinations of truth values are possible for the two propositions (e.g., one possibility is that p is false and q is true; list the other possibilities)?

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RESPONSE -->

P Q

T T

T F

F T

F F

confidence assessment: 3

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16:48:27

It is possible that p is true and q is true.

Another possibility is that p is true and q is false.

A third possibility is that p is false and q is true.

A fourth possibility is that p is false and q is false.

These possibilities can be listed as TT, TF, FT and FF, where it is understood that the first truth value is for p and the second for q.

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RESPONSE -->

OK

self critique assessment:

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16:49:36

`q002. For each of the for possibilities TT, TF, FT and FF, what is the truth value of the compound statement p ^ q ?

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RESPONSE -->

p q P and Q

T T T

T F F

F T F

F F F

confidence assessment: 3

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16:49:43

p ^ q means 'p and q', which is only true if both p and q are true.

In the case TT, p is true and q is true so p ^ q is true.

In the case TF, p is true and q is false so p ^ q is false.

In the case FT, p is false and q is true so p ^ q is false.

In the case FF, p is false and q is false so p ^ q is false.

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RESPONSE -->

OK

self critique assessment: 3

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16:50:30

`q003. Write the results of the preceding problem in the form of a truth table.

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RESPONSE -->

p q p^q

T T T

T F F

F T F

F F F

confidence assessment: 3

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16:50:38

The truth table must have headings for p, q and p ^ q. It must include a line for each of the possible combinations of truth values for p and q. The table is as follows:

p q p ^ q

T T T

T F F

F T F

F F F.

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RESPONSE -->

OK

self critique assessment: 3

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16:51:31

`q004. For each of the possible combinations TT, TF, FT, FF, what is the truth value of the proposition p ^ ~q?

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RESPONSE -->

p q p or not q

T T T

T F T

F T F

F F T

confidence assessment: 3

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16:52:01

For TT we have p true, q true so ~q is false and p ^ ~q is false.

For TF we have p true, q false so ~q is true and p ^ ~q is true.

For FT we have p false, q true so ~q is false and p ^ ~q is false.

For FF we have p false, q false so ~q is true and p ^ ~q is false.

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RESPONSE -->

OK

self critique assessment: 3

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16:52:20

`q005. Give the results of the preceding question in the form of a truth table.

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RESPONSE -->

p q ~q p^~q

T T F F

T F T T

F T F F

F F T F

confidence assessment: 3

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16:52:26

The truth table will have to have headings for p, q, ~q and p ^ ~q. We therefore have the following:

p q ~q p^~q

T T F F

T F T T

F T F F

F F T F

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RESPONSE -->

OK

self critique assessment: 3

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16:53:07

`q006. Give the truth table for the proposition p U q, where U stands for disjunction.

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RESPONSE -->

Assuming that U means ""or""

p q p U q

T T T

T F T

F T T

F F F

confidence assessment: 3

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16:53:15

p U q means 'p or q' and is true whenever at least one of the statements p, q is true. Therefore p U q is true in the cases TT, TF, FT, all of which have at least one 'true', and false in the case FF. The truth table therefore reads

p q p U q

T T T

T F T

F T T

F F F

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RESPONSE -->

OK

self critique assessment: 3

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16:58:07

`q007. Reason out the truth values of the proposition ~(pU~q).

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RESPONSE -->

TT

p is true and q is true, so ~q is false. p or~q is true, since p is true. So ~(p U ~q) is false.

TF

p is true and q is false, so ~q is true. p U ~q is true, since p is true (as is q). So ~(p U ~q) is false.

FT

p is false and q is true, so ~q is false. p U ~q is false, since neither p nor ~q is true. So ~(p U ~q) is true.

FF p is false and q is false, so ~q is true. p U ~q is true, since ~q is true. So ~(p U ~q) is false.

confidence assessment: 3

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16:58:21

In the case TT p is true and q is true, so ~q is false. Thus p U ~q is true, since p is true. So ~(p U ~q) is false.

In the case TF p is true and q is false, so ~q is true. Thus p U ~q is true, since p is true (as is q). So ~(p U ~q) is false.

In the case FT p is false and q is true, so ~q is false. Thus p U ~q is false, since neither p nor ~q is true. So ~(p U ~q) is true.

In the case FF p is false and q is false, so ~q is true. Thus p U ~q is true, since ~q is true. So ~(p U ~q) is false.

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RESPONSE -->

OK

self critique assessment: 3

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16:59:17

`q008. Construct a truth table for the proposition of the preceding question.

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RESPONSE -->

p q ~q pU~q ~(pU~q)

T T F T F

T F T T F

F T F F T

F F T T F

confidence assessment: 2

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16:59:30

We need headings for p, q, ~q, p U ~q and ~(p U ~q). Our truth table therefore read as follows:

p q ~q pU~q ~(pU~q)

T T F T F

T F T T F

F T F F T

F F T T F

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RESPONSE -->

Awesome! OK

self critique assessment: 3

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