14 replies to this topic

[bonanova]

Choose the best answer and say why.

Which of the even prime numbers [excluding 2] are [evenly] divisible by 5?

[1] all
[2] some
[3] none
[4] the question has no defensible answer.

p.s. Martini: I assert this is a logical not mathematical question.

Edited for "clarity". [meaning that I originally screwed up what I wanted to ask .. ]

I do not choose [4].

[Writersblock]

All of them. There is only one, which is 2. All other even numbers are divisible by 2 and therefore not prime. The question doesn't specify that the solution must be evenly divisible by 5, therefore ALL numbers are divisible by five, including 2.

[bonanova]

All of them. There is only one, which is 2. All other even numbers are divisible by 2 and therefore not prime. The question doesn't specify that the solution must be evenly divisible by 5, therefore ALL numbers are divisible by five, including 2.

Hmmmm.. well, ok.

But I've edited the question now, hopefully asking what I had intended to ask.
See if your answer changes ...

[Writersblock]

Well obviously it changes my answer now that 2 is excluded. I don't see where you are going though. 4 is the obvious answer, so not the right one for this forum. However, 1,2,&3 are excluded as those are words that only apply to some definable set. Since the set of answers you have defined is null, you can't apply "all" "some" or "none" to that set.

[cpotting]

I propose that the answer could be given as [1], [2] or [3]. But the answer is definitely not [4].

The set of even prime numbers excluding 2 is an empty set. Being an empty set, it simultaneously meets the criteria that all, some and none of its members are divisible by 5.

However, since answers are being chosen and reasons for those answers are being given, then by definition, there are defensible answers to the questions. Thus [4] would be incorrect.

[bonanova]

I would take issue with the defenses put forth so far.
In fairness, tho, I have changed my mind as to which answer[s] I would defend.

Today, these clues; tomorrow, my picks.

[1] Existential import.
[2] Boolean logic - as opposed to Aristotelian logic.

[bonanova]

My picks:

Spoiler for ...

[1] and [3] --- originally I picked only [1] but I changed my mind, making the problem a little less satisfying. <!-- s:oops: --><!-- s:oops: -->

[4] is out ... some answers have been defended, and
[2] is out ... because of existential import.

Let's examine [1]-[3] by restating them as categorical propositions.
Big words ... they just mean statements that relate members of categories of things.
The categories in the statements are their Subjects [S] and their Predicates [P].

Let S = numbers that are even primes, excluding 2
Let P = numbers that are [evenly] divisible by 5.

Then the options become

[1] All S is P
[2] Some S is P
[3] No S is P

It's been noted that S has no members. S is an empty set.

That means we have to eliminate [2]. Why?
In Boolean logic, the categorical proposition Some S is P carries the assertion that S has at least one member.
That is, in Boolean logic, the word Some means at least one.
Logicians call this "existential import" [EI].

But [1] and [3] are OK.
"All" and "None" do not assert the existence of even one member of the category. They don't have EI.
It makes logical sense to say

[1] All five-headed women have two toes.
[3] No five-headed women have two toes.

Why do these statement make sense? Because their logically contradictory statements are false.
[1a] Some five-headed women do not have two toes. This is false because of EI.
[3a] Some five-headed women do have two toes. This is also false because of EI.

Originally, I thought I could exclude [3] by saying "None" meant "Not one"
and then asserting that "Not one" implied there was at least one, simply
because we talked about it.

Nah ... upon reflection, can't do that.

So my picks are ... [1] and [3] - equally sensible.

[BoilingOil]

All even numbers [excluding 2] are evenly divisable by 2, and therefor not prime. Therefor there are NO EVEN PRIME NUMBERS if 2 is excluded. Hence no even prime numbers which would be evenly divisable by 5!


There you have it!

[bonanova]

All even numbers [excluding 2] are evenly divisable by 2, and therefor not prime. Therefor there are NO EVEN PRIME NUMBERS if 2 is excluded. Hence no even prime numbers which would be evenly divisable by 5!


There you have it!

If your choice is [3], it's correct; but ...

Why is it the best answer?
Why did you pick it over [1] All even prime numbers [excluding 2] are divisible by 5?

Aren't they all divisible by 5? Show me one that is not.

[red text edited]

[Benson]

that's must be 4 because divide by zero.
1. there's no even prime numbers excludeing 2. so, the solution set is none {}.
2. "evenly" is the x^2/sizeof(x). however, the size of x is zero.

It's just like customers never visit the resturant, you can't tell which one of them love the food or not.

PS. If you choose 4, it works for your spoiler, too.