logic question

[eric beans]

If the claim "All students in my class speak at least three different languages" is false, then it is certainly true that in my class

A there is a student who speaks no more than two different languages
B there is a student who speaks at least three different languages
C there is a student who speaks two different languages
D all students speak more than three different languages
E all students speak no more than two different languages

Could someone please help me with this? I never took logic so whenever I run across something like this, I never know how to tackle it and I get flustered and end up giving up. I'm assuming there is some method or some special notations used to figure something out like this?

 

[HallsofIvy]

The only statement that is clearly true is A:
A there is a student who speaks no more than two different languages
If this were not true then it would be true that "All students in my class speak at least three different languages" which we are told is false.

B there is a student who speaks at least three different language
C there is a student who speaks two different languages
D all students speak more than three different languages
E all students speak no more than two different languages
It might be that, in fact, all students in the class speak only one language so these four might be false.

 

[Steven G]

How can it be that All students in my class speak at least three different languages is false?

If all students.... is NOT true, then there must be at least one student that makes the statement false. Why? Because there are students who do NOT speak at least three different language. That is, there are some students who speaks no more than two languages.

Think about this. You are told that all the students in your class speak at least three different languages. Now how would you determine that is NOT true? It really is simple when you think about it. You just need to find a student (or more) that says that they speak less than three languages.

Is that clear?

 

[eric beans]

How can it be that All students in my class speak at least three different languages is false?

If all students.... is NOT true, then there must be at least one student that makes the statement false. Why? Because there are students who do NOT speak at least three different language. That is, there are some students who speaks no more than two languages.

Think about this. You are told that all the students in your class speak at least three different languages. Now how would you determine that is NOT true? It really is simple when you think about it. You just need to find a student (or more) that says that they speak less than three languages.

Is that clear?

I see that scenario can fulfill the requirements. Then how about choice "C there is a student who speaks two different languages ". That students speaks 2 languages too and fulfills the requirement. right?


But also, what if "speak at least three different languages" is false? if all students don't speak 3 languages, that's also fulfills the requirements does it not? which is why choice "E all students speak no more than two different languages " would also fulfill the requirement?

 

[pka]

If the claim "All students in my class speak at least three different languages" is false, then it is certainly true that in my class

There are four basic formal propositional forms:
\(A\text{. universal positive: all P is Q . }\left( {\forall x} \right)\left[ {P(x) \to Q(x)} \right]\)
\(E\text{. universal negative: no P is Q . }\left( {\forall x} \right)\left[ {P(x) \to \neg Q(x)} \right]\)
\(I\text{. existential positive: some P is Q . }\left( {\exists x} \right)\left[ {P(x) \wedge Q(x)} \right]\)
\(O\text{. existential negative: some P is Q . }\left( {\exists x} \right)\left[ {P(x) \wedge \neg Q(x)} \right]\)
This is called the square of opposition because the \(A~\&~O\) propositions are the negations of each other.
As are the \(E~\&~I\) propositions the negations of each other.
You were give an A propitiation, "every student speaks at least three languages".
So that its negation is an \(O\) propitiation, "Some student does not speak at least three languages".

 

[Steven G]

I see that scenario can fulfill the requirements. Then how about choice "C there is a student who speaks two different languages ". That students speaks 2 languages too and fulfills the requirement. right?


But also, what if "speak at least three different languages" is false? if all students don't speak 3 languages, that's also fulfills the requirements does it not? which is why choice "E all students speak no more than two different languages " would also fulfill the requirement?

Yes you are correct if there is a person who lives up to those conditions.
The problem stated: If the claim "All students in my class speak at least three different languages" is false, then it is certainly true that in my class ...

It is NOT certainly true that there is a student who speaks at least three different languages
It is NOT certainly true that there is a student who speaks two different languages
It is NOT certainly true that all students speak more than three different languages
It is NOT certainly true that all students speak no more than two different languages

It IS certainly true that there is a student who speaks no more than two different languages

 

[eric beans]

There are four basic formal propositional forms:
\(A\text{. universal positive: all P is Q . }\left( {\forall x} \right)\left[ {P(x) \to Q(x)} \right]\)
\(E\text{. universal negative: no P is Q . }\left( {\forall x} \right)\left[ {P(x) \to \neg Q(x)} \right]\)
\(I\text{. existential positive: some P is Q . }\left( {\exists x} \right)\left[ {P(x) \wedge Q(x)} \right]\)
\(O\text{. existential negative: some P is Q . }\left( {\exists x} \right)\left[ {P(x) \wedge \neg Q(x)} \right]\)
This is called the square of opposition because the \(A~\&~O\) propositions are the negations of each other.
As are the \(E~\&~I\) propositions the negations of each other.
You were give an A propitiation, "every student speaks at least three languages".
So that its negation is an \(O\) propitiation, "Some student does not speak at least three languages".

Wow, so there IS some sort of organized symbols to make concrete those thoughts and diagram them to get a visual handle on them. What are those symbols you're using called? Can you explain those symbols you're using in a simple to understand way?

 

[pka]

There are four basic formal propositional forms:
\(A\text{. universal positive: all P is Q . }\left( {\forall x} \right)\left[ {P(x) \to Q(x)} \right]\)
\(E\text{. universal negative: no P is Q . }\left( {\forall x} \right)\left[ {P(x) \to \neg Q(x)} \right]\)
\(I\text{. existential positive: some P is Q . }\left( {\exists x} \right)\left[ {P(x) \wedge Q(x)} \right]\)
\(O\text{. existential negative: some P is Q . }\left( {\exists x} \right)\left[ {P(x) \wedge \neg Q(x)} \right]\)
This is called the square of opposition because the \(A~\&~O\) propositions are the negations of each other.
As are the \(E~\&~I\) propositions the negations of each other.
You were give an A propitiation, "every student speaks at least three languages".
So that its negation is an \(O\) propitiation, "Some student does not speak at least three languages".

\(\left( {\forall x} \right)\left[ {P(x) \to Q(x)} \right]\) translates for all x, if x has property P then x has property Q in other words "Every P is a Q"
\(\left( {\forall x} \right)\left[ {P(x) \to \neg Q(x)} \right]\) translates as for all x, if x has property P then x does not haves property Q in other words "No P is a Q"
Now I have no intention of writing these out. I you have access to a mathematics library then look for Symbolic Logic, by Copi or The Grammar of Mathematics by Durst of any text by A.H.Lightstone. Eslewise talk to the person assigning this question, asking WHY?

 

[eric beans]

\(\left( {\forall x} \right)\left[ {P(x) \to Q(x)} \right]\) translates for all x, if x has property P then x has property Q in other words "Every P is a Q"
\(\left( {\forall x} \right)\left[ {P(x) \to \neg Q(x)} \right]\) translates as for all x, if x has property P then x does not haves property Q in other words "No P is a Q"
Now I have no intention of writing these out. I you have access to a mathematics library then look for Symbolic Logic, by Copi or The Grammar of Mathematics by Durst of any text by A.H.Lightstone. Eslewise talk to the person assigning this question, asking WHY?

I'm assuming that A and E refers to answer choices A and E. But what do the letters I and O refer to? What do they stand for?

 

[Dr.Peterson]

Did you try searching for the name pka used, "Square of Opposition"? Your question (and more) is answered in

Square of opposition - Wikipedia

en.wikipedia.org

Note that A and E, like I and O, are part of this system, and do not refer to your problem!