A simple law of syllogism logic problem

[startover]

My little brother (9th grade) took a math quiz and one of the questions says:

Which conjecture is valid by the Law of Syllogism for the following scenario?
"If Joe buys Nancy a card, he will give it to her. If Nancy is having a birthday, then Joe will buy her a card."

A. If Nancy is having a birthday, then Joe will give her a card.
B. If Joe buys Nancy a card, then she is having a birthday.
C. If Joe gives Nancy a card, then he bought the card.
D. If Joe gives Nancy a card, then she is having a birthday.

My little brother chose A which I believe is the 100% correct answer because from "Nancy having a birthday"(p) to "Joe buys her a card" (q) and "Joe buys her a card" (q) to "Joe will give it(the card) to her" (r) we get p->r which is exactly what A states.

However his teacher marked the answer wrong and upon further inquries she *insisted* that the correct answer is D, which to me doesn't make any sense and is logically incorrect.

This seems to be a very simple question and yet the teacher's response makes me seriously question her skills.

What do you guys think?

 

[startover]

Your brother's "teacher" is simply an idiot who, we can hope, will soon be promoted to administration.

LoL that cracked me up even though it's a bit harsh. Honestly I think it's more of a case of irresponsibility rather than stupidity. The teacher most likely scored the quiz against an answer kit which had an error and didn't even bother reading into the problem at all when questioned by her student (my little brother mentioned that the teacher only took a few seconds before confirming the answer with him). But if she really came up with that answer after giving it careful consideration, then I fully agree with you JeffM on the "promoted to administration" part.

This incident makes me worried because it's not the first time it happened. I noticed several errors and inaccuracies in my little brother's previous tests and work sheets, some of which the teacher acknowledged without admitting any wrongdoing on her part. My stance is that everybody makes errors especially on math, but they must be corrected immediately upon discovery to avoid any confusions for the students. It is the teacher's responsibility to properly proofread the test sheets and study/homework materials. Such simple and basic errors should not have been hard to correct, and having them pop up at such a frequency is definitely not normal.

At this point I'm thinking about submitting an anonymous complaint to the school principle/district. The problem is that my little brother is a very nice kid and doing so would make him feel "guilty" (he won't even tell me the teacher's name). But still I think education is a serious matter and scientific subjects like math require even more discipline and precision. I simply do not see how a teacher with this kind of attitude can be qualified to properly teach his students, and I feel bad for all the other kids in my little brother's class who probably did not have someone to consult with and were totally mislead by the teacher's "correct' answers. Coming from a Chinese educational background (known for its strictness and higher difficulty K12 math/science level), I find this kind of errors and attitude very unacceptable. Without any intention to bring up a debate, I wonder if it's common to run into teachers like this here in the US high schools?

Anyway I started this thread mainly to prove to my little brother that his teacher is indeed wrong in case he doesn't fully trust my judgement. I'd also appreciate your opinions on the teacher and my planned actions against her. Thank you for all the replies.

 

[Chencha]

Okay the teacher is not wrong, and the correct answer is D.

First, you have to make two conditionals using the two statements.
Conditional #1: If Joe buys Nancy a card, then he will give it to her.
Conditional #2: If Nancy is having a birthday, then Joe will buy her a card.

Second, seeing as you can't use the "if" part of the first statement and the "then" part of the second statement, you reverse it and use the "then" part of the first statement and the "if" part of second statement to come to a conclusion.
Conclusion: If Joe gives Nancy a card, then Nancy is having a birthday.

Not sure if you understand, but this is what I would have gotten by using the Law of Syllogism correctly.

 

[pka]

Okay the teacher is not wrong, and the correct answer is D.
First, you have to make two conditionals using the two statements.
Conditional #1: If Joe buys Nancy a card, then he will give it to her.
Conditional #2: If Nancy is having a birthday, then Joe will buy her a card.
Second, seeing as you can't use the "if" part of the first statement and the "then" part of the second statement, you reverse it and use the "then" part of the first statement and the "if" part of second statement to come to a conclusion.
Conclusion: If Joe gives Nancy a card, then Nancy is having a birthday.
Not sure if you understand, but this is what I would have gotten by using the Law of Syllogism correctly.

Let \(\displaystyle \mathcal{B}\) stand for "Joe buys Nancy a card".
Let \(\displaystyle \mathcal{G}\) stand for "Joe gives Nancy a card".
Let \(\displaystyle \mathcal{H}\) stand for "Nancy Has a birthday".

Do yo agree that this what we have given:
\(\displaystyle \text{If }\mathcal{B}\text{ then }\mathcal{G}.\)
\(\displaystyle \text{If }\mathcal{H}\text{ then }\mathcal{B}.\)??

Now the only way to have a valid argument form with two hypothetical statements is to have a distributed middle.
So \(\displaystyle \mathcal{B}\) must be that middle term.
So only \(\displaystyle \text{If }\mathcal{H}\text{ then }\mathcal{G}\) works as the conclusion.

 

[soroban]

Hello, startover!

My little brother (9th grade) took a math quiz and one of the questions says:

Which conjecture is valid by the Law of Syllogism for the following scenario?
. . If Joe buys Nancy a card, he will give it to her.
. . If Nancy is having a birthday, then Joe will buy her a card.
. . Therefore:

A. If Nancy is having a birthday, then Joe will give her a card.
B. If Joe buys Nancy a card, then she is having a birthday.
C. If Joe gives Nancy a card, then he bought the card.
D. If Joe gives Nancy a card, then she is having a birthday.

My little brother chose A which I believe is the 100% correct answer
because from "Nancy having a birthday"(p) to "Joe buys her a card" (q)
and "Joe buys her a card" (q) to "Joe will give it (the card) to her" (r),
we get p->r which is exactly what A states.

However his teacher marked the answer wrong and upon further inquries,
she *insisted* that the correct answer is D,
which to me doesn't make any sense and is logically incorrect.

This seems to be a very simple question and yet the teacher's response
makes me seriously question her skills.

What do you guys think?

I think teachers like that will bring back public flogging.


\(\displaystyle \text{If} \underbrace{\text{Nancy is having a birthday}}_p,\;\underbrace{\text{ then }}_{\to}\,\underbrace{\text{Joe will buy her a card.}}_q\)

\(\displaystyle \text{If} \underbrace{\text{Joe buys Nancy a card}}_q,\;\underbrace{\text{ then }}_{\to}\,\underbrace{\text{ he will give it to her.}}_r\)


The form of the argument is: .\(\displaystyle \begin{array}{ccc} p \;\to\;q \\ q\;\to\;r \\ \hline \end{array}\)

The only logical conclusion is: .\(\displaystyle p \to r\) .**

. . A. If Nancy is having a birthday, then Joe will give her a card.


The other choices just don't cut it . . .

\(\displaystyle \text{B. If} \underbrace{\text{Joe buys Nancy a card}}_q,\,\underbrace{\text{then}}_{\to}\,\underbrace{\text{she is having a birthday.}}_p\)

\(\displaystyle \text{C. If} \underbrace{\text{Joe gives Nancy a card}}_r,\,\underbrace{\text{then}}_{\to}\,\underbrace{\text{he bought the card.}}_q\)

\(\displaystyle \text{D. If} \underbrace{\text{Joe gives Nancy a card}}_r,\,\underbrace{\text{then}}_{\to}\,\underbrace{\text{she is having a birthday.}}_{_p\)


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Some know-it-all will have to point out that \(\displaystyle \sim\!r \to\: \sim\!p\)
. . the contrapositive, is also a logical conclusion.

This is true, but so are: .\(\displaystyle \sim\!p \vee r\,\text{ and }\,\sim(p\:\wedge \sim\!r)\),
. . none of which is among the answer choices.
So what's your point?