Math Is Fun Forum

It’s a table displaying every multiplication result in the set. For example, for

For

So do you know how to construct a multiplication table for

Then it’s 2 3 5 – the first three prime numbers. That takes us to the first 18 decimal places: e = 2.718281828459045235…

An equilateral triangle is one possibility. But there might be others, of course.

I’m sorry, but I don’t really like it.

What your proof needs is a bit more organization of thought. First, you focus on what are you trying prove. What are you trying to prove? Namely, that if S is a member of P(A) then S is a subset of P(A). Then you think of how to go about proving it. For this particular problem, it’s absolutely straightforward. Namely, you assume S ∊ P(A), and then you try, by a series of deductions, to arrive at S ⊆ P(A).

In general, when you want to prove a statement of the form P ⇒ Q, you can use one of three methods. (i) The most straightforward way is to assume P and then, by a series of deductions, try and arrive at Q. (ii) You can also assume that Q is false and show that this would lead to P being false. This is known as a contrapositive proof. (iii) The third method is proof by contradiction. Assume that P is true but Q is false, then show that this would lead to a contradiction.

For this particular problem, method (i) is the one to use. What’s more I’ve already shown you how to do it (see above).

EDIT: Maybe I’ll explain part of my proof in more detail. Up to S ⊆ A should be straightforward. Now S ⊆ A means S is a set of members of A. But A has property that each of its members is also a subset of A. Since the members of S are members of A, we see that the members of S are all subsets of A. So S is a set of subsets of A. Now carry on with the proof.

I like both mathematics and language.

A statistic like top 20% or top 10% would be much more relevant when there’s a much larger group (say, 100 or more). 6 is really too small a group.

At the moment, I’m listening to the first three movements of Mahler’s Symphony № 3 in D Minor.

Oh geez, I misread the question.

I’m not very sure of my answer but that was what I got.

Correct.

You’re welcome. Can you follow the proof all right?