#$&* course mth151
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Given Solution: The proposition will be true in every case except the one where p is true and q is false, which is the TF case. The truth table therefore reads as follows: p q p -> q T T T T F F F T T F F T &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. Reason out, then construct a truth table for the proposition ~p -> q. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: T T F T T F F T F T T T F F T T confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: This proposition will be false in the T -> F case where ~p is true and q is false. Since ~p is true, p must be false so this must be the FT case. The truth table will contain lines for p, q, ~p and ~p -> q. We therefore get p q ~p ~p -> q T T F T since (F -> T) is T T F F T since (F -> F) is T F T T T since (T -> T) is T F F T T since (T -> F) is F &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. Reason out the truth value of the proposition (p ^ ~q) U (~p -> ~q ) in the case FT (i.e., p false, q true). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: both are false so make it all false confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To evaluate the expression we must first evaluate p ^ ~q and ~p -> ~q. p ^ ~q is evaluated by first determining the values of p and ~q. If p is false and q true, then ~q is false. Thus both p and ~q are false, and p ^ ~q is false. ~p -> ~q will be false if ~p is true and ~q is false; otherwise it will be true. In the FT case p is false to ~p is true, and q is true so ~q is false. Thus it is indeed the case the ~p -> ~q is false. (p ^ ~q) U (~p -> ~q ) will be false if (p ^ ~q) and (~p -> ~q ) are both false, and will otherwise be true. In the case of the FT truth values we have seen that both (p ^ ~q) and (~p -> ~q ) are false, so that (p ^ ~q) U (~p -> ~q ) is false. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. Construct a truth table for the proposition (p ^ ~q) U (~p -> ~q ). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: T T F F F T T T F F T T T T F T T F F F F F F T T F T T confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We will need headings for p, q, ~p, ~q, (p ^ ~q), (~p -> ~q ) and (p ^ ~q) U (~p -> ~q ). So we set up our truth table p q ~p ~q (p ^ ~q) (~p -> ~q ) (p ^ ~q) U (~p -> ~q ) T T F F F T T T F F T T T T F T T F F F F F F T T F T T To see the first line, where p and q are both T, we first see that ~p and ~q must both be false. (p ^ ~q) will therefore be false, since ~q is false; (~p -> ~q) is of the form F -> F and is therefore true. Since (~p -> ~q) is true, (p ^ ~q) U (~p -> ~q ) must be true. To see the second line, where p is T and q is F, we for see that ~p will be F and ~q true. (p ^ ~q) will therefore be true, since both p and ~q are true; (~p -> ~q) is of the form F -> T and is therefore true. Since (p ^ ~q) and (~p -> ~q ) are both true, (p ^ ~q) U (~p -> ~q ) is certainly true. To see the fourth line, where p is F and q is F, we for see that ~p will be T and ~q true. (p ^ ~q) will be false, since p is false; (~p -> ~q) is of the form T -> T and is therefore true. Since (~p -> ~q ) is true, (p ^ ~q) U (~p -> ~q ) is true. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. If we have a compound sentence consisting of three statements, e.g., p, q and r, then what possible combinations of truth values can occur? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: TTT, TFT, FTT, FFT, TFF, FTF, FFF, TTF confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A compound statement with two statements p and q has four possible combinations of truth values: TT, TF, FT, FF. Here we also have r, which can be either T or F. So we can append either T or F to each of the possible combinations for p and q. If r is true then we have possible combinations TT T, TF T, FT T, FF T. If r is false we have TT F, TF F, FT F, FF F. This gives us 8 possible combinations: TTT, TFT, FTT, FFT, TTF, TFF, FTF, FFF. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. Evaluate the TFT, FFT and FTF lines of the truth table for (p ^ ~q) -> r. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: T F T T T T F F T T F T F T F F F T confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We would need column headings p, q, r, ~q, (p^~q) and (p^~q) -> r. The truth table would then read p q r ~q (p^~q) (p^~q) -> r T F T T T T F F T T F T F T F F F T &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q007. Construct a truth table for the statement ~p -> q YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: TTFTF FFTTF TFFTF TFTFF
#$&* course mth151 nov 3rd8:00pm
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Given Solution: `a** 'Milk contains calcium' can be put into p -> q form as 'if it's milk then it contains calcium'. The converse of p -> q is q -> p, which would be 'if it contains calcium then it's milk' The inverse of p -> q is ~p -> ~q, which would be 'if it's not milk then it doesn't contain calcium'. The contrapositive of p -> q is ~q -> ~p, which would be 'if it doesn't contain calcium then it's not milk'. Note how the original statement and the contrapositive say the same thing, and how the inverse and the converse say the same thing. NOTE ON ANOTHER STATEMENT: If the statement is 'if it ain't broke don't fix it: Converse: If you don't fix it, then it ain't broke Inverse: If it's broke, then fix it. Contrapositive: If you fix it, then it's broke. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery 3.4.18 state the contrapositive of 'if the square of the natural number is even, then the natural number is even.' Using examples decide whether both are truth or false. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: IT WOULD BE FALSE confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The statement is of the form p -> q with p = 'square of nat number is even' and q = 'nat number is even'. The contrapositive of p -> q is ~q -> ~p, which in this case would read 'if a natural number isn't even then its square isn't even'. STUDENT RESPONSE WITH SOMEWHAT PICKY BUT IMPORTANT INSTRUCTOR CORRECTION: if the natural number isn't even , then the square of a natural numbewr isn't even Good. More precisely: if the natural number isn't even , then the square of THAT natural number isn't even. To say that the square of a natural number isn't even doesn't necessarily refer to the given uneven natural number. COMMON ERROR WITH INSTRUCTOR COMMENT: The natural number is not even, if the square of a natural number is not even. ex.-3^2=9,5^2=25 This statement is true. ** You have stated the inverse ~p -> ~q. It doesn't matter that the 'if' is in the second half of your sentence, the 'if' in your statement still goes with ~p when it should go with ~q. COMMON ERROR WITH INSTRUCTOR COMMENT: If the natural number is not even, then the square of the natural number is not even. This statement does not involve square roots. It addresses only squares. And 26 isn't the square of a natural number. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qExplain how you used examples to determine whether both statements are true or both false. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: BOTH STATEMENTS WOULD BE TRUE confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The first statement said that if the square of a natural number is even then the natural number is even. For example, 36 is the square of 6, 144 is the square of 12, 256 is the square of 16. These examples make us tend to believe that the statement is true. The contrapositive says that if the natural number is even then its square isn't even. For example, the square of the odd number 7 is 49, which is not an even number. The square of the odd number 13 is 169, which is not an even number. This and similar examples will convince us that this statement is true. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qExplain why either both statements must be true, or both must be false. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: THEY BOTH HAVE TO BE TRUE OR THEY BOTH HAVE TO BE FALSE. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The reason is that the truth tables for the statement and its contrapositive are identical, so if one is true the other is true and if one is false the other must be false. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery 3.4.24 write 'all whole numbers are integers' in form 'if p then q'. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: if it's a whole number then it's an integer confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** p could be 'it's a whole number' and q would then be 'it's an integer'. The statement would be 'if it's a whole number then it's an integer'. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery 3.4.30 same for ' principal hires more only if board approves YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: IT IS NOT EQUIVILENT TO THE STATEMENT confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aCOMMON ERROR WITH INSTRUCTOR COMMENT: If the principal will hire more teachers, then the school board would approve. INSTRUCTOR COMMENT: p only if q is the same as if p then q; should be 'if the principle hires, the school board approved' ** STUDENT COMMENT I switched the two because I thought the 'only if' meant that was the p part. I thought that it made more sense that the teacher hiring was dependent on the board approving. INSTRUCTOR RESPONSE To say that 'the teacher hiring was dependent on the board approving' would be correct, and would have the same meaning as the instructor's stated solution. However the statement 'the teacher hiring was dependent on the board approving' is not equivalent to your statement 'If board approves then the principal hires more', which is not equivalent to the given statement. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery 3.4.48 true or false: 6 * 2 = 14 iff 9 + 7 neg= 16. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: BOTH ARE TRUE ALTHOUGH COMPOUND STATEMENT IS FALE confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Both statments are false, but the compound statement is true. The compound statement 'p if and only if q' is equivalent to 'if p then q, AND if q then p'. This compound statement is true because p and q are both false, so 'if p then q' and 'if q then p' are both of form F -> F and therefore true ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery 3.4.55 contrary or consistent: ' this number is an integer. This number is irrational.' YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: THE NUMBER CANT BE BOTH SO CONTRARY confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a**Any integer n can be expressed in the form p / q as n / 1. So all integers are rational. Irrational numbers are defined as those numbers which are not rational. So the statements are indeed contrary-it is impossible for a number to be both an integer and irrational. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!