jowens5017emailvccsedu

course Mth 151

I am having a difficult time understanding the circuits. Any insights to the last problem would be greatly appreciated.

ý›Ä…»Ÿ†o¼yÜr»áü}ôØ”ÝxÐÔö}üassignment #014

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014. `query 14

College Algebra

06-28-2007

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00:57:01

3.3.6 rewrite using if then ' all marines love boot camp '.

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RESPONSE -->

If you are a marine, then you love boot camp.

confidence assessment: 3

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00:58:52

** The statement is equivalent to 'If it's a Marine, it loves boot camp' or equivalent.

The statement is not equivalent to 'if it is boot camp, then all Marines love it', which is the converse of the original statement. **

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RESPONSE -->

I wrote ""If you are a marine, then you love boot camp."" I understand the statement that it is not equivalent to but is my statement correct or should I have wrote it just like yours?

self critique assessment: 2

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01:00:23

3.3.18 ~p false q false p -> q true

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RESPONSE -->

This statement would be false because ~p is false so p is true and p->q would be false.

confidence assessment: 3

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01:00:39

** Since ~p is false then p is true.

Since q is false it follows that p -> q is of the form T -> F, which is false.

The conditional is false when, and only when, the antecedent is true and the consequent false. **

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RESPONSE -->

I understand this.

self critique assessment: 3

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01:01:21

Query 3.3.36 write in symbols 'If play canceled, then it does not rain.'

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RESPONSE -->

p->~r

confidence assessment: 3

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01:01:35

** If p stands for 'play canceled' and r for 'it rains' then the statement would be p -> ~q. **

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RESPONSE -->

I understand this.

self critique assessment: 3

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01:04:11

Query 3.3.48 q true, p and r false, evaluate and (-r U p) -> p

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RESPONSE -->

~r would be true and p would be false so (~rVp) would be True and p is false so (~rVp)->p is false.

confidence assessment: 3

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01:04:38

** The antecedent (~r U p ) would be true, since ~r true and p false.

The consequent p would be false.

Since the antecedent is true and the consequent false, the conditional is false. **

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RESPONSE -->

I understand this.

self critique assessment: 3

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01:07:48

Query 3.3.60 truth table for (p ^ q) -> (p U q)

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RESPONSE -->

p q (p^q) (pVq) (p^q)->(pVq)

T T T T T

T F F T T

F T F T T

F F F F F

confidence assessment: 2

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01:09:06

** The headings would be p, q ,(p^q), (pUq), (p^q)->(pUq)

Row 1 would read T T T T T

Row 2 would read T F F T T

Row 3 would read F T F T T

Row 4 would read F F F F T

The common sense of this is that whenever both p and q are true, then the statement 'p or q' must be true. That's what means to say (p ^ q) -> (p U q).

The fact that this statement is true is indicated by the last column of the truth table, which has True in every possible case. **

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RESPONSE -->

I slightly messed up on this one. I think it was in the last column but I do understand and think I just messed up on one of the disjunctions.

self critique assessment: 2

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01:10:22

Query 3.3.72 negation of ' if loving is wrong then I don't want to be right'

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RESPONSE -->

Loving you is wrong and I don't want to be right.

confidence assessment: 2

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01:11:27

** The negation has to have the exact opposite truth values of the original statement.

It is difficult and confusing to try to negate a conditional. It is much easier to translate the conditional to a disjunction then negate the disjunction. It is easy to negate the disjunction using deMorgan's Laws.

Since p -> q is identical to ~p U q, the negation of p -> q is ~ ( ~p U q), which by de Morgan's Law is ~ ~p ^ ~q, or just p ^ ~q.

So the negation would ge 'loving you is wrong AND I want to be right.

COMMON ERROR AND NOTE: If loving you is wrong, then I want to be right.

INSTRUCTOR COMMENT:

The negation of a conditional can't be a conditional (a conditional is false in only one case so its negation would have to be false in three cases). **

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RESPONSE -->

I wrote ""Loving you is wrong and I don't want to be right. I should have said ""I want to be right. I just got in a hurry.

self critique assessment: 2

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01:13:56

Query 3.3.90 circuit q parallel [ p in series with ( q par. ~p) ].

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RESPONSE -->

I honestly don't know were to start with these problems. I read the section in the book but am still confused.

confidence assessment: 0

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01:17:20

** If I'm looking at the same circuit you are the translation would be q U [ p ^ [q U ~p ]] .

Since intersection and union each distribute over the other, this would simplify to

q U [ ( p ^ q) U ( p ^ ~p) ]. Since p ^ ~p can't be true, this is the same as

q U ( p ^ q).

Since q U ( p ^ q) is true whenever q is true and false whenever q is false expression is true exactly when q is.

So the circuit finally simplifies to just q.

You should be able to see by looking at the circuit that current will flow whenever all the q switches are closed. **

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RESPONSE -->

I don't understand this at all. It is so foreign to me that I don't even know what questions to ask to help understand it. I'll keep reading the text over to try to get it.

self critique assessment: 1

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"

Don't worry too much about the circuits right now. However if you email me Monday, I can post some pictures of actual circuits and some explanation.

Note that there are no questions on circuits on the test, so this isn't required; however once you see what is going on the idea is simple and well worth a little effort, and it also helps solidify your understanding of the logic.