Ok, so going your route:

You need to show that (n+1)^n / n^n >= 2 for n = 1, 2, ...

My method for this:
1) Show that it's true when n=1.
2) Consider f(x) = (1+1/x)^x, and show that its derivative is positive for x > 1 [hence it is increasing].
2a) You may wish to note that the Taylor Series for ln(1+x) = x - x^2/2 + ..., which is alternating, hence: ln(1+x) >= x - x^2/2 for x > 0.

Once you've proved this fact, you can say (as you did in a backwards way in your LaTeX above):
(n+1)^n / n^n >= 2
so (n+1)^n >= 2*n^n
so ((n+1)/2)^n >= 2*(n/2)^n >= n! by the inductive hypothesis.

Now you can multiply both sides by n+1 and clean up the left to finish off your proof.

I thought about that and it's pretty obvious but we haven't talked about functions yet, so I tried to do it via induction.

Thanks everyone

I goofed. phi(4) = 2 => 2 elements order 4. Also -1 has order 2, which is the 4th root you're looking for!

Would be great if someone could help me with this assignment:

Be X a countable Set. Prove that {Y ⊆ X : Y finite} is countable.

Thanks in advance!

You can use a coding of finite sets of numbers (for example the Godel beta function). This is a bijective function from the finite powerset of N to N.
You can also use the Cantor pairing function repeatedly, sending a subset {n_1,..., n_k} to <k,<n_1,<n_2,...<n_{k-1},n_k>...>
where <x,y> is the pairing of x and y. This is less elegant, but you only need an injective function to prove the statement.

|X| is countable
Y c X => |Y|=< |X| => |Y| countable

That wasn't the question. The question is whether the set of finite subsets of N is countable. (More generally, the cardinality of the finite powerset of X is equal to the cardinality of X.)

You've severely misread the problem.

I don't really understand this proof yet. Do you mind further elaborating on it?

Thank you.

Have you seen the bijective Cantor pairing function NxN-->N?

(I'm happy to elaborate but first I need to know where you're at.)

No, sorry, I haven't seen the bijective Cantor pairing function.

What class is this for? What have you been learning about? What kind of examples similar to this have you done?

This is a homework assignment, so I'm assuming you're not just supposed to redevelop arguments Cantor made without at least a push in the right direction.

Ok, I'll avoid that for now but it's probably useful to look at because it is a simpler version of your problem.

What you need is an injective function from P_f(N) to N, where P_f(N) is the set of finite subsets of N. This means that you need a way of taking a finite subset A={n_1,...,n_k} of N and associate to it a natural number phi(A) in such a way that you associate different numbers to different subsets.

There are many ways of achieving this, but here is one (modulo a few little technical details): write v(i) for p_i^{n_i}, where p_i is the i-th prime number. Then let

phi(A)=\prod_{0 \leq i \leq k} v(i)

be the product of these v(i).

For example, the set of numbers A={4,3,8} will get sent to 2^4 x 3^3 x 5^8.
The reason this works well is because we can factor any number n as
n=\prod_i p_i^{n_i}
uniquely.

It is therefore straightforward to check that this phi is injective. (You do need to extend the definition to include the empty set though, but that is easy.)

yep realized it now.
english is not my first language and i am especially bad at math vocabulary in english

We can also prove #1 directly.


with * because of (a^2+b^2)/2 >= ab
∎

Simple stats question that IDK how to complete. I am trying to test a hypothesis and am given the following. N=48, sample mean=57%, alpha=.05, population mean=76%. z score is 2.1 on a two tailed normal distribution curve. I can do this with integers and a giving standard deviation, but didn't know wtf to do with these percentage values. Thanks.

I'm supposed to factor the difference of two squares

81x^4-1
=(9x^2-1)(9x^2+1)

I don't think that is the correct answer but what am I doing wrong?

It's correct; it's just that there's a *more* correct answer . Hint: one of the two terms you have can be factored again in exactly the same way.

81x^2 - 1 = (9x - 1)(9x + 1)

Rearranging shows this is equivalent to:

(1 + 1/n)^n >= 2.

To prove this, just use the binomial theorem to expand the LHS. The first two terms are 1 + 1, and every other term is positive.

aaah I got it. There's a difference of a square in the first term.

= (3x-1)(3x+1)(9x^2+1)

Ooh, very nice.

9x^2=30x-25
9x^2-30x+25=0

I can't figure out how to factor from here.

2 ways: One is pick factors of 9 (say 9 & 1 or 3 & 3) and write out (9x+a)(x+b) or (3x+a)(3x+b) and find an a & b that work.

The other is to use the quadratic formula, find the roots (a, b) and write this as 9(x-a)(x-b).

I feel like this is a good response, but that it assumes I know more than I do.

The problem I am having here is that the way I learned it i multiply 9*25 (first number and last number) and find the factors of that number that have a sum of 30 (the middle number). This is no problem for me most of the time, for example:

4x^2-12x+9 = 0
4x^2-6x-6x+9 =0
2x(2x-3) -3(2x-3) = 0
(2x-3)(2x-3) = 0
(2x-3)^2=0
2x-3=0
2x=3
x=3/2

The reason I can't do this problem is that the product of 9*25 doesn't have factors who's sum is 30.

I assume that what you wrote up there shows me how to find what I am looking for, but I really don't understand it. Specifically what A and B are in this equation. 9(x-3)(x-5) Is what seems intuitive based on what you wrote and what I see in the equation. I don't know why the 9 is singled out though, and I don't know how the 30x from my equation is equated for with the expression 9(x-3)(x-5)


edit to add: -15 * -15 = 225 -15 + -15 = -30. So I'll be able to do it now, but I still would like a little clarification about your post.