# Tag Info

33

I don't think there's anything wrong with the wording; it's clear what is being asked. Your example with the three dollars is also not always the way we speak in everyday language. If you ask someone with three children if they have two children, they're unlikely to say "yes" and leave it at that. Getting more silly, a bicycle isn't a unicycle ...

32

As you've noticed, there are (at least) three potential ways of proving an implication $p \Rightarrow q$: Assume $p$, and conclude $q$. Assume $\neg q$, and conclude $\neg p$. Assume both $p$ and $\neg q$, and derive a contradiction. If I understand you right, you're asking for a proof which is of the third kind. Moreover, you're asking for a proof which ...

20

When we describe counts in natural language, there's almost always an implicit "exactly" when phrasing like this. We use phrases like "at least 4" when we want a more general description. Most children who have reached a development level where this quiz would be reasonable will probably already have learned this. In fact, this is why ...

16

Perhaps "shows" instead of "has". If you asked me to show you 4 apples, I can't think of a logical argument in favor of me grabbing 5 apples and smiling smugly.

11

Nearly every test like this includes instructions to choose the "best answer" to cover exactly this scenario. This looks like it's part of a test of basic counting skills, and in that context, the best answer is B. While one could make an argument for either C or D, I can't imagine an argument for either of those being the best answer when B is ...

9

I think you are overlooking the fact that proof by contradiction must invoke the tautology $(P\ \hbox{or}\ \neg P)$, called the law of excluded middle. To prove $P\Rightarrow Q$ by contradiction, we show that $(\neg Q\Rightarrow\neg P)$. The next step, which is where we deduce the conclusion $Q$, is where we must invoke the assumption that $P$ is "...

8

No, I suspect this situation never occurs. Here is why: If $P$ really implies $Q$, then we know logically that $\neg Q$ implies $\neg P$. Thus if you assume $\neg Q$, you will be able to deduce $\neg P$. In this way, you never "need" the assumption $P$.

8

I showed this question to my three-year old son. His response - because he counted the apples one by one in each picture, passing "4" each time - was B, C and D. Hence, we need to take into account how children arrive at their conclusion, since they do not apply formal logic. The thought process is very different from the abstract approach a ...

2

If you allowed this, how do you grade answers? Suppose the same maths test said "John has 2 apples and Lucy has 3 apples. How many apples do they have in total?" By your logic, the child could say "1" and be entirely correct. If you have 5 apples and someone asks you "do you have an apple?", the answer of course is "yes&...

2

I think a lot of it has to do with the age of the child, and what the goal of the question is. If this is for children just learning their numbers, like say 4 or 5 years old, then I think B is the correct answer as thy are not being asked to stretch their logical capabilities, but to simply recognize and call out the difference between 4 of something, or 5 ...

2

Expanding my comment into an answer: So you want the umbrella term for a predicate (propositional function), like the triangle or Cauchy–Schwarz inequality, that's always true, but not necessarily so regardless of interpretation. So, ‘validity’ and ‘tautology’ are ruled out. If the object is an equality instead of inequality, I'd have suggested ‘identity’ (a ...

2

Claim: Compact ($P$) metric space implies sequentially compact ($Q$). Proof. Assume $P\land \lnot Q$. We use $\lnot Q$ to construct an infinite sequence with no limit points. Because there are no limit points, each point in our space has an open neighbourhood containing only finitely many points of the sequence. These neighbourhoods form a cover and we use $... 2 I think you may be looking at this the wrong way. There are three distinct logical rules which are all equivalent. Each rule holds for all propositions$P$and$Q$. Rule 1:$\neg \neg P \implies P$This is the classic "proof by contradiction" rule. Rule 2:$(\neg P \implies \neg Q) \implies (Q \implies P)$This is "proof by contrapositive"... 1 Re (1), you should lower your standards. As previously discussed, the population for Math55 in high school is tiny. (Add onto that the difficulty of finding them and it just makes little sense to try to develop this.) In addition, you really lack the math knowledge OR the practical pedagogical experience to develop Math55 for high school (even the time, ... 1 For (2), I'd suggest creating the lessons first as blog posts or youtube videos and then sharing them. I think the math reddit might be a good place, but I'm not active there. These can then in turn serve as advertisements for any active curricula you'd like to implement. For (3): Start with an overall structure of the course. Make a list of course-level ... 1 Your self-answer is wrong, because a tautology must be true under every interpretation, but your example is certainly not so. There is no single word for what you want, but the standard terminology is that your formula is true under standard/intended interpretation. And you can note that it is conventional to say just "true" to mean that when ... 1 If you will eventually be teaching the basic methods of proof (conditional proof, proof by contradiction, etc.) in your course for math majors, you might consider starting with the truth tables for NOT, AND and OR and postpone the truth table for IMPLIES until they understand some of those basic methods of proof. Then they should be able to understand a ... 1 In my opinion, the truth-table definition of material implication is disturbing because "if ..., then ..." is used in mathematics in two distinct ways (and no similar distinction exists for the other connectives like NOT, AND, and OR): "If$p$, then$q$" means that you can start out with$p$, make some deductions, and end up with$q\$. As ...

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